An electron (mass m=9.11×10−31kg) is accelerated from the rest in the uniform field E⃗ (E=1.45×104N/C) between two thin parallel charged plates. The separation of the plates is 1.90 cm . The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, with what speed does it leave the hole?

Answer:

18.7 m

Explanation:

[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s

θ = angle of launch from the horizontal = 47 deg

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 1.80 m

t = time of travel

Using the kinematics equation

[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]

- 1.80 = (9.5) t + (0.5) (- 9.8) t²

t = 2.11 s

Consider the motion along the horizontal direction

x = horizontal displacement of the shot

[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s

[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.11 s

Using the kinematics equation

[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]

x = (8.87) (2.11) + (0.5) (0) (2.11)²

x = 18.7 m

The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.

A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.

Given to us

The initial velocity of the shotput, u = 13 m/s

The angle from the horizontal, θ = 47.0°

The height from which shotput was thrown h = 1.80 m

As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,

[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]

Substitute the values,

[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]

R = 18.7 m

Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.

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Answer:

4.7 x 10⁵ J

Explanation:

m = mass of the car = 925 kg

d = distance traveled parallel to incline = 207 m

θ = angle of the slope = 14.5 deg

Force applied to move the car up is given as

F = mg Sinθ                                               eq-1

minimum work needed is given as

W = F d

using eq-1

W =  mgd Sinθ

inserting the values

W = (925)(9.8)(207) Sin14.5

W = 4.7 x 10⁵ J

Answer:

Liquid to a gas

Explanation:

When the molecules in a body move with increased speed, it's possible that the body will change from liquid to gas. The speed is increasing means they have more kinetic energy. The molecules of gas are very far apart from each other. They have much space between them so that they can move freely.

So, when the molecules move with increased speed, the body will change from liquid to gas. Hence, the correct option is (d) " liquid to gas".

Answer:

6 x 10⁻¹⁵ J

Explanation:

d = distance between the plates = 1.5 cm = 0.015 m

E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m

q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C

Force on the electron due to electric field is given as

F = q E

F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)

F = 4 x 10⁻¹³ N

KE₀ = initial kinetic energy of electron at negative plate = 0 J

KE = final kinetic energy of electron at positive plate = ?

Using work-change in kinetic energy

F d = KE - KE₀

(4 x 10⁻¹³) (0.015) = KE - 0

KE = 6 x 10⁻¹⁵ J

An electron released from rest towards the positive plate of a capacitor, which is separated by a distance of 1.5 cm and an electric field of 2.5 x 10^6 V/m, attains a kinetic energy of 37.5 keV when it reaches the positive plate.

The subject you're studying is called electrical potential energy, specifically its conversion into kinetic energy in the context of a parallel plate capacitor. In the case of an electron released from rest towards the positive plate of a capacitor, it is said to be moving through an electrical potential difference. This difference, coupled with the charge of the electron, provides the electron with energy, accelerating it.

Given that the electric field (E) is 2.5 x 106 V/m and the distance between the plates (d) is 1.5 cm or 0.015 m, we can use the formula E = V/d to calculate the potential difference (V). Substituting the given values, the potential difference is 2.5 x 106 V/m * 0.015 m = 37,500 V or 37.5 kV.

Furthermore, as per the relation that an electron accelerated through a potential difference of 1 V attains an energy of 1 electron-volt (eV), an acceleration through 37.5 kV will grant an energy of 37.5 keV. Since its initial kinetic energy was zero (as it was released from rest), this 37.5 keV is the kinetic energy of the electron just as it reaches the positive plate.

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B = 17.5mT.

A solenoid is a coil formed by a wire (usually copper) wound into a cylindrical spiral shape capable of creating a magnetic field that is extremely uniform and intense inside, and very weak outside.

To calculate the magnetic field generated inside the solenoid through which a current flows is given by the equation:

B = μ₀nI

Where  μ₀ is the constant of magnetic proportionality of the vacuum (4π x 10⁻⁷T.m/A), n is the relation between the number of turns of wire and its length given by N/L and I is the current flowing through the solenoid.

Given a long solenoid of length 0.67m, 1700.00 turns of wire and a current flowing through the wire of 5.50A. Calculate the magnetic field inside the solenoid.

B = (4π x 10⁻⁷T.m/A)(1700turns/0.67m)(5.50A)

B = 0.0175T

B = 17.5mT

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

[tex]U=\dfrac{1}{2}kx^2[/tex]...(I)

Using gravitational potential energy

[tex]U' =mgh[/tex]....(II)

Using energy of conservation

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

[tex]U_{i}+U'_{i}=U_{f}+U'_{f}[/tex]

[tex]\dfrac{1}{2}kx^2+0=0+mgh[/tex]

[tex]h=\dfrac{kx^2}{2mg}[/tex]

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

[tex]h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}[/tex]

[tex]h=11.653\ m[/tex]

Hence, The maximum height above the point of release is 11.653 m.

Answer:

Answer to the Question:

Explanation:

In this case, the total work done by the cable is zero, since in the aforementioned problem, the work depends only on the starting and ending point, these two being equal. Keeping its gravitational potential energy equal.

Answer:

The total work done is zero.

Explanation:

Given;

mass of the crate, m = 70.0 kg

height above ground through which the crate is lifted, h = 12.0 m

The only work associated in lifting and lowering this crate is gravitational potential energy.

Potential Energy during lifting = - mgh

                                                  = - 70 x 9.8 x 12

                                                  = - 8232 J

Potential Energy during lowering = mgh

                                                       =  70 x 9.8 x 12

                                                       =  8232 J  

Total total work done by the force = - 8232 J + 8232 J  = 0

Therefore, the total work done is zero.

answers:

m2 = 1.05 kg and m1 = 4.75 kg

Explanation:

the gravitational force is given by:

             Fg = Gm1×m2/(r^2)

9.20×10^-9 = [(6.67×10^-11)×(m1×m2)]/[(19×10-2)^2]

    (m1×m2) = 4.98 kg^2

             m1 = 4.98/m2 kg

but we given that:

m1 + m2 = 5.80 kg

4.98/m2 + m2 = 5.80

4.98 + (m2)^2 = 5.80×m2

(m2)^2 - 5.80×m2 + 4.98 = 0

by solving the quadratic equation above:

m2 = 4.75 kg or m2 = 1.05 kg

due to that from the information, m2 has a lighter mass, then m2 = 1.05 kg.

then m1 = 5.80 - 1.05 = 4.75 kg.

Answer:

6.30 L

Explanation:

P1 = P, V1 = 4.20 L, T1 = T

P2 = P/3, V2 = ?, T2 = T/2

Where, V2 be the final volume.

Use ideal gas equation

[tex]\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}[/tex]

[tex]V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}[/tex]

By substituting the values, we get

V2 = 6.30 L

Answer:

The two balls meet in 1.47 sec.

Explanation:

Given that,

Height = 25 m

Initial velocity of ball= 0

Initial velocity of another ball = 17 m/s

We need to calculate the ball

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2+h[/tex]

Where, u = initial velocity

h = height

g = acceleration due to gravity

Put the value in the equation

For first ball

[tex]s_{1}=0-\dfrac{1}{2}gt^2+25[/tex]....(I)

For second ball

[tex]s_{2}=17t-\dfrac{1}{2}gt^2+0[/tex]....(II)

From equation (I) and (II)

[tex]-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0[/tex]

[tex]t=\dfrac{25}{17}[/tex]

[tex]t=1.47\ sec[/tex]

Hence, The two balls meet in 1.47 sec.

The two balls will meet at a height of approximately 16.465 meters above the ground.

To find the height at which the two balls meet, we need to consider their respective equations of motion and solve for the height when they intersect.

For the tennis ball thrown from the building:

[tex]\[ h_1(t) = 25 - \frac{1}{2} g t^2 \][/tex]

For the ball thrown from the ground:

[tex]\[ h_2(t) = 17t - \frac{1}{2} g t^2 \][/tex]

To find the time when they meet, we'll set:

[tex]\[ 25 - \frac{1}{2} g t^2 = 17t - \frac{1}{2} g t^2 \]\[ 25 = 17t \]\[ t = \frac{25}{17} \][/tex]

Now, substitute this value of \( t \) into either equation to find the height at which they meet:

[tex]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \left(\frac{25}{17}\right)^2 \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 9.8 \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 17.07 \]\[ h_1\left(\frac{25}{17}\right) = 25 - 8.535 \]\[ h_1\left(\frac{25}{17}\right) = 16.465 \][/tex]

Answer:

the motorcycle travels 42.4 meters until it stops.

Explanation:

Vi= 18 m/s

Vf= 0 m/s

t= 4.7 sec

Vf= Vi - a*t

deceleration:

a= Vi/t

a= 18m/s  / 4.7 sec =>  a=-3.82 m/s²

x= Vi*t - (a * t²)/2

x= 42.4m

Answer:

5706.5

Explanation:

B = 7.171 T, i = 1000 A, μ0 = 1.25664 x 10^-6 T m/A, l = 1 m

Let the number of turns be N.

The magnetic field due to a current carrying solenoid is given by

B = μ0 x N x i / l

B x l / (μ0 x i) = N

N = 7.171 x 1 / (1.25664 x 10^-6 x 1000)

N = 5706.5

To design a superconducting solenoid that generates an interior field of 7.171 T with a maximum current of 1000 A, approximately 5709 turns per meter are required on a 1-meter length of the solenoid.

To determine the number of windings required on a 1-meter solenoid to achieve a magnetic field of 7.171 T with a maximum current of 1000 A, we can use the formula for the magnetic field inside a solenoid, which is given by:

B = μ_0 * n * I

Where:

Rearranging the formula to solve for n, we get:

n = B / (μ_0 * I)

Substituting the given values:

n = 7.171 T / (1.25664 × 10⁻⁶ T · m/A × 1000 A)

Now, calculating the value:

n = 7.171 / (1.25664 × 10⁻⁶ × 1000)

n = 7.171 / (1.25664 × 10⁻³)

n = 5708.57 turns per meter

To achieve the desired magnetic field, the solenoid must have approximately 5709 turns per meter to accommodate the rounding to the nearest whole number, since we can't have a fraction of a turn.

Answer:

Heat energy required = 687.96 kJ

Explanation:

Heat energy required, H = mCΔT.

Mass of cooking oil, m = 2.34 kg = 2340 g

Specific heat of cooking oil, C = 1.75 J/(g⋅°C)

Initial temperature = 23 °C

Final temperature = 191 °C

Change in temperature, ΔT = 191 - 23 = 168 °C

Substituting values

            H = mCΔT

            H = 2340 x 1.75 x 168 = 687960 J = 687.96 kJ

Heat energy required = 687.96 kJ

Answer:

1.84 m from the initial point (3.16 m from the ceiling)

Explanation:

According to the law of conservation of energy, the initial kinetic energy of the ball will be converted into gravitational potential energy at the point of maximum height.

Therefore, we can write:

[tex]\frac{1}{2}mv^2 = mg\Delta h[/tex]

where

m = 2 kg is the mass of the ball

v = 6 m/s is the initial speed of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\Delta h[/tex] is the change in height of the ball

Solving for [tex]\Delta h[/tex],

[tex]\Delta h = \frac{v^2}{2g}=\frac{6^2}{2(9.8)}=1.84 m[/tex]

So, the ball raises 1.84 compared to its initial height.

Therefore:

- if we take the initial position of the ball as reference point, its maximum height is at 1.84 m

- if we take the ceiling as reference point, the maximum height of the ball will be

5 m - 1.84 m = 3.16 m from the ceiling

Answer:

Angle between the magnetic field and the wire's velocity is 15.66 degrees.

Explanation:

It is given that,

Potential difference or emf, V = 35 mV = 0.035 V

Length of wire, l = 12 cm= 0.12 m

Magnetic field, B = 0.27 T

Speed, v = 4 m/s

We need to find the angle between the magnetic field and the wire's velocity. We know that emf is given by :

[tex]\epsilon=Blv\ sin\theta[/tex]

[tex]sin\theta=\dfrac{\epsilon}{Blv}[/tex]

[tex]sin\theta=\dfrac{0.035\ V}{0.27\ T\times 0.12\ m\times 4\ m/s}[/tex]

[tex]sin\theta=0.25[/tex]

[tex]\theta=15.66^{\circ}[/tex]

So, the angle between the magnetic field and the wire's velocity is 15.66 degrees.

The angle between the wire's velocity and the magnetic field, when they are perpendicular to each other, is 90 degrees. This impacts the force exerted on the wire in the magnetic field.

The question pertains to the interaction of a moving conductor wire in a magnetic field - a fundamental concept in electromagnetism. From the question, the velocity of the wire is perpendicular to the magnetic field. This is the key because the magnetic force on a moving charge within a magnetic field depends on the angle between the charge's velocity and the direction of the magnetic field.

According to the fundamentals of physics, when velocity is perpendicular to the magnetic field, the angle between them is 90 degrees. The force exerted on the wire due to the magnetic field is then given by F = qvBsinθ, where q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. With θ being 90 degrees, the sin(90°) equals 1, and this simplifies the calculation. So the angle developed between the magnetic field and the wire's velocity is 90 degrees in this case.

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Answer:

6.889 ft/s^2

Explanation:

1 m = 3.28 feet

So, 2.1 m/s^2 = 2.1 × 3.28 ft/s^2

= 6.889 ft/s^2

Explanation:

a)

The circumference of the path is:

C = 2πr

C = 2π (0.235 m)

C = 1.48 m

Velocity = displacement / time

v = 1.48 m / 1.95 s

v = 0.757 m/s

b)

1 rev / 1.95 s = 0.513 Hz

c)

1 rev / 1.95 s × (2π rad / rev) = 3.22 rad/s

(a) The translational speed of the object is 0.76 m/s.

(b) The frequency of the object's motion is 0.51 Hz.

(c) The angular speed of the object is 3.22 rad/s.

The angular speed of the object is calculated as follows;

ω = 1 rev/ 1.95 s = 0.51 rev/s

ω =  0.51 rev/s x 2π rad

ω = 3.22 rad/s

The frequency of the object's motion is determined from the angular speed as shown below;

ω = 2πf

f = ω/2π

f = (3.22)/2π

f = 0.51 Hz

The translational speed of the object is calculated as follows;

v = ωr

v = 3.22 x 0.235

v = 0.76 m/s

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Answer:

a) 45.87 m

b) 2.08 N

Explanation:

Mass of projectile=0.85 kg=m

Velocity of projectile=30 m/s=u = initial velocity

Final velocity =0

g= acceleration due to gravity=9.81 m/s²

h=maximum height of the projectile

a) In this case loss of Kinetic energy (K.E.) = loss in potential energy (P.E.)

ΔK.E.=ΔP.E.

[tex]\frac{1}{2}mu^2-\frac{1}{2}mv^2=mgh[/tex]

[tex]\frac{1}{2}m\times u^2=mgh\\\Rightarrow h=\frac {u^2}{2\times g}\\\Rightarrow h=\frac {30^2}{2\times 9.81}\\\therefore h=45.87\ m[/tex]

b) h'=height of the projectile=36.7 m

F=Average force due to air resistance

There will be a loss of P.E. due to air resistance

ΔP.E.=mg(h-h')

F×h'=mg(h-h')

F×36.7=0.85×9.81(45.87-36.7)

[tex]F=\frac{0.85\times 9.81(45.87-36.7)}{36.7}[/tex]

∴ F=2.08 Newton

Answer:

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]

here we know that

[tex]F = 7.83 \times 10 N/m[/tex]

[tex]d = 7.0 cm = 0.07 m[/tex]

[tex]i_1 = 28 A[/tex]

now we will have

[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]

[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Answer:

Specific heat of the granite rock = 3387.05 Jkg⁻¹°C⁻¹

Explanation:

We have heat required, H = mcΔT

Mass of granite, m = 0.500 kg

Specific heat of granite, c = ?

Change in temperature, ΔT = 21°C

Heat energy, H = 8.5 kcal = 8500 x 4.184 = 35564 J

Substituting

         H = mcΔT

         35564 = 0.500 x c x 21

          c = 3387.05 Jkg⁻¹°C⁻¹

Specific heat of the granite rock = 3387.05 Jkg⁻¹°C⁻¹

Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

The average acceleration of the car is 2.45 m/s^2. If the car maintains this acceleration, it would be moving at a speed of 117.65 km/h 4.0 seconds later.

(a) To find the average acceleration, we can use the formula: average acceleration (a) = (final velocity - initial velocity) / time. Converting the velocities to m/s gives us 13.47 m/s and 22.28 m/s, respectively. Plugging in the values, we get: a = (22.28 - 13.47) m/s / 3.6 s = 2.45 m/s^2. Therefore, the average acceleration of the car is 2.45 m/s^2.

(b) Since the car is maintaining the same acceleration, we can use the kinematic equation: final velocity = initial velocity + acceleration * time. Converting the initial velocity to m/s, we have 22.28 m/s. Plugging in the values, we get: final velocity = 22.28 m/s + 2.45 m/s^2 * 4.0 s = 32.68 m/s. Converting back to km/h gives us 32.68 m/s * 3.6 km/h/m = 117.65 km/h.

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Final answer:

The average acceleration of the sports car is approximately 2.446 m/s^2, and if it maintained this acceleration for an additional 4.0 seconds, it would be moving at roughly 115.4 km/h.

Explanation:

Calculating the Average Acceleration and Future Velocity

To find the average acceleration (a) of the car, we use the formula a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time taken for the change in velocity. First, we need to convert the velocities from km/h to m/s by multiplying by (1000 m/km) / (3600 s/h).

v_i = 48.5 km/h = (48.5 * 1000) / 3600 m/s ≈ 13.472 m/s

v_f = 80.2 km/h = (80.2 * 1000) / 3600 m/s ≈ 22.278 m/s

Now, we can calculate the average acceleration:

a = (22.278 m/s - 13.472 m/s) / 3.6 s ≈ 2.446 m/s²

To find the velocity 4.0 seconds after the car reaches 80.2 km/h, we use the formula v = v_f + a * t:

v = 22.278 m/s + 2.446 m/s² * 4.0 s ≈ 32.062 m/s

To convert this back to km/h:

v = 32.062 m/s * (3600 s/h) / (1000 m/km) ≈ 115.4 km/h

Answer:

[tex]\frac{dB}{dt}[/tex] = 591.45 T/s

Explanation:

i = induced current in the loop = 0.367 A

R = Resistance of the loop = 117 Ω

E = Induced voltage

Induced voltage is given as

E = i R

E = (0.367) (117)

E = 42.939 volts

[tex]\frac{dB}{dt}[/tex] = rate of change of magnetic field

A = area of loop = 7.26 x 10⁻² m²

Induced emf is given as

[tex]E = A\frac{dB}{dt}[/tex]

[tex]42.939 = (7.26\times 10^{-2})\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt}[/tex] = 591.45 T/s

Answer:

29.3 rad/s

Explanation:

Moment of inertia in straight position, I1 = I

He takes 2 rotations in 1.5 second

So, Time period, T = 1.5 / 2 = 0.75 second

w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s

Moment of inertia in tucked position, I2 = I / 3.5

Let the new angular speed is w2.

By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.

L1 = L2

I1 w1 = I2 w2

I x 8.37 = I / 3.5 x w2

w2 = 29.3 rad/s

To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. The moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. By applying the conservation of angular momentum equation, we can calculate the angular speed in the straight position.

To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. According to the conservation of angular momentum, the product of the initial moment of inertia and initial angular speed is equal to the product of the final moment of inertia and final angular speed.

Let's assume the initial angular speed in the tuck position is denoted by w' and the final angular speed in the straight position is denoted by w. We are given that the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. Therefore, the final moment of inertia (I) is 3.5 times greater than the initial moment of inertia (I').

The conservation of angular momentum equation can be written as:

I' * w' = I * w

Since the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position, we have:
w = w' * (I' / I)

Using the given values, we can calculate the angular speed in the straight position.

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Answer:

0.35 T

Explanation:

Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s

Use the Faraday's law of electromagnetic induction

e = rate of change of magnetic flux

Let b be the strength of magnetic field.

e = dФ / dt

e = d ( B A) / dt

e = B x dA / dt

0.0271 = B x 0.0785

B = 0.35 T

The magnitude of the magnetic field is 3.45 × 10^-4 T.

The magnitude of the induced emf in a square, single-turn coil can be calculated using the equation: emf = -N * A * (d(B)/dt), where N is the number of turns, A is the area of the coil, and d(B)/dt is the rate of change of the magnetic field. In this case, the area of the coil is decreasing at a rate of 0.0785 m2/s, and the emf is given as 27.1 mV. We can rearrange the equation to solve for the magnitude of the magnetic field (B):

B = -emf / (N * d(A)/dt) = -27.1 mV / (1 * 0.0785 m2/s) = -345.222 T/s = 3.45 × 10^-4 T/s.

Therefore, the magnitude of the magnetic field is 3.45 × 10^-4 T.

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Answer:

[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

[tex]\lambda = \frac{q}{L}[/tex]

now the electric field due to long line charge at a distance d from it is given as

[tex]E = \frac{2k\lambda}{d}[/tex]

[tex]E = \frac{q}{2\pi \epsilon_0 d}[/tex]

now the force on the other charge in this electric field is given as

[tex]F = QE[/tex]

[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]

Explanation:

It is given that,

Mass of the object, m = 3.66 kg

Charge, q = -19 μC = -19 × 10⁻⁶ C

It is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground such that,

[tex]F_g=F_e[/tex]

[tex]F_g\ and\ F_e[/tex] are gravitational and electrostatic forces respectively

[tex]mg=qE[/tex]

[tex]E=\dfrac{mg}{q}[/tex]

[tex]E=\dfrac{3.66\ kg\times 9.8\ m/s^2}{-19\times 10^{-6}\ C}[/tex]

E = −1887789.47 N/C

[tex]E=-1.89\times 10^6\ N/C[/tex]

Negative sign shows that the electric field is in the opposite direction of the electric force. Since, the weight of the object is in downward direction and its electric force (which is balancing its weight) is in upward direction. So, we can say that the electric field is in downward direction.

Answer:

0.273 m/s

Explanation:

Momentum is conserved:

mu = (m + M) v

(0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v

v = 0.273 m/s

The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.The final velocity will be 0.273 m/sec.

The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.

The given data in the problem is;

m is the mass of clay model before collision= 0.200 Kg

u is the sliding velocity before collision=0.75 m/sec

M is the mass after collision= 0.350 Kg

v is the velocity after collision=?

On applying the law of conservation of momentum we get;

[tex]\rm mu = (m + M) v \\\\\ (0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v \\\\ v = 0.273 m/s[/tex]

To learn more about the law of conservation of momentum refer ;

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To calculate the change in entropy during the approach to equilibrium, use the formula ΔS = mcΔT. For the aluminum block, ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum. For the water, ΔS_water = m_water * c_water * ΔT_water. Set ΔT_aluminum equal to ΔT_water and solve for T_water. Add the change in entropy for the aluminum and water to find the total change in entropy. Simplify the equation by converting cal to J and kg to g. Set the total change in entropy equal to zero and solve for T_aluminum to find the initial temperature of the aluminum.

To calculate the change in entropy during the approach to equilibrium, we need to use the formula for entropy change:

ΔS = mcΔT

where ΔS is the change in entropy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the aluminum block is dropped into the water, so the final temperature will be the same for both the aluminum and the water. We can calculate the change in entropy for each substance separately and then add them together to find the total change in entropy.

For the aluminum block:

ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum

ΔS_aluminum = 2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)

For the water:

ΔS_water = m_water * c_water * ΔT_water

ΔS_water = 5.00 kg * 1 cal/(g∙K) * (T_water - 20.0 °C)

Since the final temperature is the same for both substances, we can set ΔT_aluminum equal to ΔT_water and solve for T_water:

50.0 °C - T_aluminum = T_water - 20.0 °C

70.0 °C - T_aluminum = T_water

Substituting this value into the equation for ΔS_water:

ΔS_water = 5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C]

Now we can add the change in entropy for the aluminum and water:

ΔS_total = ΔS_aluminum + ΔS_water

ΔS_total = (2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

To simplify the equation, we can convert cal to J by multiplying by 4.184 and divide both sides by 1000 to convert kg to g:

ΔS_total = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

We can now solve for T_aluminum by setting ΔS_total equal to 0:

0 = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

Simplifying the equation:

0 = (8.368 J/(g∙K) * (50.0 °C - T_aluminum)) + (20.92 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))

Combining like terms:

0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))

Simplifying further:

0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum) - 104.6 J/(g∙K) * (70.0 °C - T_aluminum)

Expanding the equation:

0 = (29.288 J/(g∙K) * 50.0 °C - 29.288 J/(g∙K) * T_aluminum) - (104.6 J/(g∙K) * 70.0 °C - 104.6 J/(g∙K) * T_aluminum)

Simplifying the equation further:

0 = (1464.4 J - 29.288 J/(g∙K) * T_aluminum) - (7322 J - 104.6 J/(g∙K) * T_aluminum)

Combining like terms:

0 = 1464.4 J - 29.288 J/(g∙K) * T_aluminum - 7322 J + 104.6 J/(g∙K) * T_aluminum

Simplifying the equation:

0 = -5857.6 J + 75.312 J/(g∙K) * T_aluminum

Isolating T_aluminum:

5857.6 J = 75.312 J/(g∙K) * T_aluminum

T_aluminum = (5857.6 J) / (75.312 J/(g∙K))

T_aluminum ≈ 77.7 °C

Therefore, the initial temperature of the aluminum was approximately 77.7 °C.