I need someone that knows physics. I have a test in 10 hrs and Im not good at it. Can anyone help me?

Answers

Answer 1
Hi there! I can help!

What grade of physics?
Answer 2

Answer:

I can help! What level of physics is it and what are your main topics?


Related Questions

Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.


Alguien que me lo hagaaaaaaa

Answers

Answer:

To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.

The image attached shows these forces.

Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.

A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s,vy=110m/sv x = 90m/s,v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s,vy=40m/sv x =−170m/s,v y=40m/s.
(a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.

Answers

The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or

[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]

which can be split into the [tex]x[/tex] and [tex]y[/tex] components as

[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]

[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]

The magnitude of this average acceleration is

[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]

and its direction is [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]

which corresponds to a direction of about 15.1º South of West.

If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%

Answers

Answer:

  More than 48%

Explanation:

If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...

  (1 +4%)^12 -1 = 60.1% . . . .  more than 48%

The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%

Calculation of Annual Interest rate

The formula used to calculate annual Interest rate =

[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]

where i= nominal interest rate = 4%

n= number of periods= 12 months

Annual Interest rate=

[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]

= (1+0.333)^12 -1

= (1.333)^12-1

= 31.56 - 1

= 30.56%

Therefore, the effective annual or yearly interest rate would be= 30.56%

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help yall 13 points!!

Answers

Answer:

Explanation:

12.)

A. Opposite poles attract

B. Same poles repel

13.)

IDK

0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.

Answers

Answer:

The final volume of R-134a is 0.212m³

Explanation:

Using one of the general gas equation to find the final volume of the R-134a.

According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.

VαT

V = kT

k = V/T

V1/T1 = V2/T2 = k

Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K

V2 = ? at T = 100°C = 100+273 = 373K

On substituting this values for T2;

0.14/246.6 = V2/373

373*0.14 = 246.6V2

V2 = 373*0.14 /246.6

V2 = 0.212m³

The final volume of R-134a is 0.212m³

When a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder

Answers

Answer:

Vf = 0.0017 m³ = 1.7 L

Explanation:

The work done by the system on the surrounding at constant pressure is given by the following formula:

W = PΔV

W = P(Vf - Vi)

where,

W = Work done = 288 J

P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa

Vf = Final Volume f Cylinder = ?

Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³

Therefore,

288 J = (202650 Pa)(Vf - 0.00025 m³)

Vf = 288 J/202650 Pa + 0.00025 m³

Vf = 0.0017 m³ = 1.7 L

Assuming 100% efficient energy conversion, how much water stored behind a 50
centimeter high hydroelectric dam would be required to charge the battery?

Answers

Answer:

Explanation:

The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h

Potential energy possessed by water at that height = mgh

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 3600 = 180,000 C

V = 12 V

qV = 180,000 × 12 = 2,160,000 J

Energy = 2,160,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (2,160,000)

4900V = 2,160,000

V = (2,160,000/4900)

V = 440.82 m³

Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.

Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.

Hope this Helps!!!

Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.

Answers

t\12 and the parties are spreading ever

Explanation:

my point is that you can get sick if

you sont wash your ha

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save

Which formation is one feature of karst topography?


Answers

Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.

What is karst topography?

Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.

The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.

When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.

Hence, sinkholes  formation is one feature of karst topography

To learn more about the karst topography, refer to the link;

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#SPJ2

Answer: A) Caves

Explanation:

A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.

Answers

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg

Answers

Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.

HELP, END OF SCHOOL YEAR, 30 POINTS Unit 9 lesson 15 astronomy unit test answers I ONLY HAVE ONE MORE DAY
1 As evidence supporting the Big Bang theory, what does the redshift of light from galaxies indicate?
The universe is mainly hydrogen.

The universe is 13.8 billion years old.

The universe is cooling off.

The universe is expanding.

2 Which evidence supports the idea that Cosmic Microwave Background radiation is a remnant of the Big Bang?(1 point)

Its temperature is uniform.

Its mass fluctuates greatly.

Its temperature fluctuates greatly.

Its mass is uniform.

3 Which of these items provide evidence supporting the Big Bang theory? Select the two correct items.(1 point)

rate of star formation

composition of matter in the universe

sizes and shapes of distant galaxies

cosmic background radiation

4 How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory?(1 point)

The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

5 How does weak background radiation coming from every direction in the sky support the Big Bang Theory?(1 point)

It provides evidence of the universe's increasing mass.

It provides evidence of universe expansion.

It provides evidence of universe contraction.

It provides evidence of the universe's decreasing mass.

6 Which statements describe ways that nuclear fission is different than nuclear fusion? Select the two correct answers.(1 point)

Nuclear fission is used to produce electricity at nuclear power plants.

Nuclear fission involves one large atom splitting into two smaller atoms.

Nuclear fission takes place in the nucleus of an atom.

Nuclear fission releases a huge amount of energy.


7 Blueshift is observed when(1 point)

a distant luminous object travels rapidly away from an observer.

a distant luminous object travels rapidly towards an observer.

a luminous object travels alongside an observer.

a luminous object is stationary compared to an observer.


8 Which statements about nuclear fusion are false? Select the two correct answers.(1 point)

The fuel for nuclear fusion is often uranium.

Nuclear fusion is used to generate electricity at nuclear power plants.

Nuclear fusion releases large amounts of energy.

Nuclear fusion takes place in the cores of stars.


9 Which of the following statements provide evidence to support the big bang theory? Select the two correct answers.

The ratios of hydrogen and helium in the universe match those of the early universe.

The universe began as a very high density singularity.

Dark matter makes up the majority of matter in the galaxy.

Small spiral galaxies become larger elliptical galaxies when they collide.


10 Which represents a correct match between ideas related to the formation of the universe? Select the two correct answers.(1 point)

accelerating expansion — dark energy

structures forming in the early universe — dark matter

greatest percent of mass of universe — dark matter

glowing nebulae — dark energy

11 How is dark energy related to the theory of the Big Bang?(1 point)

It causes the expansion of the universe to accelerate.

It causes the universe to expand.

It seeded the formation of galaxies and star clusters.

It causes the spinning of galaxies.

Answers

Answer:

1. The Universe is Expanding

2. It’s temperature is it’s uniform

3. Cosmic background radiation

4. I will give a hint for this one, since I don’t know, the hint is the universe is cooling.

5. It provides evidence of universe expansion.

6. Sorry I don’t know the rest

Explanation:

The universe is the collection of every item in space and time as well as the contents of those items

The correct options are as follows;

1. The universe is expanding

2. Its temperature is uniform

3. Composition of matter in the universe, cosmic background radiation

4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

5. It provides evidence of universe expansion

6. Nuclear fission involves one large atom splitting into two smaller atoms

Nuclear fission takes place in the nucleus of an atom

7. A distant object travels rapidly towards an observer

8. The fuel in nuclear fusion is often uranium

Nuclear fusion is used to generate electricity at nuclear power plants

9. The universe began as a very high density singularity

Dark matter makes up majority of the universe

10.  Acceleration expansion — Dark energy

Structure forming in the early universe — Dark matter

11. It causes the expansion of the universe to accelerate

The reasons for selecting the above options are as follows;

1. The universe is expanding

The redshift of light from galaxies indicates that that are moving away

2. Its temperature is uniform

The uniform temperature of the microwave background suggest a common source

3. Composition of matter in the universe, cosmic background radiation

The matter present in the universe are characteristically similar in their origins

The cosmic background provides evidence of the existence of a singularity

4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Based on the Big Bang Theory, the temperature of the universe is reducing as the universe expands, compared to the initial temperature

5. It provides evidence of universe expansion

The background radiation coming from a single source as the rest of the universe is expected to spread throughout the universe

6. Nuclear fission involves one large atom splitting into two smaller atoms

Nuclear fission takes place in the nucleus of an atom

Nuclear fusion involves the joining of small atoms to form a larger atom

7. A distant object travels rapidly towards an observer

The redshift is the opposite, indicating that the object is moving further away

8. The fuel in nuclear fusion is often uranium

Nuclear fusion is used to generate electricity at nuclear power plants

Nuclear fusion usually consists of joining small atoms together. It has not been used for commercial energy production

9. The universe began as a very high density singularity

According to the Big Bang Theory, the universe started from the dense, high temperature singularity

Dark matter makes up majority of the universe

10.  Acceleration expansion — Dark energy

Dark energy causes expansion

Structure forming in the early universe — Dark matter

Dark matter is instrumental to the formation of structures in the universe

11. It causes the expansion of the universe to accelerate

Dark energy is seen as the cause of the accelerating expansion of the universe

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A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position

Answers

Answer:

v = 3.33 m/s

Explanation:

In the position of 45 degrees, all the energy of the rock is gravitational, then we have:

E = m*g*L*cos(angle)

and in the vertical position of the string, all the energy is kinetic, so we have:

E = m*v^2/2

If there is no dissipation, both energies are equal, so we have:

m*g*L*cos(45) = m*v^2/2

9.81 * 0.8 * 0.7071 * 2 = v^2

v^2 = 11.0986

v = 3.33 m/s

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)

Answers

Answer:

d = 80 m

its vertical displacement (in m) after 4 s is 80 m

Explanation:

From the equation of motion;

d = vt + 0.5at^2 ......1

Where;

d = displacement

v = initial velocity = 0 (dropped with no initial speed)

t = time of flight = 4s

a = g = acceleration due to gravity = 10 m/s^2

Substituting the given values into equation 1;

d = 0(4) + 0.5(10 × 4^2)

d = 0.5(10×16)

d = 80 m

its vertical displacement (in m) after 4 s is 80 m

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action–reaction pairs. (The horizontal truck bed is not frictionless.)

Answers

Answer:

The description of that same situation has been listed throughout the explanation segment below.

Explanation:

When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.

So that the above seems to be the right answer.

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground

Answers

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

[tex]v_y^2=v_{oy}^2+2gh[/tex]   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]

vx is calculated by using the information about the horizontal range of the ball:

[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex]    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]

The velocity of the ball just before it touches the ground is 46.99 m/s

Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 m on a side, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. Part A How much thermal energy is added to the air by the drag force

Answers

Answer:

13.9 kJ

Explanation:

Given that

Length of the side, l = 1.8 m

Drag coefficient, C(d) = 1.4

Distance of run, d = 200 m

Velocity of run, v = 5 m/s

Density, ρ = 1.23

Using the Aerodynamics Drag Force formula. We have

F(d) = 1/2.ρ.A.C(d).v²

The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,

F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²

F(d) = 139.482/2

F(d) = 69.74

recall that, energy =

W = F * d

W = 69.74 * 200

W= 13948

W = 13.9kJ

Therefore, the thermal energy added to the air by the drag force is 13.9kJ

If a metal rod is moved through magnetic field, the charged particles will feel a force, and if there is a complete circuit, a current will flow. We talk about the induced emf of the rod. The rod essentially acts like a battery, and the induced emf is the voltage of the battery. A magnetic field with a strength of 0.732 T is pointing into the page and a metal rod L=0.362 m in length is moved to the right at a speed v of 15.1m/s.

Required:
a. What is the induced emf in the rod?
b. Suppose the rod is sliding on conducting rails, and a complete circuit is formed. If the load resistance is 5.74Ω , what is the magnitude and direction (clockwise or counterclockwise) of the current flowing in the circuit?

Answers

Answer:

a.  4 V

b. 0.697 A

Explanation:

Magnetic field strength B =  0.732 T

length of rod l = 0.362 m

velocity of rod v = 15.1 m/s

a.  EMF can be calculated as

E = Blv = 0.732 x 0.362 x 15.1 = 4 V

b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω

current I = V/R = 4/5.74 = 0.697 A

the current flows in a clockwise direction

A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?

Answers

Explanation:

We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of the pendulum

g is acceleration due to gravity on which it is placed

It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.

Answers

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

b) The length of the rod:

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]

Given:

d = 1.5 mλ = 2.5 nC/m

Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.

Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]

How to calculate Electric Field?

E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]

Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis

[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.

Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]

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Friction is a force that acts in an ___________ direction of movement.
a) similar
b) opposite
c) parallel
d) west

Answers

Answer:

the answer is opposite.

plz mark brainliest

Explanation:

A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?

Answers

Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

E = 389 MeV

Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?

Answers

Answer:

Explanation:

1 ) Since it is a isochoric process , heat energy passed into gas

= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .

= 7.4 x 12.47 x ( 500 - 300 )

= 18455.6 J.

2 ) Since there is no change in volume , work done by the gas is constant.

3 ) from  , gas law equation

PV = nRT

P = nRT / V

= 7.4 x 8.3 x 500 / .74

= .415 x 10⁵ Pa.

4 ) Average kinetic energy  of gas molecules after attainment of final temperature

= 3/2 x R/ N x T

= 1.5 x 1.38  x 10⁻²³ x 500

= 1.035 x 10⁻²⁰ J

1/2 m v² = 1.035 x 10⁻²⁰

v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶

= 1.49 x 10⁶

v = 1.22 x 10³ m /s

5 )  In this process , pressure remains constant

gas is cooled from 500 to 300 K

heat will be withdrawn .

heat withdrawn

= n Cp dT

= 7.4 x 20.79 x 200

= 30769.2 J .

6 )

gas will have reduced volume due to cooling

reduced volume = .74 x 300 / 500

= .444 m³

change in volume

= .74 - .444

= .296 m³

work done on the gas

= P x dV

pressure x change in volume

= .415 x 10⁵ x .296

= 12284 J.


Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)​

Answers

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = [tex]\sqrt[4]{P/Ae}[/tex]

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio

Answers

Answer:[tex]8.062\ m/s[/tex]

Explanation:

Given

masss of football player [tex]M=110\ kg[/tex]

Velocity of football player [tex]u_1=8\ m/s[/tex]

mass of football [tex]m=0.41\ kg[/tex]

velocity of football [tex]u_2=25\ m/s[/tex]

Final velocity will be given by applying conservation of linear momentum

After catching the ball Player and ball moves with same velocity

[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]

[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]

[tex]\Rightarrow 880+10.25=110.41\times v[/tex]

[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]

So, final velocity will be [tex]8.062\ m/s[/tex]

A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.

Answers

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now

Answers

Answer:

1.92 Watt lost

Explanation:

Power rating of each toaster = 600 Watts

Current that flows = 5 Amperes

Wasted power = 1 Watt

Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R

where I = current

and R = Resistance

600 = [tex]5^{2}[/tex] x R

R = 600/25 = 24 Ohms.

According to joules loss due to heating of wire

Power loss P ∝ [tex]I^{2}[/tex]R

imputing values,

1 ∝ [tex]5^{2}[/tex] x 24

1 ∝ 600

to remove the proportionality sign, we introduce a constant k

1 = 600k

k = 1/600 = 0.00167

For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.

The resistance across the wire becomes

1200 = [tex]10^{2}[/tex]R

R = 1200/100 = 12 Ohms

power loss P = k x [tex]I^{2}[/tex]R

P = 0.0016 x [tex]10^{2}[/tex] x 12

P = 1.92 Watt lost

This question involves the concepts of power, current, and resistance.

The power wasted by the wires in the home for two units will be "4 watt".

POWER WASTAGE

The power wasted by the wires can be given in terms of current and resistance by the following formula:

[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]

where,

P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 A

Therefore,

[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]

P₂ = 4 watt

Learn more about power here:

https://brainly.com/question/7963770

How can socialism
impact populations?

Answers

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

hope this helped!

It represents a fundamental difference by gaining friends and losing friends or gaining jobs and losing jobs etc

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward

Answers

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex]  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.

Answers

Answer:

31.4 m/s

44.4°

Explanation:

Momentum is conserved in the horizontal direction:

pₓᵢ = pₓ

m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ

vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ

17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ

vₓ = 22.5 m/s

Momentum is conserved in the vertical direction:

pᵧᵢ = pᵧ

2m vᵢ₁ sin θ = (m + 2m) vᵧ

2 vᵢ₁ sin θ = 3 vᵧ

2 (41.5 m/s) (sin -52.7°) = 3 vᵧ

vᵧ = -22.0 m/s

The speed is:

v = √(vₓ² + vᵧ²)

v = √((22.5 m/s)² + (-22.0 m/s)²)

v = 31.4 m/s

The direction is:

θ = atan(vᵧ / vₓ)

θ = atan(-22.0 m/s / 22.5 m/s)

θ = -44.4°

The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.

Since momentum is conserved horizontally;

17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx

vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3

vx =  22.5 m/s

Also, momentum is conserved vertically hence;

2 (41.5 m/s) (sin -52.7°) = 3 vy

vy = 2 (41.5 m/s) (sin -52.7°) /3

vy =  -22.0 m/s

The effective speed therefore, is;

v = √((22.5 m/s)² + (-22.0 m/s)²)

v = 31.4 m/s

The direction of this effective speed is;

θ = tan-1(22.0 m/s / 22.5 m/s)

θ = 44.4°

Learn more: https://brainly.com/question/13322477

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