A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85x10-12 C2/N.m2 . Find the energy U1 of the dielectric-filled capacitor.

The ball's position vector has components

[tex]x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ t[/tex]

[tex]y=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2[/tex]

where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. The ball hits the ground when [tex]y=0[/tex]:

[tex]0=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2\implies t=1.22\,\mathrm s[/tex]

The ball's velocity vector has components

[tex]v_x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ[/tex]

[tex]v_y=\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ-gt[/tex]

so that after 1.22 s, the velocity vector is

[tex]\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]

and the magnitude is

[tex]\|\vec v\|=\sqrt{6.13^2+(-6.79)^2}\,\dfrac{\rm m}{\rm s}=\boxed{9.14\dfrac{\rm m}{\rm s}}[/tex]

(a) [tex]2.56\cdot 10^4 J[/tex]

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

[tex]W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J[/tex]

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

[tex]E=2.56\cdot 10^4 J[/tex]

Therefore we can set up a simple proportion

[tex]1 cal : 4186 J = x : 2.56\cdot 10^4 J[/tex]

to find the equivalent energy in calories:

[tex]x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal[/tex]

The cheetah needs 25,972.35 Joules or 6.2 Calories of work to reach its top speed of 31.7 m/s.

The cheetah accelerates from a state of rest to its top speed. This involves work done on the cheetah which we can compute using the formula for kinetic energy, as work done is equal to change in kinetic energy. The initial speed of the cheetah is 0, so initial kinetic energy is also 0. The final kinetic energy when the cheetah is at its top speed is ½ m v^2, where m is the mass and v is the speed.

(a) So, the work done, W = ½ * 51.0kg * (31.7 m/s)^2 =  25972.35 J,

(b) To convert Joules into Calories, we use 1 Calorie = 4186 J, so the Calories of work done = 25972.35 J / 4186 = 6.2 Calories. Note that due to inefficiencies in the energy conversion process in bodies, the actual energy produced by the cheetah's body will be more than 6.2 Calories to reach this speed.

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3. The nuclear potential that binds protons and neutrons in the nucleus of an atom

is often approximated by a square well. Imagine a proton confined in an infinite

square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength

and energy associated with the photon that is emitted when the proton undergoes a

transition from the first excited state (n = 2) to the ground state (n = 1). In what

region of the electromagnetic spectrum does this wavelength belong?

Answer 3

We are given that,

Length of square well = L = 10−5

nm = 10−14 m.

Energy of proton in state n is given by,

En =

π

2n

2~

2

2mpL2

,

where L is the width of the square well.

⇒ E1 =

π

2~

2

2mpL2

E2 =

4π

2~

2

2mpL2

·

Answer:

U = 4.39 J

Explanation:

Electric field energy stored in the medium or vacuum is given as

[tex]U = \frac{1}{2}\epsilon_0 E^2 V[/tex]

here we know that

[tex]\epsilon_0 = 8.85 \times 10^{-12} [/tex]

E = 81.1 V/m

V = volume

[tex]V = (0.0253)(speed \times time)[/tex]

[tex]V = (0.0253)(3\times 10^8 \times 19.9)[/tex]

[tex]V = 1.51 \times 10^8 m^3[/tex]

now from above formula we have

[tex]U = \frac{1}{2}(8.85 \times 10^{-12})(81.1)^2(1.51 \times 10^8)[/tex]

[tex]U = 4.39 J[/tex]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

[tex]v_x = v cos \theta[/tex]

where

v = 12.0 m/s is the initial velocity

[tex]\theta=51.0^{\circ}[/tex] is the angle between the direction of v and the horizontal

Substituting,

[tex]v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s[/tex]

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

[tex]d = v_x t = (7.55 m/s)(2.08 s)=15.7 m[/tex]

Final answer:

The shot putter threw the shot approximately 15.71 meters by using the horizontal component of the initial velocity and the time of flight.

Explanation:

To calculate the distance the shot putter threw the shot, we need to break down the velocity into its horizontal and vertical components. Given that the shot was released with a velocity of 12.0 m/s at an angle of 51.0 degrees above the horizontal, we can use trigonometry to find these components.

The horizontal velocity (vx) is given by vx = v * cos(θ) = 12.0 m/s * cos(51.0) = 7.55 m/s. The vertical velocity (vy) is vy = v * sin(θ) = 12.0 m/s * sin(51.0) = 9.35 m/s.

Since there's no acceleration in the horizontal direction (ignoring air resistance), this component of the velocity will remain constant until the shot hits the ground. Thus, the horizontal distance (range) the shot travels is simply the product of the horizontal velocity and the time it's in the air, given by Range = vx * t = 7.55 m/s * 2.08 s = 15.71 meters. So, the shot putter threw the shot approximately 15.71 meters.

Answer:

D. a very small positive charge with little s

Explanation:

A test charge is a very small charge with positive value which do not disturb the electric field exist in the region

So test charge is to find out the strength of electric field that exist in the region.

If the magnitude of test charge is large then it will change the strength of the existing electric field and due to this the value of the force will be altered.

So here in this case the test charge must be small as well as it must be positive nature

Answer:

1.39 W

Explanation:

The volume of blood is

[tex]V=7300 L = 7.3 m^3[/tex]

the density is

[tex]\rho = 1050 kg/m^3[/tex]

So the total mass of blood lifted in one day is

[tex]m=\rho V=(1050 kg/m^3)(7.3 m^3)=7665 kg[/tex]

So the total work done is:

[tex]W=mgh=(7665 kg)(9.8 m/s^2)(1.6 m)=1.2\cdot 10^5 J[/tex]

The total time taken is one day, so

[tex]t=24 h = 86400 s[/tex]

So the power output is

[tex]P=\frac{W}{t}=\frac{1.2\cdot 10^5 J}{86400 s}=1.39 W[/tex]

Final answer:

The sprinter's total time in the 100 m dash is 10 seconds.

Explanation:

In this problem, the sprinter reaches his top speed of 11.2 m/s in 2.14 seconds.

To find the total time, we need to consider two parts: the time it takes to reach the top speed and the time it takes to cover the remaining distance at that speed.

First, calculate the time it takes to reach the top speed using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values: 11.2 = 0 + a(2.14), we can solve for a to find a = 5.23 m/s².

Now, to find the time it takes to cover the remaining distance at the top speed, we can use the equation:

t = d / v

where d is the distance and v is the velocity.

Since the total distance is 100 m, subtracting the distance covered during the acceleration phase, we get the remaining distance as 100 - (0.5)(5.23)(2.14)² = 88.2 m.

Dividing this distance by the top speed of 11.2 m/s, we find t = 88.2 / 11.2 = 7.86 s.

To find the total time, we add the time it takes to reach the top speed and the time it takes to cover the remaining distance, so the sprinter's total time is 2.14 s + 7.86 s = 10 s.

Answer:

option (A)

Explanation:

electric flux is defined as the number of electric field lines which crosses through any area.

It is given by

Ф = E . A (It is the dot product of electric field vector and area vector)

According to the question, the angle between electric filed vector and area vector is θ.

So, electric flux

Ф = E x π R^2 Cosθ

The electric flux through a disk in a uniform electric field is represented by E(πR^2)cosΘ, so the correct answer is C: E(πR²)cosΘ.

The question is asking about the electric flux through a disk when the disk's normal is oriented at an angle Θ with respect to a uniform electric field. Electric flux is given by the product of the electric field strength, the area through which the field is passing, and the cosine of the angle between the field and the normal to the surface. The formula for the electric flux through a surface is Φ = E * A * cos(Θ), where E is the electric field strength, A is the area of the surface, and Θ is the angle between the electric field and the normal to the surface. For a disk with radius R, the area is πR². Thus, the correct answer for the electric flux through the disk is C: E(πR²)cosΘ.

The rock strikes the ground after approximately 3.67 seconds. The speed of the rock just before it strikes the ground is approximately 17.0 m/s.

To find the time it takes for the rock to strike the ground, we can use the equation for vertical motion. Assuming negligible air resistance, the initial velocity is 18.0 m/s and the vertical displacement is 50.0 m.

Using the equation y = yo + vyo*t - 1/2*g*t^2, where y is the final position, yo is the initial position, vyo is the initial velocity, g is the acceleration due to gravity, and t is the time, we can solve for t. Plugging in the values, we get:

50.0 = 0 + 18.0*t - 1/2*9.8*t^2

After rearranging the equation and solving the quadratic equation, we find that t = 3.67 seconds.

To find the speed of the rock just before it strikes the ground, we can use the equation for vertical motion. The final velocity v is equal to the initial velocity vyo - g*t. Plugging in the values, we get:

v = 18.0 - 9.8*3.67

Calculating the value, we find that the speed of the rock just before it strikes the ground is approximately 17.0 m/s.

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(a) 168.2 ft/s

The vertical position of the ball is given by

[tex]s = 144t - 16t^2[/tex]

where t is the time.

By differentiating this expression, we find the velocity:

[tex]v = 144-32 t[/tex]

The maximum height is reached when the velocity is zero, so:

[tex]0 = 144 - 32 t[/tex]

From which we find

[tex]t = \frac{144}{32}=4.5 s[/tex]

And substituting this value into the equation for s, we find the maximum height:

[tex]s = 144(4.5 s)-16(4.5 s)^2=324 ft[/tex]

(b) 16 ft/s

We want to find the velocity of the ball when the position of the ball is

s = +320 ft

Substituting into the equation for the position,

[tex]320 = 144t-16t^2[/tex]

[tex]16t^2 -144t +320 = 0[/tex]

Solving for t, we find two solutions:

t = 4 s

t = 5 s

The first one corresponds to the instant in which the ball is still on its way up: Substituting into the equation for the velocity, we find the velocity of the ball at that time

[tex]v = 144 - 32 t=144- 32(4 s)=16 ft/s[/tex]

(c) -16 ft/s

Now we want to find the velocity of the ball when the position of the ball is

s = +320 ft

but on its way down. In the previous part, we found

t = 4 s

t = 5 s

So the second time corresponds to the instant in which the ball is at s = 320 ft but on the way down.

Substituting t = 5 s into the equation for the velocity, we find:

[tex]v = 144 - 32 t=144- 32(5 s)=-16 ft/s[/tex]

And the negative sign means the direction is downward.

The answers for the ball thrown vertically upward with a velocity of 144 ft/s and with a height after t seconds of s = 144t - 16t² are:

a) The maximum height reached by the ball is 324 ft.

b) The velocity of the ball when it is 320 ft above the ground on its way up is 16 ft/s.

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s.

a) The maximum height reached by the ball can be calculated with the given equation:

[tex] s = 144t - 16t^{2} [/tex]   (1)

Where:

s: is the height

t: is the time

We can find the time with the following equation:

[tex] v_{f} = v_{i} - gt [/tex]   (2)

Where:

[tex] v_{f} [/tex]: is the final velocity = 0 (at the maximum height)

[tex] v_{i} [/tex]: is the initial velocity = 144 ft/s

g: is the acceleration due to gravity = 32 ft/s²    

Solving equation (2) for t and entering into equation (1), we can find the maximum height:

[tex]s = 144t - 16t^{2} = 144(\frac{v_{i}}{g}) - 16(\frac{v_{i}}{g})^{2} = 144(\frac{144 ft/s}{32 ft/s^{2}}) - 16(\frac{144 ft/s}{32 ft/s{2}})^{2} = 324 ft[/tex]  

Hence, the maximum height is 324 ft.

b) To find the velocity of the ball when it is 320 ft above, we can use the following equation:                       

[tex] v_{f}^{2} = v_{i}^{2} - 2gs [/tex]

[tex]v_{f}^{2} = (144 ft/s)^{2} - 2*32 ft/s^{2}*320 ft[/tex]    

The above equation has two solutions:

[tex]v_{f_{1}} = 16 ft/s[/tex]

[tex]v_{f_{2}} = -16 ft/s[/tex]

Since the question is for the velocity of the ball on its way up and considering the way up as the positive direction, the answer is the positive value [tex]v_{f_{1}} = 16 ft/s[/tex].                      

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s (we take the negative value calculated above, [tex] v_{f_{2}}[/tex]). We consider the way down as the negative direction.

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(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

[tex]I= \Delta p = m \Delta v = m (v-u)[/tex]

where

m is the mass of the object

[tex]\Delta v[/tex] is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

[tex]I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s[/tex]

(b) 155 N

The impulse can also be rewritten as

[tex]I=F \Delta t[/tex]

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

[tex]\Delta t[/tex] is the duration of the collision

In this situation, we have

[tex]\Delta t = 0.06 s[/tex]

So we can re-arrange the equation to find the magnitude of the average force:

[tex]F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N[/tex]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

[tex]v^2 - u^2 = 2ad[/tex]

To find the acceleration of the car, a:

[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2[/tex]

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

[tex]F=(1100 kg)(-2.23 m/s^2)=-2451 N[/tex]

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) [tex]-1.53\cdot 10^5 N[/tex]

We can use again the equation

[tex]v^2 - u^2 = 2ad[/tex]

To find the acceleration of the car. This time we have

[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2[/tex]

So now we can find the force exerted on the car by using again Newton's second law:

[tex]F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N[/tex]

As we can see, the force is much stronger than the force exerted in part a).

Answer:

The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

[tex]\phi=BA\costheta[/tex]

Where, B = magnetic field

A = area

Put the value into the formula

[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}[/tex]

[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927[/tex]

[tex]\phi=6.6\times10^{-4}\ T-m^2[/tex]

Hence, The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].

Answer:

4 s

Explanation:

u = 19.6 m/s, g = 9.8 m /s^2

Let the time taken to reach the maximum height is t.

Use first equation of motion.

v = u + at

At maximum height, final velocity v is zero.

0 = 19.6 - 9.8 x t

t = 19.6 / 9.8 = 2 s

As the air resistance be negligible, is time taken to reach the ground is also 2 sec.

So, total time taken be the ball to reach at original point = 2 + 2 = 4 s

The ball takes 4 seconds to return to its original launch point.

To find the time it takes for the ball to return to its original launch point, we need to determine the total time it takes for the ball to reach its maximum height and come back down. The ball is shot straight up, which means its initial velocity is positive and its final velocity is negative (when it returns to the launch point). We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed. In this case, the final velocity is -19.6 m/s (negative because it's going down), the initial velocity is 19.6 m/s, and the acceleration is -9.8 m/s^2 (due to gravity). Plugging the values into the equation and solving for t:

v = u + at

-19.6 = 19.6 - 9.8t

-39.2 = -9.8t

t = -39.2 / -9.8

t = 4 seconds

It takes 4 seconds for the ball to return to its original launch point.

Answer:

Option (D)

Explanation:

According to the Gauss theorem in electrostatics, the electric flux passing through any surface is equal to the one divided by epsilon note ties the total cahrge enclosed in the surface.

As the flux is zero it means the enclosed charge is zero. It means the sum of the total cahrge inside is zero.

Answer:

41.4 m/s

Explanation:

Consider downward direction of motion as negative

v₀ = initial velocity of the package as it is dropped over the side = 12 m/s

v = final velocity of the package just before it hits the ground

y = vertical displacement of the package = - 80 m

a = acceleration = - 9.8 m/s²

Using the kinematics equation

v² = v₀² + 2 a y

v² = 12² + 2 (- 9.8) (- 80)

v² = 144 + 1568

v = - 41.4 m/s

The negative sign indicates the downward direction of motion.

Hence the speed of package comes out to be 41.4 m/s

Answer:

The angle the wire make with respect to the magnetic field is 30°

Explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

[tex]F=ILB\ sin\theta[/tex]

[tex]\theta=sin^{-1}(\dfrac{F}{ILB})[/tex]

[tex]\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})[/tex]

[tex]\theta=30^{\circ}[/tex]

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

Final answer:

The crate will hit the road before your car hits the crate. The horizontal speed of the crate is the same as that of the truck.

Explanation:

(a) If you are driving directly behind a pickup truck at the same speed and neither brake nor swerve, the crate will hit the road before your car hits the crate. This is because the crate and your car are both traveling at the same horizontal speed, and the crate will have a shorter distance to fall than your car would have to travel to reach the crate.

(b) During the fall, the horizontal speed of the crate is the same as that of the truck. This is because both the truck and the crate are moving at the same speed horizontally, and gravity acts only vertically on the falling crate.

Answer:

10.9 m

Explanation:

We can solve the problem by using the law of conservation of energy.

The initial mechanical energy is just the kinetic energy of the ball:

[tex]E = K_i = \frac{1}{2}mu^2[/tex]

where m is the mass of the ball and u = 16.9 m/s the initial speed.

At a height of h, the total mechanical energy is sum of kinetic energy and gravitational potential energy:

[tex]E=K_f + U_f = \frac{1}{2}mv^2 + mgh[/tex]

where v is the new speed, g is the gravitational acceleration, h is the height of the ball.

Due to the conservation of energy,

[tex]\frac{1}{2}mu^2 = \frac{1}{2}mv^2 +mgh\\u^2 = v^2 + 2gh[/tex] (1)

Here, at a height of h we want the speed to be 1/2 of the initial speed, so

[tex]v=\frac{1}{2}u[/tex]

So (1) becomes

[tex]u^2 = (\frac{u}{2})^2+2gh\\\frac{3}{4}u^2 = 2gh[/tex]

So we can find h:

[tex]h=\frac{3u^2}{8g}=\frac{3(16.9 m/s)^2}{8(9.8 m/s^2)}=10.9 m[/tex]

Final answer:

To find the height where the ball has one-half its initial speed, we can use the equations vf = v0 + gt and d = v0t - 0.5gt2.

Explanation:

To find the height above its initial position where the ball has one-half its initial speed, we need to use the fact that the initial velocity (v0) of the ball is 16.9 m/s. At the highest point of the ball's trajectory, the velocity will be zero. We can use the formula vf = v0 + gt, where vf is the final velocity, g is the acceleration due to gravity, and t is the time it takes for the ball to reach its highest point.

By substituting vf = 0 and v0 = 16.9 m/s, we can solve for t. Once we have the value of t, we can use the equation d = v0t - 0.5gt2 to calculate the height (d) above the initial position where the ball will have one-half its initial speed.

By substituting the calculated value of t into the equation, we can find the value of d.

Answer: it should be cut with a chainsaw

Explanation:

A bolt cutter is usually the preferred tool to use to cut through a chain link fence quickly, taking into account the thickness and hardness of the chain link fence. Safety precautions should be taken while using such tools.

To cut through a chain link fence quickly without undue strain, the preferred tool is typically a bolt cutter. Bolt cutters possess the strength and design needed to snip through metal links easily. They come in various sizes, and the size needed would depend on the thickness and hardness of the chain link fence. Ideally, a medium-sized bolt cutter would be used for a standard fence. However, it's advisable to wear protective gear while using such tools, as the cut metal links might be razor-sharp and could cause injuries.

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Answer:

Explanation:

Lenz law is used to find the direction of induced emf in the coil.

It state taht the direction of induced emf in the coil is such that it always opposes the change due to which it is produced.

Suppose there is a coil and a north pole of the magnet comes nearer to the coil. Due to changing magnetic flux an induced emf is developed in the coil whose direction is such that the north pole moves away. That means this face of the coil behaves like a north pole and the current flows at this face is in anticlockwise wise direction.

Final answer:

Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by a changing magnetic field.

Explanation:

Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. This means that when there is a change in the magnetic field through a circuit, the induced current will create a magnetic field that acts against the change.

For example, if a magnet is brought near a wire loop, the induced current will flow in such a way that it creates a magnetic field that opposes the motion of the magnet. This is because the changing magnetic field induces an emf in the wire loop, and Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by the magnet.

Answer:

Displacement of the car, XY = 81.92 km

Explanation:

From the attached figure,

OX = 64.9 km (in north direction)

OY = 50 km (in west direction)

We have to find the displacement of the car. The shortest path covered by a car is called displacement of the car. Here, XY shows the displacement of the car :

Using Pythagoras equation as :

[tex]XY^2=OX^2+OY^2[/tex]

[tex]XY^2=(64.9\ km)^2+(50\ km)^2[/tex]

XY = 81.92 km

Hence, the displacement of the car is 81.92 km.

The magnitude of the car's displacement is calculated using the Pythagorean theorem by treating the westward and northward movements as the sides of a right-angled triangle. The displacement, representing the hypotenuse of that triangle, is found to be approximately 81.93 km.

The student has asked to find the magnitude of displacement of a car that moves 50.0 km West and then 64.9 km North.

Calculating Displacement

To calculate the resultant displacement, we treat the distances as vector quantities and use the Pythagorean theorem. The car's westward and northward movements are at right angles to each other, so we can draw this as a right-angled triangle where the westward distance is one side (50.0 km), the northward distance is the other side (64.9 km), and the hypotenuse will be the displacement.

Using the Pythagorean theorem:

The magnitude of the car's displacement, therefore, is approximately 81.93 km.

Answer:

Focal length, f = 16 cm

Explanation:

Image distance, v = 24 cm

Object distance, u = -48 cm

We need to find the focal length of the lens. It can be determined using the lens formula as :

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex]

[tex]\dfrac{1}{24\ cm}-\dfrac{1}{(-48\ cm)}=\dfrac{1}{f}[/tex]

f = 16 cm

So, the focal length of the converging lens is 16 cm. Hence, this is the required solution.

Answer:

f = 16 cm

Explanation:

If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left of a lens, the focal length of the lens is 16 cm.

Answer: [tex]1.0052m/s^{2}[/tex]

Explanation:

Assuming there is only force in the y-component, the total net force [tex]F_{y}[/tex] acting on the parachute and the sky diver is:

[tex]F_{y}=F_{D}-W[/tex]   (1)

Where:

[tex]F_{D}=1032N[/tex] is the drag force acting upwards

[tex]W=936N[/tex] is the weight of the sky diver acting downwards, hence with negative sign

Then:

[tex]F_{y}=1032N-936N=96N[/tex]   (2) This is the total net force excerted on the system parachute-sky diver, and the fact it is positive means is upwards

Now, according Newton's 2nd Law of Motion the force is directly proportional to the mass [tex]m[/tex] and to the acceleration [tex]a[/tex] of a body:

[tex]F_{y}=m.a[/tex] (3)

Where [tex]m=95.5kg[/tex] is the mass of the diver.

Substituting the known values and finding [tex]a[/tex]:

[tex]a=\frac{F_{y}}{m}[/tex] (4)

[tex]a=\frac{96N}{95.5kg}[/tex] (5)

Finally:

[tex]a=1.0052m/{s^{2}}\approx 1m/s^{2}[/tex]  This is the acceleration of the sky diver. Note it has a positive sign, which means its direction is upwards.

(a) 756.9 J

The work done by a force when moving an object is given by:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement of the object

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

In this situation,

F = 165 N

d = 5.60 m

[tex]\theta=35.0^{\circ}[/tex]

so the work done by the tension is

[tex]W=(165 N)(5.60 m)cos 35.0^{\circ}=756.9 J[/tex]

(b) 0.646

The crate is moving at constant speed: this means that the acceleration of the crate is zero, so the net force on the crate is also zero.

There are only two forces acting on the crate along the horizontal direction:

- The horizontal component of the tension, [tex]Tcos \theta[/tex], forward

- The frictional force, [tex]-\mu N[/tex], backward, with [tex]\mu[/tex] being the coefficient of kinetic friction, N being the normal reaction of the surface on the crate

The normal reaction is the sum of the weight of the crate + the vertical component of the tension, so

[tex]N=mg-T sin \theta = (31.0 kg)(9.8 m/s^2) - (165 N)sin 35.0^{\circ}=209.2 N[/tex]

Since the net force is zero, we have

[tex]T cos \theta-\mu N =0[/tex]

where

T = 165.0 N

[tex]\theta=35.0^{\circ}[/tex]

N = 209.2 N

Solving for [tex]\mu[/tex], we find the coefficient of friction:

[tex]\mu = \frac{T cos \theta}{N}=\frac{(165.0 N)(cos 35^{\circ})}{209.2 N}=0.646[/tex]

Final answer:

In this physics problem, we determine the work done by a tension force inclining at an angle above the horizontal and find the coefficient of kinetic friction when a crate moves at a constant speed.

Explanation:

A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate at a distance of 5.60 m on a rough surface.

(a) Work done by the tension force: To calculate work, use the formula W = Fd cos(θ), where F is the force, d is the distance moved, and θ is the angle between the force and displacement. Here, W = 165 N * 5.60 m * cos(35.0°).

(b) Coefficient of kinetic friction: Since the crate moves at a constant speed, the work done by the tension force equals the work done against friction. Find the frictional force using the work done formula and then the coefficient of kinetic friction.

Answer:

Tension, T = 1736 N

Explanation:

It is given that,

Mass of bricks, m = 175 kg

A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :

T - mg = ma

T = ma + mg

T = m(a + g)

T = 175 kg ( 0.12 m/s² + 9.8 m/s² )

T = 1736 N

Hence, the tension in the wire is 1736 N.

Answer:

The tension in the rope is 1736 N.

(e) is correct option.

Explanation:

Given that,

Mass of bricks = 175 kg

Acceleration = 0.12 m/s²

Let T is the tension in the rope.

A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.

Using equation of balance

[tex]T-mg=ma[/tex]

[tex]T=mg+ma[/tex]

[tex]T=175(9.8+0.12)[/tex]

[tex]T= 1736\ N[/tex]

Hence, The tension in the rope is 1736 N.

Answer:

The temperature in degrees Fahrenheit in terms of the Celsius temperature is given by .

As we know by the linear relation

[tex]\frac{^o F - 32}{212 - 32} = \frac{^o C - 0}{100 - 0}[/tex]

now we have

[tex]^o F - 32 = \frac{180}{100} ^o C[/tex]

so we have

[tex]^o F = \frac{9}{5} ^o C + 32[/tex]

The temperature in degrees Celsius in terms of the Kelvin temperature is given by

As we know by the linear relation

[tex]\frac{^o C - 0}{100 - 0} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]C = K - 273[/tex]

the temperature in degrees Fahrenheit in terms of the Kelvin temperature .

As we know by the linear relation

[tex]\frac{^o F - 32}{212 - 32} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]^o F - 32 = \frac{180}{100} (K - 273)[/tex]

so we have

[tex]^o F = \frac{9}{5} (K - 273) + 32[/tex]

Answer:

The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]

Explanation:

Given that,

Mass [tex]m = 1.81\times10^{-3}\ kg[/tex]

Velocity [tex]v = (3.00\times10^{4}\ m/s)j[/tex]

Charge [tex]q = 1.22\times10^{-8}\ C[/tex]

Magnetic field [tex] B= (1.63\hat{i}+0.980\hat{j})\ T[/tex]

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

[tex]F = ma=q(v\times B)[/tex]

[tex]a =\dfrac{q(v\times B)}{m}[/tex]

We need to calculate the value of [tex]v\times B[/tex]

[tex]v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})[/tex]

[tex]v\times B=4.89\times10^{4}[/tex]

Now, put the all values into the acceleration 's formula

[tex]a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}[/tex]

[tex]a= -0.3296\ \hat{k}\ m/s^2[/tex]

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]

The magnitude and direction of the particle’s acceleration produced by a uniform magnetic field  [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^?

A charged particle is a particle with an electric charge. Whereas electric charge is the matter physical property that causes to experience a force when placed in an electromagnetic field. Uniform magnetic field is the condition when magnetic field lines are parallel then magnetic force experienced by an object is same at all points in that field

From Newton's second law, the force is given by:

[tex]F=ma[/tex]

Magnetic force is

[tex]F= qv \times B[/tex]

[tex]ma = qv \times B[/tex]

[tex]a = \frac{qv \times B}{m}[/tex]

Subsituting with the givens above we get

[tex]a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

Therefore the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field  [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

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Answer:most likely E

Explanation:

Why would anybody do something after design there done with there work after that