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)

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:

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 .

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!

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:

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

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.

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:

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:

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:

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:

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

t\12 and the parties are spreading ever

Explanation:

my point is that you can get sick if

you sont wash your ha

nds or be

save

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.

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.

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

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#SPJ2

Answer: A) Caves

Explanation:

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)

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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:[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 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:

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

Answer:

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

Explanation:

So that the above seems to be the right answer.

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: 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:

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:

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:

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:

the answer is opposite.

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Explanation:

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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