A common technique in analysis of scientific data is normalization. The purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. The goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. In this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. Suppose that in Part 2.2.2 of the experiment, students obtain a signal value of 185 mV at a distance of 4 cm and a value of 82 mV at a distance of 6.2 cm. Normalize the students' data to the value obtained at 4 cm. (Divide the signal value by 185.) Then calculate the theoretically expected (normalized) value at 6.2 cm.___________________ Normalized experimental value at 6.2 cm____________________Theoretically expected normalized value at 6.2 cm

Answer:

m = (3/5)*M

Explanation:

Given:-

- The angular speed of both hollow and solid sphere = w

- The diameter of solid & hollow sphere = D

- The mass of the solid sphere = M

- Both rotate about their common axis with similar rotational kinetic energy.

Find:-

The mass of hollow sphere (m) ?

Solution:-

- The formula for rotational kinetic energy (K.E) of any rigid body is:

                                    K.E = 0.5*I*w^2

Where,

                  I : Moment of inertia of rigid body

- The rotational kinetic energies of both hollow sphere and solid sphere are same:

                             0.5*I_solid*w^2 = 0.5*I_shell*w^2

                             I_solid = I_shell

                             0.4*M*D^2 / 4 = (2/3)*m*D^2 / 4

                             (2/5)*M = (2/3)*m

                             m = (3/5)*M

Answer:

The velocity of mass 2m is  [tex]v_B = 0.67 m/s[/tex]

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = [tex]v_1[/tex] =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = [tex]v_A[/tex]

Let The finial speed  of the billiard ball  B = [tex]v_B[/tex]

      According to the law of conservation of Energy

                 [tex]\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]

              Substituting values  

                [tex]\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]

Multiplying through by [tex]\frac{1}{2}m[/tex]

                [tex]1 =v_A^2 + 2 v_B ^2 ---(1)[/tex]

    According to the law of conservation of Momentum

            [tex]mv_1 + 2m(0) = mv_A + 2m v_B[/tex]

    Substituting values

            [tex]m(1) = mv_A + 2mv_B[/tex]

Multiplying through by [tex]m[/tex]

           [tex]1 = v_A + 2v_B ---(2)[/tex]

making [tex]v_A[/tex] subject of the equation 2

            [tex]v_A = 1 - 2v_B[/tex]

Substituting this into equation 1

         [tex](1 -2v_B)^2 + 2v_B^2 = 1[/tex]

         [tex]1 - 4v_B + 4v_B^2 + 2v_B^2 =1[/tex]

          [tex]6v_B^2 -4v_B +1 =1[/tex]

          [tex]6v_B^2 -4v_B =0[/tex]

Multiplying through by [tex]\frac{1}{v_B}[/tex]

          [tex]6v_B -4 = 0[/tex]

            [tex]v_B = \frac{4}{6}[/tex]

            [tex]v_B = 0.67 m/s[/tex]

The stationary billiard ball of mass 2m will have a final velocity of 1/3 m/s after the head-on collision with a ball of mass m moving at 1 m/s due to conservation of momentum.

In this problem, we are dealing with a one-dimensional collision. The principle of conservation of momentum is key in solving this problem. Momentum before the collision is the same as momentum after the collision. The initial momentum is given by mass * velocity of moving ball = m * v1, while the stationary ball has 0 momentum.

So, we can write this as m * 1 = (m + 2m) * v', where v' is the final velocity of the stationary ball after the collision. Solving for v', we get v' = m / 3m = 1/3 m/s. Hence, the stationary ball (of mass 2m) has a final velocity of 1/3 m/s after the collision.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity of the both spheres is [tex]v=4.62 m/s[/tex]

Explanation:

The free body diagram of this question is shown on the second uploaded image

     Looking at this diagram we can deduce that there is no impulse force along the horizontal direction so

      The mathematical equation for the impulse force along the horizontal axis is  

           [tex]\int\limits {\sum F_s} \, dt = 0[/tex]

 The mathematical equation for the impulse force along the horizontal axis is

           [tex]\int\limits {\sum F_y} \, dt = \Delta I_y[/tex]

  Where [tex]I_y[/tex] is the impulse  momentum along the y-axis and this is mathematically given as

         [tex]\Delta I_y = 2mv_y[/tex]

substituting 9.8 N.s for [tex]\Delta I_y[/tex] and 1.5kg for m (mass of sphere) and the making [tex]v_y[/tex] the subject

        [tex]v_y = \frac{9.8}{2 *1.5}[/tex]

            [tex]=3.267 m/s[/tex]

The sum of the moment about the point I is mathematically represented as

         [tex]\int\limits {\sum M_I } \, dt =\Delta H_I[/tex]

From the free body diagram  [tex]\int\limits {\sum M_I } \, dt = I * 0.3 = 9.8 *0.3=2.94N.m[/tex]

[tex]\Delta H_I[/tex] is the angular momentum along the horizontal axis  given as

                  [tex]\Delta H_I[/tex] [tex]= 2mv_x(0.3)[/tex]

substituting parameters into above equation

                  [tex]2.94 = 2 * 1.5 * v_x * 0.3[/tex]

Making [tex]v_x[/tex] the subject

                 [tex]v_x = \frac{2.94}{2*1.5*0.3}[/tex]

                      [tex]=3.267 m/s[/tex]

the velocity of the sphere is mathematically represented as

                 [tex]v = \sqrt{(v_x)^2 + (v_y)^2}[/tex]

Now substituting values

                [tex]v = \sqrt{3.267^2 + 3.267^2}[/tex]

                   [tex]v=4.62 m/s[/tex]

The physics question pertains to the concept of impulse and momentum. The velocity of each sphere after the impulse is imparted can be calculated by dividing the change in momentum by twice the mass of one sphere.

The Physics question is relevant to impulse and momentum in the domain of mechanics. From the problem, we know that a force F acting on the sphere imparts an impulse of 9.8 N·s. Impulse, denoted as J, in physics, is the product of force and the time for which it is applied and is equivalent to the change in momentum of the body. Thus, the change in momentum, Δp is equal to the impulse imparted, which is 9.8 N·s.

Now considering the system of two spheres rigidly connected, it is an isolated system (external force F is not considered as it acts for a very short time) and therefore, the total momentum of the system remains conserved. Therefore the momentum imparted to one sphere is equally distributed to both the spheres. Therefore, the final velocity v of each sphere can be calculated by using the formula v = Δp / 2m (m being the mass of each sphere).

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Answer:

A- Equality

Explanation:

Answer: tsunamis,El Nino,and volcanic eruptions

Explanation:

the correct answer  for e d g e n u i t y

The examples of short-term environmental change are Tsunamis, El Niño, and Volcanic eruptions.

Tsunamis are huge ocean waves that are triggered by undersea disturbances like earthquakes or landslides. They can inflict major damage to coastal regions, however they usually happen in a short period of time.

El Nio is a climatic trend characterised by higher-than-normal equatorial Pacific ocean temperatures.

It can cause changes in weather patterns, such as greater rainfall in some areas and droughts in others, although the impacts are usually transient and endure for a few months to a couple of years.

Volcanic eruptions spew lava, ash, and gases into the atmosphere, which can have direct effects on the environment, including as changes in air quality and local weather patterns.

Thus, these effects, however, are often short-lived and localized.

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The question is not complete, so i have attached an image with the complete question.

Answer:

A) speed of block a before it hits block B;v = √2gR

B) speed of combined blocks after collision;V_ab = ½√2gR

C) kinetic energy lost = ½MgR

D) temperature change; ∆t = ½gR/c

E) Additional thermal energy;W_f = 2μMgL

Explanation:

A) From conservation of energy, potential energy = kinetic energy.

Thus; P.E = K.E

So, mgr = ½mv²

m will cancel out to give;

gr = v²

So, v = √2gR

B) from conservation od momentum, momentum before collision = momentum after collision.

Thus;

M_a•V_a = M_ab•V_ab

Where;

M_a is the mass of block A

V_a is speed of block A

M_ab is the mass after collision

V_ab is speed after collision

So, making V_ab the subject, we have;

V_ab = M_a•V_a/M_ab

Now from answer in a above,

V_a = V = √2gR

Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.

Thus;

V_ab = (M√2gR)/(2M)

M will cancel out to give;

V_ab = ½√2gR

C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.

Thus;

W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.

We have seen that:

M_a = M and M_ab = 2M

V_a = √2gR and V_ab = ½√2gR

Thus;

∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)

∆K = ½MgR - MgE

∆K = -½MgR

Negative sign means loss of energy.

Thus kinetic energy lost = ½MgR

D) The formula for heat energy is given by;

Q = m•c•∆t

Thus, change in temperature is;

∆t = Q/Mc

Q is the heat energy, thus Q = ½MgR

Thus;∆t = ½MgR/(Mc)

M will cancel out to give;

∆t = ½gR/c

E) Additional thermal energy is gotten from;

Work done by friction;W_f = F_f x d

Where;

F_f is frictional force given by μF_n. F_n is normal force

d is distance moved

Thus;

W_f = μF_n*d

Mass after collision was 2M,thus, F_n = 2Mg

We are told to express distance in terms of L

Thus;

W_f = μ2Mg*L

W_f = 2μMgL

The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).

The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.

After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.

The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).

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Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

=  I x π x²  / π R²

= I x²  /  R²

B x 2π x = μ₀  x current enclosed

B x 2π x = μ₀  x  I x²  /  R²

B =  μ₀   I x  / 2π R²

Maximum magnetic B₀ field  will be when x = R

B₀ = μ₀I   / 2π R

Given

B = B₀ / 3

μ₀   I x  / 2π R² = μ₀I   / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I   / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I   / 2π R x 3

x = 3R .

The location inside the wire where the magnetic field equals a third of its maximum value is at r/R = 1/3. There is no point outside the wire where the magnetic field reaches a third of its maximum value because it monotonically decreases.

To determine at what location inside the wire the magnetic field produced by the current is equal to a third of its largest value, we need to apply Ampère's Law. Inside a conductor carrying uniform current, the magnetic field B increases linearly with the distance r from the center of the wire due to the proportion of current enclosed. Thus, the magnetic field is given by B = μ_0 J r / 2, where J is the current density and μ_0 is the permeability of free space. Since the value at the surface (r=R) will be maximum (B_max), for B to be a third of B_max, we must have (1/3)B_max = (μ_0 J r / 2). Solving for r, we find r/R = 1/3.

For locations outside the wire, Biot-Savart Law or Ampère's Law show that the magnetic field decreases with 1/r. However, since the field is maximum at the surface, and we do not have an equation that varies outside the wire, we cannot directly calculate when the field will be a third without additional information on how the field varies with r beyond the wire's surface. Normally, the largest value of B is at the surface, and it decreases monotonically outside, never increasing again to allow for a location at which it is a third of its largest value.

The magnetic field for points inside the wire is proportional to the distance from the center, and for points outside the wire, it is inversely proportional to the distance from the center. However, the magnetic field does not reach a third of its largest value at any given point outside the cylindrical wire since it monotonically decreases.

Answer:nuclear energy is not derived from the sun

Explanation:

Nuclear energy is not derived from the sun.nuclear energy comes from the energy released when atoms are split apart and some mass is converted to energy

Answer:

T = 1.07 s

a = 1.034 m/s2

Explanation:

Metric unit conversion

m = 145 g = 0.145 kg

x = 3 cm = 0.03 m

Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation

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

[tex]T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s[/tex]

The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m

The spring force at that point would be

[tex]F_s = kx = 5*0.03 = 0.15 N[/tex]

According to Newton's 2nd law, the acceleration at this point would be

[tex]a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2[/tex]

Answer:

nail will stop after traveling 1000 m    

Explanation:

We have given mass m = 50 gram = 0.05 kg

Frictional force which is used to stop the mass F = 62.5 kN

Initial velocity is given u = 50 m/sec

From newton's law force is equal to F = ma, here m is mass and a is acceleration

So [tex]a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2[/tex]

As finally nail stops so final velocity v = 0 m/sec

From third equation of motion [tex]v^2=u^2+2as[/tex]

So [tex]0^2=50^2-2\times 1.25\times 10^6\times s[/tex]

s = 1000 m

So nail will stop after traveling 1000 m

Answer:

The object will continue spinning as it has been.

Explanation:

When an object is spinning in a closed system with no unbalance

forces and no external force or torques are applied to it, it will have no change in angular momentum.

According to the law of conservation of angular momentum which states that when no external torque acts on an object, no change of angular momentum will occur

If the change in angular momentum is zero, then the angular momentum is constant; therefore, the object will continue spinning as it has been.

The same scenario is applicable to the rotation and the spinning of the earth.

Answer: The object will continue spinning as it has been.

Explanation: Unbalanced forces refers a force that changes the position, speed or direction of the object to which it is applied. While torque refers to otational or twisting effect of a force; a moment of force, defined for measurement purposes as an equivalent straight line force multiplied by the distance from the axis of rotation.

Any object that spins in place with no unbalanced forces or torques acting upon it will continue spinning as it has been. An example can be seen in the rotation of planets.

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

Answer:

a) W = 1815 J

b) W = 135 J

Explanation:

We need to model the force using Hooke's law

We first get the elastic constant, k

A force of 60 N causes an extension of 0.5 m

F = 60 N, e = 0.5 m

F = k * e

60 = 0.5 k

k = 60/0.5

k = 120 N/m

Therefore the force in terms of the extension, x is:

F(x) = 120x

a) Work done in stretching the spring 5.5 m from its equilibrium position

b = 0 m, a = 5.5 m

[tex]W = \int\limits^a_b {F(x)} \, dx[/tex]

[tex]W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5[/tex]

[tex]W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ][/tex]

[tex]W = 1815 J[/tex]

b) Work required to compress the spring 1.5 m from its equilibrium position

[tex]W = \int\limits^a_b {F(x)} \, dx[/tex] b = 0, a = -1.5

[tex]W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]\\[/tex]

W = 135 J

The work done to compress the spring 1.5 m from its equilibrium position is 135 J

Answer:1.513 rps

Explanation:

Given

Diameter of cylindrical space [tex]d=780\ m[/tex]

When the station rotates it creates centripetal acceleration which is given by

[tex]a_c=\omega ^2r[/tex]

Now it must create the effect of gravity so

[tex]g=\omega ^2r[/tex]

[tex]\omega =\sqrt{\frac{g}{r}}[/tex]

[tex]\omega =0.158\ rad/s[/tex]

and [tex]\omega =\frac{2\pi N}{60}[/tex]

Thus [tex]N=\frac{4.755}{3.142}[/tex]

[tex]N=1.513\ rps[/tex]

Answer:oops

Explanation:

Options:

a) more than 0.8 m .

b) equal to 0.8 m .

c) between 0.5 m and 0.8 m .

d) less than 0.5 m .

Answer:

b) equal to 0.8 m .

Explanation:

Note:

An upside down image = Inverted Image

An image that appears in front of the mirror = Real image

An image that appears behind the mirror = Virtual image

Let the object distance from the pole of the mirror be u

When the child stands 1.2 m from the mirror:

u = 1.2 m ( Real and Inverted image of the child is formed)

When the child stands about 0.8 m from the mirror:

u = 0.8 m (Virtual, erect and magnified image of the child is formed)

When the child stands about 0.5 m from the mirror:

u = 0.5 m ( Virtual and erect image of the child is formed)

Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.

Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m

The approximate focal length of the mirror will be "0.8 m". A complete solution is provided below.

According to the question,

When a child stands 1.2 m, the real as well as inverted image will be formed, then

When a child stands 0.8 m, the virtual, erect as well as magnified image will be formed, then

When a child stands 0.5 m, the virtual as well as erect image will be formed, then

Since,

Whenever the image is formed from real to virtual, the focal length will become 0.8 m.

Thus the answer above is right.

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Answer:

the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

the internal energy generated in crossing the rough patch is 820.26 J

Explanation:

Given that,

Mass, m = 62 kg,

Initial speed,  = 6.90 m/s

Length of rough patch, L = 4.50 m,

coefficient of friction,  = 0.3

Height of inclined plane, h = 2.50 m

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

According to energy conservation equation,

Part (b)

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

[tex]F_f=\mu mg[/tex]

Therefore, the work done by friction is:

[tex]W=-F_f d =-\mu mg d[/tex]

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

[tex]W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J[/tex]

So, the internal energy generated in crossing the rough patch is 820.26 J

Part (a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

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

After crossing the rough patch, the new mechanical energy is

[tex]E'=E+W[/tex]

where

W = -820.26 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

[tex]E' = K_f = \frac{1}{2}mv^2[/tex]

where v is the final speed.

According to the law of conservation of energy, we can write:

[tex]E+W=E'[/tex]

So we find v:

[tex]\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s[/tex]

Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

The question repeated the units attached the the values

So in order of their appearance in the question, the correct units are;

62kg, 6.9m/s, 4.5m, 0.30, 2.5m

Answer:

A) Speed at bottom of hill = 8.38 m/s

B) Internal energy generated in crossing the patch = 820.26J

Explanation:

From the question, it's clear that there are 2 distinct stages of of her skis. The first stage is when she is on the rough patch while the second stage is when she skis down the hill.

A) For the first stage when she is on the rough patch;

If we apply Newton's second law of motion to the skis along the vertical direction, we obtain;

ΣF_y = N - mg = 0

Thus,N = mg

m = 62kg,thus,N = 62 x 9.8 = 607.6N

Now, let's find the kinetic friction. It's given by; f_k = μN

Where μ is coefficient of kinetic friction

Thus, f_k = 0.3 x 607.6 = 182.28N

Now, let's calculate the work done by this frictional force,

So, W_fk = f_k x distance = 182.28 x

4.5 = 820.26J

This work is done in a direction opposite to the displacement and it will have a negative sign. Thus,

W_fk = - 820.26J

Now, since the skier skies horizontally and perpendicular to the gravitational force, the work done due to gravity is zero. Thus,

W_grav = 0

Let's calculate the kinetic energy at the beginning of the rough patch.

K1 = (1/2)m(v1)²

K1 = (1/2) x 62 x (6.9)²= 1475.91J

Also,let's calculate the kinetic energy at end of rough patch.

K2 = (1/2)m(v2)²

where v2 is final velocity at end of rough patch

K2 = (1/2)(62)(v2)² = 31(v2)²

Now, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K2 - K1

In this case, W_other is W_fk

Thus,

- 820.26J + 0 = 31(v2)² - 1475.91J

31(v2)² = 1475.91J - 820.26J

31(v2)² = 655.65

(v2)² = 655.65/31

v2 =√21.15 = 4.6 m/s

Now, for the second stage when she skis down the hill;

In this case the only force acting is gravity, thus, W_other = 0

Work done by gravity = mgh

W_grav = 62 x 9.8 x 2.5 = 1519 J

Now K2 will be kinetic energy at top of hill while K3 will be kinetic energy at bottom of hill.

Thus,

K2 = (1/2)m(v2)²

K2 = (1/2)(62)(4.6)²

K2 = 655.96J

Similarly,

K3 = (1/2)m(v3)²

Where v3 is velocity at bottom of hill.

Thus,

K3 = (1/2)(62)(v3)²

K3 = 31(v3)²

Again, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K3 - K1

So,

W_other is zero.

Thus,

1519J = 31(v3)² - 655.96J

31(v3)² = 1519J + 655.96J

31(v3)² = 2174.96J

(v3)² = 2174.96/31

v3 = √70.16

v3 = 8.38 m/s

B) Workdone by non-conservative forces manifests itself as changes in the internal energy of bodies. Now, frictional force experienced in the first stage in crossing the patch is a non - conservative force. Thus,

Internal energy = W_fk = 820.26J

Answer:

Current is half

Voltage is half

Explanation:

according to Ohms law V = IR

Case 1

I₁ = V / R

I₂ = V / (R+R) = V / 2R

∴ I₁ / I₂ = (V/R) / (V/2R)

=> I₁ / I₂ = (V/R) * (2R/V)

=> I₁ / I₂ = (2VR / RV)

I₁ / I₂ = 2

I₂ = I₁ / 2

The current through each resistor in the two-resistor circuit is half the current through the resistor in the one-resistor circuit (the circuit in Part A).

The voltage across each resistor in the two-resistor circuit is half the voltage across the resistor in the one-resistor circuit.

The current through each resistor in a two-resistor series circuit is the same as the current through a one-resistor circuit. However, the voltage across each resistor in the two-resistor circuit is half of the voltage across the single resistor due to how voltages distribute in series circuits.

In a series circuit, such as the two-resistor circuit you have mentioned, the current through each resistor is the same as the current through the single resistor in the one-resistor circuit. This is due to Kirchhoff's Current Law which states that the current entering a junction or a node must equal the current leaving it. However, the voltage accross each resistor in the two-resistor circuit is half of the voltage across the single resistor in the one-resistor circuit. This is because in a series circuit, the total voltage is the sum of the individual voltage drops across each resistor, according to Ohm’s law.

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Answer:

(a). [tex]\dfrac{f_3'}{f_3} =\sqrt{5}.[/tex]

(b). The wavelength remains unchanged.

Explanation:

The speed [tex]v[/tex] of the waves on the string with tension [tex]T_i[/tex] is given by

[tex]v = \sqrt{\dfrac{T_iL}{m} }[/tex]

And if the string is vibrating, its fundamental wavelength is [tex]2L[/tex], and since the frequency [tex]f[/tex] is related to the wave speed and wavelength by

[tex]f = v/\lambda[/tex]

the fundamental frequency [tex]f_1[/tex] is

[tex]f_1 =\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}[/tex]

and since the frequency of the third harmonic is

[tex]f_3 = 3f_1[/tex]

[tex]f_3 = 3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L},[/tex]

and the wavelength is

[tex]\lambda_3 = \dfrac{2L}{3}.[/tex]

(a).

Now, if we increase to the string tension to

[tex]T_f = 5.0T_i[/tex]

the third harmonic frequency becomes

[tex]f_3' = 3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L},[/tex]

The ratio of this new frequency to the old frequency is

[tex]\dfrac{f_3'}{f_3} = \dfrac{3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L}}{3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}}[/tex]

[tex]\boxed{\dfrac{f_3'}{f_3} =\sqrt{5}.}[/tex]

(b).

The wavelength of the third harmonic remains unchanged because [tex]\lambda_3 = \dfrac{2L}{3}.[/tex] depends only on the length of the string

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

[tex]I(t) = I_o(1-e^{-\frac{t}{T}})[/tex]

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

[tex]I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A[/tex]

Therefore, the current in the circuit 7 ms later is 0.2499 A

The current later after 7 ms is 0.2499 A

The current after a time (t) is given by:

[tex]I(t)=I_o(1-e^{-\frac{t}{\tau} })\\\\I_o=initial\ current\ at\ t=0,\tau=time \ constant\\\\Given\ that\ L=45mH=45*10^{-3}H,R=60\ ohm,V = 15V,t=7\ ms\\\\\tau=\frac{L}{R}=\frac{45*10^{-3}}{60} =7.5*10^{-4} \ s\\\\I_o=\frac{V}{R}=\frac{15}{60}=0.25\ A \\\\\\I(t)=I_o(1-e^{-\frac{t}{\tau} })=0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}} })\\\\\\I(t)=0.2499\ A[/tex]

The current later after 7 ms is 0.2499 A

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Answer:

a

The potential of the earth surface is  [tex]V_E= - 9.6*10^9 V[/tex]

b

when the potential of the earth is zero the charge is choose to be negative this implies that  the potential at infinity would be [tex]V_\infty=+9.6*10^9V[/tex] and (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge the potential at the surface increase and the electric field remains the same

Explanation:

From the question we are told that

      The magnitude  electric field is  [tex]E = 150V/m[/tex]

The potential of the earth surface when  V = 0 and  [tex]r = \infty[/tex] is mathematically represented as

                [tex]V_E = - \frac{q}{4 \pi \epsilon_o r} = - E *R[/tex]  

Where q is the charge on the surface of the earth  which is negative

           R is the radius of the earth

            [tex]V_E = 150 * 64 *10^6[/tex]

                 [tex]V_E= - 9.6*10^9 V[/tex]

when the potential of the earth is zero the charge is choose to be negative

Then the potential at infinity would be [tex]+9.6*10^9V[/tex]

Answer: 4.821m/s

Explanation:

initial momentum = final momentum 

(80 x 4.1) + (85 x 5.5) = (80 + 85)u

328 + 467.5 = 165u  =  795.5 / 165 = 4.821m/s

Answer:  4.0024 x 10^ -11 m or 0.040024 nm

Explanation:

λ = h c/ΔE

λ = wave lenght

h = 6.626 x 10 ^ -34  m² kg /s  = planck constant

ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)

c = velocity of light = 3 x 10⁸ m/s

substitute gives

λ  =    6.626 x 10 ^ -34  m² kg /s x 3 x 10⁸ m/s  = 4.0024 x 10^ -11 m

                 31 x 1.6021x10^-16 J

Explanation:

Given that,

Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]

Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]

The kinetic energy of the ejected electron in terms of work function is given by :

[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]

The formula of kinetic energy is given by :

[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]

Hence, this is the required solution.

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes, then the voltage is impressed across it would be 2 volts .

The rate of doing work is known as power. The Si unit of power is the watt .

The power dissipated in the circuit  = work / time

As given in the problem we have to find out what voltage is impressed across the circuit If part of an electric circuit dissipates energy at 6 Watts  when it draws a current of 3 Amperes,

The power dissipated across the circuit  = Voltage × current

                                              6 watts  = Voltage × 3 ampere

                                              Voltage  = 6 / 3

                                              Voltage = 2 volts

Thus , the voltage across the circuit would be  2 volts .

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Answer:

A) Kinetic

Explanation:

Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.

Answer:

24.609375 kgm²

Explanation:

I = Moment of inertia of the merry go round

[tex]N_1[/tex] = Initial speed = 51 rpm

[tex]N_2[/tex] = Final speed = 17 rpm

m = Mass of child = 31.5 kg

r = Radius = 1.25 m

In this system the angular momentum is conserved

[tex]IN_1=(I+mr^2)N_2\\\Rightarrow I\times 51=(I+31.5\times 1.25^2)17\\\Rightarrow I\dfrac{51}{17}-I=31.5\times 1.25^2\\\Rightarrow 2I=49.21875\\\Rightarrow I=\dfrac{49.21875}{2}\\\Rightarrow I=24.609375\ kgm^2[/tex]

The moment of inertia of the merry go round is 24.609375 kgm²

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

Answer:

37.5N

Explanation:

According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e

F ∝ e

F = k x e         -------------------------(i)

where;

k = proportionality constant known as the spring constant.

From the question;

k = 750N/m

e = 5.0cm = 0.05m

Substitute these values into equation (i) as follows;

F = 750 x 0.05

F = 37.5N

Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N

Answer:

The resulting time constant will be [tex]\bf{5.09~ms}[/tex].

Explanation:

Given:

the time constant of the RC circuit, [tex]\tau = 8~ms[/tex]

The value of the capacitor in the circuit, [tex]C_{1} = 4~\mu F[/tex]

The value of addition capacitor added to the circuit, [tex]C_{2} = 7~\mu F[/tex]

The value of the time constant for a series RC circuit is give by

[tex]\tau = RC[/tex]

So the value of the resistance in the circuit is

[tex]R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega[/tex]

When the capacitor [tex]C_{2}[/tex] is added to the circuit, the net value of the capacitance in the circuit is

[tex]C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F[/tex]

So the new time constant will be

[tex]\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms[/tex]

Final answer:

Kinetic energy is the energy caused by the vibration of particles in a medium. It is calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.

Explanation:

Kinetic energy is the energy an object has because of its motion. It is caused by the vibration of particles in a medium such as steel, water, or air. Kinetic energy is calculated as one-half the product of the mass of the particle and the square of its speed.

For example, when a rock is thrown into a pond or when a swimmer splashes the water's surface, the kinetic energy of the vibrating water particles is generated. Similarly, mechanical sound waves also have kinetic energy due to the movement of air particles and the potential energy caused by the elasticity of the material through which the sound propagates.

In summary, kinetic energy arises from the motion of particles in a medium and can be calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.

Complete question:

A rectangular loop of wire with dimensions 2.0 cm by 10.0 cm and resistance 1.0 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude of 2.0 T and is directed into the plane. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

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Answer:

The magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

Explanation:

Given;

resistance of the wire; R = 1.0 Ω

magnitude of magnetic field strength, B = 2.0 T

speed of the loop, v = 3.00 m/s

Induced emf is given as;

ε = IR

[tex]I = \frac{emf}{R} = \frac{VBL}{R}[/tex]

magnitude of the force that the magnetic field exerts on the loop:

F = BIL

Substitute in the value of I

[tex]F = \frac{VB^2L^2}{R}[/tex]

where;

L is the displacement vector between the initial and final end of the portion of the wire inside the field = 2.0 cm

Substitute the given values and solve for F

[tex]F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N[/tex]

Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

Complete Question:

A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance 0.700 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.20 T and is directed into the plane of.

At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

Answer:

F = 0.133 N

Explanation:

Magnitude of the magnetic field, B = 2.20 T

Length of the loop = 1.5 cm = 0.015 m

The speed of the loop, v = 3.00 m/s

The emf induced in the loop , e = Blv

e = 2.20 * 0.015 * 3

e = 0.099 V

Current induced in the loop, I = e/R

I = 0.099/0.7

I = 0.1414 A

The magnitude of the force is given by, F = I *l *B sin90

F = 0.1414 * 0.015 * 2.20

F = 0.00467 N