You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied

Answer:

9.05 m/s ,   -14.72°  (respect to x axis)

Explanation:

To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:

[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]

[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]

m1: mass of the bowling ball = 5.50 kg

m2: mass of the bowling pin = 0.850 kg

v1xi: initial velocity of the bowling ball = 9.0 m/s

v2xi: initial velocity of bowling pin = 0m/s

v1: final velocity of bowling ball = ?

v2: final velocity of bowling pin = 15.0 m/s

θ: angle of the scattered bowling pin = ?

Φ: angle of the scattered bowling ball = 85.0°

Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.

First you solve for v1cosθ in the equation for the x component of the momentum:

[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]

and also you solve for v1sinθ in the equation for the y component of the momentum:

[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]

Next, you divide v1cosθ and v1sinθ:

[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]

the direction of the bawling ball is -14.72° respect to the x axis

The final velocity of the bawling ball is:

[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]

hence, the final velocity of the bawling ball is 9.05 m/s

Answer with Explanation:

Concepts and reason

The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.

Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.

In attached figure, resistor is connected in series to the capacitor.

As we considered [tex]V_{C}[/tex] the voltage across the capacitor and [tex]V_{s}[/tex] the voltage across the source.

Voltage across a resistor In RC circuit.

[tex]V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )[/tex]

Voltage across a resistor In RL circuit.

[tex]V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )[/tex]

The sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits can be seen in the diagram attached below.

For the Pre-Laboratory exercise, based on the assumption that the RC circuit has a capacitor and a sensing resistor while the RL circuit has a sensing resistor and an inductor.

The input voltage for both circuits is regarded as the square wave and if the square wave is much larger than the time constant for each.

Therefore, we can conclude that the below diagram shows an appropriate sketch of  [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits.

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Answer:

24.83 m

Explanation:

Applying the equation of motion;

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

Where;

d = distance

v = velocity

t = time

a = acceleration

For the trooper;

v = 28 m/s

a = 2.9 m/s^2

Substituting into equation 1;

d1 = 28t + 0.5(2.9t^2)

d1 = 28t + 1.45t^2

For the red car;

v = 40 m/s

a = 0

Substituting into equation 1

d2 = 40t

The difference in distance is;

d = d2 - d1

d = 40t - (28t + 1.45t^2)

d = 12t - 1.45t^2

The maximum distance is at d(d)/dt = 0

differentiating d;

d' = 12 - 2.9t = 0

2.9t = 12

t = 12/2.9 = 4.137931034482

t = 4.138 s

Substituting t into function d;

d(max) = 12(4.138) - 1.45(4.138^2)

d(max) = 24.8275862 = 24.83 m

the maximum distance ahead of the trooper that is reached by the red car is 24.83 m

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.

Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.

While outside the shell, the electric field is given by: k(q + Q)/r²

Where;

K= is a constant which is given as, 8.99 x 10^9 N m² / C².

Q= source charge which creates the electric field

q= is the test charge which is used to measure the strength of the electric field at a given location.

r= is the radius

Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.

Answer:

[tex]52.25\times10^4W\\699.1 hp[/tex]

Explanation:

According to the energy conversation:

ΔK=[tex]-f_kd+W[/tex]

ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]

where,

[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.

[tex]v_i[/tex]= initial velocity of the system

[tex]v_f[/tex]=final velocity of the system

W= total work done on the system

[tex]f_k[/tex]= friction force

d= distance traveled

Given: [tex]v_f[/tex]=110m/s

d=400m

[tex]f_k[/tex]=1200N

[tex]v_i[/tex]=0m/s

t=7.3s

ΔK=[tex]-f_kd+W[/tex]

W= ΔK + [tex]f_kd[/tex]

  =[tex]K_f-K_i+f_kd\\[/tex]

  [tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]

[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

                         [tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]

- We will plug in the given parameters in the equation of motion given above:

                         [tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

Answer:5.35nm

Explanation:

Consider that 1 inch is = 0.0254m

we have,

1m= 1x10^9 nm  

While:

0.0254m = 2.54x10^7nm  

1/55 (2.54x10^7) = 4.6181 x 10^5nm  

1 day= 24 hrs  

= (24x60) when calculating in min  

= (24x60x60) calculating in seconds we have:

= 8.64x10⁴sec  

In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm

Therefore, the amount by which the hair grows in 1 second  will be;

= (4.6181 x 10^5)/(8.64x10^4)  

= 5.35nm  

The rate of growth will be 5.35nm

Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.

(A) The wheel has angular velocity at time [tex]t[/tex] according to

[tex]\omega=\omega_0+\alpha t[/tex]

so that after 2.50 s, the wheel will have attained an angular velocity of

[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]

(B) The angular displacement of the wheel is given by

[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]

After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to

[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]

Answer:

The width of the slit will be ".946 mm".

Explanation:

The given values are:

Wavelength = 610 × 10⁻⁹

Length, L = 3 m

As we know,

⇒  [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]

On putting the estimated values, we get

⇒  [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]

On applying cross-multiplication, we get

⇒  [tex]a=9.46\times 10^{-4}[/tex]

⇒  [tex]a = .946 mm[/tex]

Answer:

Once at a steady cruising speed of about 16,150mph (26,000kph

Explanation:

Answer:

the force we will feel is F

Explanation:

 According to the Gauss law, electric field due to very large sheet of charge is as follows.

[tex]E = \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

where,        

[tex]\sigma[/tex] =  charge per unit area

Since, it is given that there are two sheets of equal and opposite charge. Therefore, electric field between the plates will be as follows.   [tex]E = \frac{\sigma}{2 \times \epsilon_{o}} + \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

Also, we know that relation between force and electric field is as follows.

                      F = qE

Hence, force felt by the charge present inside the plates will be as follows.

                 [tex]F = q \times \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

This depicts that force is not dependent on the distance and the charge is kept from one of the plate. Therefore,  force F felt by the charge is same when it is placed at a distance d/2 and at a distance d/4 from one of the plate.

Answer:

6 A

Explanation:

Parallel connected resistors needs to be calculated as one single resistor. To do that: [tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]=[tex]\frac{3}{15}[/tex]=[tex]R^{-1}[/tex]

[tex]\frac{3}{15} ^{-1}[/tex]= 5 Ω (total resistance)

U = R* I

[tex]\frac{U}{R}[/tex]=I

[tex]\frac{30}{5}[/tex]=6 A

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]

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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Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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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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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".

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,

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

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Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

[tex]\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m[/tex]

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

[tex]d=\sqrt{(x-x_o)^2+(y-y_o)^2}[/tex]   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

[tex]d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m[/tex]

hence, the displacement of the airplane is 3.45*10^4 m

Answer:

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg

Explanation:

First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:

A = 4πr²

where,

A = Surface Area = ?

r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m

Therefore,

A = 4π(1.1 x 10⁻⁶ m)²

A = 15.2 x 10⁻¹² m²

Now, we find the volume of the cell wall. For that purpose, we use formula:

V = At

where,

V = Volume of the Cell Wall = ?

t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m

Therefore,

V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)

V = 60.82 x 10⁻²⁰ m³

Now, to find mass of cell wall, we use formula:

ρ = m/V

m = ρV

where,

ρ = density of water = 1000 kg/m³

m = Mass of Wall = ?

Therefore,

m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg

The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where

So, V = 4πR²t

Substituting the values of the variables into the equation, we have

V = 4πR²t

V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.

V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.

V = 193.6π × 10⁻²¹ m³

V = 608.21 × 10⁻²¹ m³

V = 6.0821 × 10⁻¹⁹ m³

V ≅ 6.082 × 10⁻¹⁹ m³

We know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.

Making m subject of the formula, we have

m = ρV

Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³

So, m = ρV

m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³

m = 6.082 × 10⁻¹⁶ kg

Converting to mg, we have

m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg

m = 6.082 × 10⁻¹⁰ mg

So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

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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

Answer:

Explanation:

During slowing down , initial angular velocity ω₁ = 22 rad /s

final angular velocity ω₂ = 13.5 rad /s

using the law's of motion formula for rotation

ω₂² =  ω₁² + 2 αθ  , α is angular acceleration and θ is angle in radian rotated during this period

13.5² = 22² - 2xα x 13.8

2xα x 13.8 = 484 - 182.25

α  =  10.93 rad / s²

Answer:

Explanation:

1.  [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s

3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m

4.  [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

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]

Answer:

Explanation:

capacitance of air capacitor

C = ε₀ A /  d

ε₀ is permittivity of medium , A is plate area , d is distance between plate .

C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²

= 170.19 x 10⁻¹⁴ F .

charge on the capacitor when it is charged to  potential of 255 V

= CV , C is capacitance and V is potential

= 170.19 x 10⁻¹⁴  x 255

= 4.34 x 10⁻¹⁰ C .

After it is disconnected from the source , and it is immersed in water , charge on it remains the same .

So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.

b )

When it is immersed in water its capacity increases  k times where k is dielectric constant of water which is 80 .

capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴  F

= 13615.2  x 10⁻¹⁴ F .

= 1.36 x 10⁻¹⁰ F

potential difference = charge / capacitance

= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰

= 3.2 V

c )

Energy of capacitor = 1/2 C V²

Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴

=  55.33 x 10⁻⁹ J

Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²

= .7  x 10⁻⁹ J .

decrease of energy = 54.63 x 10⁻⁹ J .

Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is

[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]

For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],

[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]

which makes B, approximately 17 s, the correct answer.

The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)

Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:

[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)

Where:

[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.

[tex]\omega[/tex] - Final angular velocity, in radians per second.

[tex]\alpha[/tex] - Angular acceleration, in radians per square second.

[tex]t[/tex] - Time, in seconds.

If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:

[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]

[tex]t = 16.667\,s[/tex]

The time interval of angular deceleration is 16.667 seconds. (Answer: B)

Please this related question: https://brainly.com/question/10708862

Responder: 12000N

Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%

Dado MA = Carga / Esfuerzo

Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)

MA = Carga / 3000

VR = 2 / 0.5 VR = 4

Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%

1 = (Carga / 3000) / 4

4 = Carga / 3000

Carga = 4 × 3000

Carga = 12000N

Esto significa que el peso máximo que se puede levantar es 12000N