13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compared to what it was at the lake surface? The area of the eardrum is 0.60 cm².

Answer:

The change in length is 3.4 mm.

Explanation:

Given that,

Length = 4 m

Diameter = 2.3 cm

Load = 70 kN

Modulus of elasticity = 200 GPa

We need to calculate the change in length

Using formula of modulus of elasticity

[tex]E=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]

[tex]\Delta l=\dfrac{Fl}{AE}[/tex]

Where, F = force

A = area

L = length

E = modulus elasticity

Put the value into the formula

[tex]\Delta l=\dfrac{70\times10^{3}\times4}{\pi\times(1.15\times10^{-2})^2\times200\times10^{9}}[/tex]

[tex]\Delta l=0.00336\ m[/tex]

[tex]\Delta l=3.4\ mm[/tex]

Hence, The change in length is 3.4 mm.

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

[tex]T = 39 days \times 24 hr \times 3600 s[/tex]

[tex]T = 3369600 s[/tex]

now the radius of the orbit is given as

[tex]r = 4.8 \times 10^8 m[/tex]

so the total path length is given as

[tex] L = 2 \pi r[/tex]

[tex]L = 2\pi (4.8 \times 10^8)[/tex]

[tex]L = 3.015 \times 10^9 [/tex]

now the speed will be given as

[tex]v = \frac{L}{T}[/tex]

[tex]v = \frac{3.015 \times 10^9}{3369600} [/tex]

[tex]v = 895 m/s[/tex]

Answer:

7.6 m/s

Explanation:

m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m

Use the conservation of energy

Potential energy of the block = Elastic potential energy of the spring

m g h = 1/2 k x^2

Where, k be the spring constant

2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2

k = 3317.88 N/m

Now, let v be the velocity of the block, when the compression is 15 cm.

Again use the conservation of energy

Potential energy of the block = Kinetic energy of block + Elastic potential  

                                                                                         energy of the spring

m g h = 1/2 m v^2 + 1/2 k x^2

2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2

103.684 = 1.15 v^2 + 37.33

v = 7.6 m/s

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, [tex]\lambda=608\ nm=608\times 10^{-9}\ m[/tex]

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

[tex]y=\dfrac{mD\lambda}{a}[/tex]

where

a = width of the slit

[tex]a=\dfrac{mD\lambda}{y}[/tex]

[tex]a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}[/tex]

a = 0.000167 m

[tex]a=1.67\times 10^{-4}\ m[/tex]

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

Answer:

final equilibrium temperature of the system is ZERO degree Celcius

Explanation:

Hear heat given by water + iron = heat absorbed by ice

so here first we will calculate the heat given by water + iron

[tex]Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1[/tex]

[tex]Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)[/tex]

now the heat absorbed by ice so that it will melt and come to the final temperature

[tex]Q_2 = m s \Delta T + mL + m s_{water}\Delta T'[/tex]

[tex]Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)[/tex]

now we will have

[tex]17377 + 209.3T = 3600 - 90T + 8372 - 209.3T[/tex]

[tex]17377 + 209.3T + 90T + 209.3T = 11972[/tex]

[tex]T = -10.6[/tex]

since it is coming out negative which is not possible so here the ice will not completely melt

so final equilibrium temperature of the system is ZERO degree Celcius

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

[tex]v = \sqrt{\frac{G M }{r}}[/tex]

[tex]v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24}  }{3.18064\times 10^{10}}}[/tex]

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years

Answer:

Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m

Explanation:

Thermal expansion is given by the expression

[tex]\Delta L=L\alpha \Delta T \\ [/tex]

Here length of Eiffel tower, L = 300 m

Coefficient of thermal expansion, α = 0.000012 per degree Celsius

Change in temperature, = 35 - (-24) = 59degrees Celsius

Substituting

[tex]\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\ [/tex]

Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m

Given:

R = 13 Ω

L = 46 mH

V = 240 V(rms)

f = 60 Hz

Formula used:

[tex]I_{L} = \frac{V}{R + jX_{L} }[/tex]

[tex]X_{L} = 2\pi fL[/tex]

Solution:

Now, using the above formula for [tex]X_{L}[/tex]:

[tex]X_{L} = 2\pi\times 60\times 46\times 10^{-3} [/tex] = 17.34 Ω

From the above formula for  [tex]I_{L}[/tex]:

[tex]I_{L} = \frac{240\angle0}{13 + j17.34 }[/tex]

[tex]I_{L}[/tex] = (6.64 - j8.86) A = [tex]11.07\angle-53.14^{\circ}[/tex] A

[tex]i_{L}(t)[/tex] = [tex]\sqrt{2}\times 11.07cos(2\pi \times 60t - 53.14)[/tex] A

[tex]i_{L}(t)[/tex] = 15.65cos(376.99t - 53.14)A

Load current in the given AC circuit can be obtained by using the concept of Impedance and Ohm's law in its AC variant where current, I = V/Z. Impedance, Z is calculated taking into account both Resistance and Reactance.

In the case of the mentioned electric dryer, we are dealing with an AC circuit that contains both resistance and inductance. In such cases, we should leverage the concept of Impedance, which is the effective resistance in an AC circuit resulting from combined effect of resistors, inductors and capacitors.

Impedance is defined by the relation Z = √ (R^2 + (XL)^2), where R is the resistance (13 Ω) and XL = 2πfL is the Reactance, f is the frequency (60Hz), and L the inductance (46mmH). However, when calculating the load current, we use Ohm's law in its AC version, I = V/Z, where I denotes the current, V the voltage supplied (240V), and Z the impedance defined earlier.

By substitizing all the values and calculating we can get the desired load current.

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Answer:

0.45 m

Explanation:

m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s

velocity is maximum when it passes through the mean position.

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi \sqrt{\frac{1.23}{37.4}}[/tex]

T = 1.139 sec

w = 2 π / T

w = 2 x 3.14 / 1.139

w = 5.51 rad / s

Vmax = w A

Where, A be the amplitude

2.48 = 5.51 A

A = 2.48 / 5.51 = 0.45 m

Final answer:

The amplitude of oscillation for the given spring oscillator is 0.58 m.

Explanation:

The amplitude of oscillation can be calculated using the equation:

A = vmax/ω

where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.

The angular frequency can be calculated using the equation:

ω = sqrt(k/m)

where k is the spring constant and m is the mass of the oscillator.

Substituting the given values into the equations:

ω = sqrt(37.4 N/m / 1.23 kg) = 4.29 rad/s

A = 2.48 m/s / 4.29 rad/s = 0.58 m

Therefore, the amplitude of oscillation is 0.58 m.

Answer:

4.78 x 10^-11 J

Explanation:

A = 1.5 x 10^-4 m^2

d = 2 mm = 2 x 10^-3 m

V = 12 V

Let C be the capacitance of the capacitor

C = ε0 A / d

C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)

C = 6.64 x 10^-13 F

Energy stored, U = 1/2 CV^2

U = 0.5 x 6.64 x 10^-13 x 12 x 12

U = 4.78 x 10^-11 J

The energy stored in a parallel-plate capacitor connected to a 12-V battery, with plate area of 1.5 x 10^-4 m^2 and separation of 2.0 mm, is calculated using the formula U = (1/2)CV^2 and is found to be 4.8 x 10^-11 J.

The capacitance C of a parallel-plate capacitor is given by the formula C = ε_0 * A / d, where ε_0 is the vacuum permittivity (ε_0 = 8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation between the plates.

Given that the area A is 1.5 x 10^-4 m^2, the separation d is 2.0 mm = 2.0 x 10^-3 m, and the voltage V is 12 V, we can plug in these values to first determine capacitance and then calculate the energy stored.

First, calculate the capacitance:

C = ε_0 * A / d = (8.85 x 10^-12 F/m)(1.5 x 10^-4 m^2) / (2.0 x 10^-3 m) = 6.6 x 10^-12 F

Next, calculate the energy stored:

U = (1/2)CV^2 = (1/2)(6.6 x 10^-12 F)(12 V)^2 = 4.8 x 10^-11 J

Therefore, the energy stored in the capacitor is 4.8 x 10^-11 J, which corresponds to option (a).

Explanation:

Momentum is conserved.

a) In the first scenario, Olaf and the ball have the same final velocity.

mu = (M + m) v

(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v

v = 0.0618 m/s

b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.

mu = mv + MV

(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v

v = 0.108 m/s

a) After Olaf catches the ball, Olaf and the ball will move at a speed of 0.062 m/s.

b) The speed of Olaf after the ball bounces off his chest is 0.11 m/s.    

a) We can find the speed of Olaf and the ball by conservation of linear momentum.

[tex] p_{i} = p_{f} [/tex]

[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]

Where:

[tex] m_{o}[/tex]: is the mass of Olaf = 70.2 kg

[tex] m_{b}[/tex]: is the mass of the ball = 0.400 kg

[tex] v_{i_{o}}[/tex]: is the intial speed of Olaf = 0 (he is at rest)

[tex]v_{i_{b}}[/tex]: is the initial speed of the ball = 10.9 m/s

[tex] v_{f_{o}}[/tex] and [tex]v_{f_{b}} [/tex]: are the final speed of Olaf and the final speed of the ball, respetively.

Since Olaf catches the ball, we have that [tex] v_{f_{o}} = v_{f_{b}} = v [/tex], so:

[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = v(m_{o} + m_{b}) [/tex]

We will take the direction of motion of the ball to the right side, and this will be the positive x-direction.  

By solving for "v" we have:

[tex] v = \frac{m_{o}v_{i_{o}} + m_{b}v_{i_{b}}}{m_{o} + m_{b}} = \frac{70.2 kg*0 + 0.400 kg*10.9 m/s}{70.2 kg + 0.400 kg} = 0.062 m/s [/tex]

Hence, Olaf and the ball will move at a speed of 0.062 m/s.

b) The final speed of Olaf after the collision can be calculated, again with conservation of linear momentum.

[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]

In this case, since the ball hits Olaf and bounces off his chest, we have that [tex] v_{f_{o}} \neq v_{f_{b}}[/tex]

[tex] 0.400 kg*10.9 m/s = 70.2 kg*v_{f_{o}} + 0.400 kg*(-8.10 m/s) [/tex]      

The minus sign of the speed of the ball is because it moves to the negative direction of motion after the collision.  

[tex] v_{f_{o}} = \frac{0.400 kg*10.9 m/s + 0.400 kg*8.10 m/s}{70.2 kg} = 0.11 m/s [/tex]

Therefore, the speed of Olaf after the collision will be 0.11 m/s in the positive x-direction.                                  

You can see another example of conservation of linear momentum here: https://brainly.com/question/22257322?referrer=searchResults

I hope it helps you!                                                                                                                    

Answer:

a) Train's nose will be present 9.19 seconds in the station.

b) The nose leaves the station at 21.62 m/s.

c) Train's end will leave after 15.33 seconds from station.

d) The end leaves the station at 20.70 m/s.

Explanation:

a) We have equation of motion s = ut + 0.5at²

   Here u = 23 m/s, a = -0.15 m/s², s = 205 m

   Substituting

        205 = 23t + 0.5 x (-0.15) x t²

        0.075t² -23 t +205 = 0

  We will get t = 9.19 or t = 297.47

  We have to consider the minimum time

  So train's nose will be present 9.19 seconds in the station.

b) We have equation of motion v= u + at

   Here u = 23 m/s, a = -0.15 m/s², t = 9.19

   Substituting

        v= 23 - 0.15 x 9.19 = 21.62 m/s

   The nose leaves the station at 21.62 m/s.

c) We have equation of motion s = ut + 0.5at²

   Here u = 23 m/s, a = -0.15 m/s², s = 205 + 130 = 335 m

   Substituting

        335 = 23t + 0.5 x (-0.15) x t²

        0.075t² -23 t +335 = 0

  We will get t = 15.33 or t = 291.33

  We have to consider the minimum time

  So train's end will leave after 15.33 seconds from station.      

d) We have equation of motion v= u + at

   Here u = 23 m/s, a = -0.15 m/s², t = 15.33

   Substituting

        v= 23 - 0.15 x 15.33 = 20.70 m/s

   The end leaves the station at 20.70 m/s.

In Mathematics and Physics, scalar is a quantity or a single number that shows the measurement of a medium in magnitude only (It does not include direction as vectors do); examples of scalars are voltage, mass, temperature, speed, volume, power, energy, and time.

Two examples of scalars I have used recently are Degrees Celsius to measure the temperature of my living room and Cubic Feet to measure the volume of my mug.

A scalar is a quantity represented only by a magnitude and devoid of direction. Examples of scalars include temperature and energy, which are significant in physics and everyday measurements. Unlike vectors, scalars do not change with coordinate system rotations.

Definition of Scalar

A scalar is a physical quantity that is represented only by a magnitude (or numerical value) but does not involve any direction. In contrast to vectors, which have both magnitude and direction, scalars are not affected by coordinate system rotations or translations. Scalars come in handy when representing physical quantities where direction is non-applicable.

Two examples of scalars that I have used recently are:

When dealing with physics problems or everyday measurements, it's crucial to understand whether you're working with a scalar or a vector to correctly interpret the quantity in question.

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = [tex]\sqrt{2gh}[/tex]

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = [tex]\sqrt{2*9,8m/s^2*0,77m}[/tex] = 3,884 m/s

We have that for the Question"A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initial speed will the water flow from the hole" it can be said that initial speed  the water flow from the hole is

From the question we are told

A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initial speed will the water flow from the hole? Please use a equation and explain every step

Generally the equation for the water flow speed   is mathematically given as

[tex]v=\sqrt(2gh)\\\\v=\sqrt{2*9.8*0.77}[/tex]

v=3.88m/s

Therefore

initial speed  the water flow from the hole is

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Answer:

1196.62 sec

Explanation:

V = electric potential difference at which teapot operates = 120 volts

i = current = 2.00 A

t = time of operation

m = mass of water = 0.891 kg

T₀ = initial temperature = 23.0 °C

T = final temperature = 100 °C

c = specific heat of water = 4186 J/(Kg °C)

Using conservation of energy

V i t = m c (T - T₀)

(120) (2.00) t = (0.891) (4186) (100 - 23.0)

t = 1196.62 sec

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 40 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

   4.9t²  + 2t - 40 = 0

   t = 2.66 s   or t = -3.06 s

So, time is 2.66 s.

Option C is the correct answer.

Answer:

v = 2917.35 m/s

Explanation:

let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.

at any point in the orbit  the forces acting on the satelite are balanced such that:

Fc = Fg

mv^2/r = GmM/r^2

v^2 = GM/r

   v = \sqrt{GM/r}  

      = \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}

      = 2917.35 m/s

Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.

Answer:

Approximately 18 volts when the magnetic field strength increases from [tex]\rm 20\; mT[/tex] to [tex]\rm 80\;mT[/tex] at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF [tex]\epsilon[/tex] that a changing magnetic flux induces in a coil is:

[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}[/tex],

where

However, for a coil the magnetic flux [tex]\phi[/tex] is equal to

[tex]\phi = B \cdot A\cdot \cos{\theta}[/tex],

where

For this coil, the magnetic field is perpendicular to coil, so [tex]\theta = 0[/tex] and [tex]A\cdot \cos{\theta} = A[/tex]. The area of this circular coil is equal to [tex]\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}[/tex].

[tex]A\cdot \cos{\theta} = A[/tex] doesn't change, so the rate of change in the magnetic flux [tex]\phi[/tex] through the coil depends only on the rate of change in the magnetic field strength [tex]B[/tex]. The size of the magnetic field at the instant that [tex]B = \rm 50\; mT[/tex] will not matter as long as the rate of change in [tex]B[/tex] is constant.

[tex]\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}[/tex].

As a result,

[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V[/tex].

Using Faraday's law, the magnitude of the induced emf in the coil at the instant the magnetic field is 50 mT is found to be 18 volts. The negative sign indicates the emf opposes the change in flux according to Lenz's law.

The question is about the calculation of the magnitude of induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT. To approach this, we can use Faraday's law of electromagnetic induction. It states that the induced emf is equal to the rate of change of the magnetic flux.  

First, let us recall the expression of the magnetic flux (Φ): Φ = B * A, where B is the magnetic field's magnitude and A is the area through which it passes. For a circular coil, A = π * (radius)^2. Applying the given radius 0.08 m, we calculate A = 0.02 m².  

Given that the magnetic field changes linearly with time from 20 mT to 80 mT in 20 ms, we can determine the rate of change of the magnetic field which is (80 mT - 20 mT) / 20 ms = 3 T/s. Therefore, the rate of change of Flux is dΦ/dt = (B * A)/dt = 3 T/s * 0.02 m² = 0.06 Wb/s.

Finally, as per Faraday's law, the induced emf = -NdΦ/dt, where N is the number of turns in the coil. Therefore, for N = 300 turns, emf = -300 * 0.06 Wb/s = -18 V. The negative sign indicates the emf would oppose the change in flux according to Lenz's law.

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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

[tex]s=ut+0.5at^2 \\ [/tex]

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

[tex]h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9 [/tex]

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

[tex]h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\ [/tex]

Solving both equations

[tex]600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\ [/tex]

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

[tex]h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\ [/tex]

Passing of B occurs at 4108.31 height.

For a DC circuit, the following equation relates the voltage, resistance, and current:

V = IR

V is the total voltage supplied, I is the total current, and R is the total resistance.

Given values:

V = 14.4V

R = 11.0Ω

Plug in the values and solve for I:

14.4 = I×11.0

I = 1.309A

Since one electron carries 1.6×10⁻¹⁹C of charge, divide the current by this number.

1.309/(1.6×10⁻¹⁹) = 8.182×10¹⁸

Round this value to 2 significant figures:

8.2×10¹⁸ electrons per second.

Answer:

3272.4 Hz

Explanation:

L = 31 mH

XL = 637 ohm

XL = 2 π f L

f = XL / (2 π L)

f = 637 / ( 2 x 3.14 x 31 x 10^-3)

f = 3272.4 Hz

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

[tex]W=4.8\ kg\times 9.8\ m/s^2\times 18\ m[/tex]

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

The energy stored by the electric field between two plates, known as a parallel-plate capacitor, is calculated using the respective formulas for voltage, capacitance and energy in a capacitor, factoring in the charge of the plates, surface area, and separation distance.

The energy stored by the electric field between two parallel plates (known as a parallel-plate capacitor) can be calculated using the formula for the energy stored in a capacitor, which is U = 0.5 * C * V^2. Here, V is the voltage across the plates and C is the capacitance of the capacitor.

To calculate V, we use the relation V = E * d, where E is the electric field, which can be found using the formula E = Q/A (Charge per Area), and d is the distance separating the plates. In this case, the charge Q is 14 nC, the area A is (8.5 cm)^2 and the distance d is 2.5 mm.

Finally, to find C, we use the formula C = permittivity * (A/d), where the permittivity of free space is approximately 8.85 × 10^-12 F/m. With these variables, C, V and U can be calculated accordingly.

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The energy stored in the electric field of a parallel-plate capacitor can be calculated using the dimensions of the plates, their separation, and charges. For this particular scenario, the energy stored would be approximately 0.44 μJ.

In this scenario, we have a parallel-plate capacitor where each plate is a square with side 8.5 cm and has a charge of 14 nC. The plates are separated by a 2.5mm gap. Using this information, we can calculate the energy stored in the capacitor.

Firstly, the surface area A of the plates can be calculated using the formula A = s², where s is the side of the square. Converting 8.5 cm to meters, we get s = 0.085 m. So, A = (0.085 m)² = 0.007225 m².

Next, we calculate the electric field E between the plates using the formula E = Q/ε₀A, where Q is the charge on one plate and ε₀ is the permittivity of free space (8.85 x 10^-12 F/m). Converting 14 nC to coulombs gives Q = 14 x 10^-9 C. Substituting these values into the formula, we find E ≈ 224.717 kV/m.

The energy U stored in an electric field can be calculated using the formula U = 0.5ε₀EA². Substituting the earlier calculated values, we find U ≈ 0.44 μJ.

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Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

[tex]\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]

[tex]\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]

[tex]{\frac{M_{B}}{M_{A}}} = 4[/tex]

[tex]{\frac{M_{B}}{4}} = M_{A}[/tex]

If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.

We know that the RMS velocity of the molecule is given as,

[tex]V = \dfrac{1}{\sqrt{M}}[/tex]

Given to us

RMS speed of the molecules of gas A is twice that of gas B, therefore, [tex]V_A = 2 V_B[/tex]

Substitute the value of RMS in the equation [tex]V_A = 2 V_B[/tex],

[tex]\dfrac{1}{\sqrt{M_A}} = \dfrac{2}{\sqrt{M_B}}\\\\\\\sqrt{\dfrac{M_B}{M_A}} = 2\\\\\\\dfrac{M_B}{M_A} = 4[/tex]

Hence, If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.

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Answer:

Total energy, [tex]TE=3.006\times 10^{-10}\ J[/tex]

Explanation:

It is given that, a proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy.  Let KE is the kinetic energy of the proton and E₀ is its rest energy. So,

[tex]KE=E_o[/tex]              

The total energy of the proton is equal to the sun of kinetic energy and the rest mass energy.

[tex]TE=KE+E_o[/tex]

[tex]TE=2E_o[/tex]

[tex]TE=2m_{proton}c^2[/tex]

[tex]TE=2\times 1.67\times 10^{-27}\ kg\times (3\times 10^8\ m/s)^2[/tex]

[tex]TE=3.006\times 10^{-10}\ J[/tex]

So, the total energy of the proton as measured by a physicist working with the accelerator is [tex]3.006\times 10^{-10}\ J[/tex]

Answer:

[tex]\omega' = 0.89\omega[/tex]

Explanation:

Rotational inertia of uniform solid sphere is given as

[tex]I = \frac{2}{5}MR^2[/tex]

now we have its angular speed given as

angular speed = [tex]\omega[/tex]

now we have its final rotational kinetic energy as

[tex]KE = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2[/tex]

now the rotational inertia of solid cylinder about its axis is given by

[tex]I = \frac{1}{2}MR^2[/tex]

now let say its angular speed is given as

angular speed = [tex]\omega'[/tex]

now its rotational kinetic energy is given by

[tex]KE = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]

now if rotational kinetic energy of solid sphere is same as rotational kinetic energy of solid sphere then

[tex]\frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]

[tex]\frac{2}{5}\omega^2 = \frac{1}{2}\omega'^2[/tex]

[tex]\omega' = 0.89\omega[/tex]

Answer:

w_cyl = ±√(4/5) ω

Explanation:

Kinetic energy

E = (1/2)Iw²

where I is the moment of inertia and w the angular frequency of rotation.

The moment of inertia of a solid sphere of mass M and radius R is:

I = (2/5)MR²,  

Solid cylinder is of mass M and radius R

I = (1/2)MR²

Equate the energies through

(1/2)×(2/5) M R²× (w_sphere)² = (1/2)× (1/2) MR² × (w_cyl)²

(w_cyl)² = (4/5)(w_sphere)²

w_cyl = ±√(4/5) ω

The energy of rotation is independent of the direction of rotation

Answer:

The weight of the astronaut is 0.4802  N.

Explanation:

Gravitational potential energy, [tex]U=2.94\times 10^6\ J[/tex]

Distance above earth, [tex]d=4\times 10^5\ m[/tex]

The gravitational potential energy is given by :

[tex]U=\dfrac{GMm}{R}[/tex]

G is universal gravitational constant

M is the mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]

m is mass of astronaut

R is the radius of earth, R = R + d

[tex]R=6.37\times 10^6\ m+4\times 10^5\ m=6770000\ m[/tex]

[tex]m=\dfrac{U(R+d)^2}{GM}[/tex]

[tex]m=\dfrac{2.94\times 10^6\ J\times (6770000\ m)}{6.67\times 10^{-11}\times 5.97\times 10^{24}\ kg}[/tex]

m = 0.049 kg

The weight of the astronaut is given by :

W = mg

[tex]W=0.049\ kg\times 9.8\ m/s^2[/tex]

W = 0.4802  N

So, the weight of the astronaut when he returns to the earth surface is 0.4802 N. Hence, this is the required solution.

Final answer:

The weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.

Explanation:

To calculate the weight of the astronaut when he returns to the Earth's surface, we can use the formula for gravitational potential energy:

PE = mgh

where PE is the gravitational potential energy, m is the mass of the astronaut, g is the acceleration due to gravity, and h is the altitude of the astronaut.

Given that the gravitational potential energy is 2.94 x 10^6 J and the altitude is 4.00 x 10^5 m, we can rearrange the formula to solve for m:

m = PE / (gh)

Substituting the values, we get:

m = (2.94 x 10^6 J) / ((9.80 m/s^2) * (4.00 x 10^5 m))

Calculating this, we find that the mass of the astronaut is approximately 7.49 kg.

Now, to find the weight of the astronaut when he returns to the Earth's surface, we can use the formula:

Weight = mg

Substituting the mass we just calculated, we get:

Weight = (7.49 kg) * (9.80 m/s^2)

Calculating this, we find that the weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.

Answer:

5.4 x 10⁶ Pa

Explanation:

h = depth to which penguins dive under seawater = 530 m

P₀ = Atmospheric pressure = 1.013 x 10⁵ pa

ρ = density of seawater = 1025 kg/m³

P = absolute pressure experienced by penguin at that depth

Absolute pressure is given as

P = P₀ + ρgh

Inserting the values

P = 1.013 x 10⁵ + (1025) (9.8) (530)

P = 5.4 x 10⁶ Pa

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = [tex]= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})[/tex]

kinetic energy = [tex]= \frac{1}{2}*4*(1.5^{2}-11^{2})[/tex] = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

The magnitude of the non-conservative force acting on the box is 236 J and it acts horizontally.

The magnitude and direction of the non-conservative force acting on the box can be determined using the work-energy theorem. The work done by the non-conservative force is equal to the change in kinetic energy of the box. From the given information, the initial kinetic energy of the box is 0.5 * 4.0 kg * (11 m/s)^2 = 242 J and the final kinetic energy is 0.5 * 4.0 kg * (1.5 m/s)^2 = 6 J. Therefore, the work done by the non-conservative force is 242 J - 6 J = 236 J. Since the box moves horizontally, the non-conservative force acts horizontally as well.

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Answer:

- 2.7 rad/s^2

Explanation:

f0 =  + 5.85 rotations per second (counter clockwise)

t = 21.3 s

f =  - 3.31 rotations per second (clockwise)

w0 = 2 x 3.14 x 5.85 = + 36.738 rad/s

w = - 2 x 3.14 x 3.31 = - 20.79 rad/s

Let α be teh angular acceleration.

α = (w - w0) / t

α = (-20.79 - 36.738) / 21.3

α = - 2.7 rad/s^2

The flywheel's average angular acceleration is -0.43 rad/s².

The average angular acceleration of the flywheel is determined by applying the following kinematic equation as shown below;

α = (ωf - ωi)/t

where;

α = (-3.31 - 5.85)/21.3

α = -0.43 rad/s²

Thus, the flywheel's average angular acceleration is -0.43 rad/s².

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