A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (μ 0 = 4π × 10-7 T · m/A

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

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

so we have

[tex]F = ma = \mu mg[/tex]

now the acceleration is given as

[tex]a = \mu g[/tex]

[tex]a = (0.32)(9.8) = 3.14 m/s^2[/tex]

Final answer:

The force that accelerates the box when the truck moves forward is the force of static friction. The maximum acceleration the truck can have before the box slides can be found using the equation: max acceleration = coefficient of friction x acceleration due to gravity.

Explanation:

When the truck accelerates forward, the force that accelerates the box is the force of static friction between the box and the surface of the truck bed. This force opposes the motion of the box and allows it to stay in place on the truck.

The maximum acceleration the truck can have before the box slides can be found using the equation:

max acceleration = coefficient of friction x acceleration due to gravity

Using the given coefficient of friction between the box and the surface, you can substitute the value and solve for the maximum acceleration.

Answer:

Option C is the correct answer.

Explanation:

We have charge stored in a capacitor

           Q = CV

C is the capacitance and V is the voltage.

A capacitor is connected to a 9 V battery and acquires a charge Q.

That is

          Q = C x 9 = 9C

What is the charge on the capacitor if it is connected instead to an 18 V battery

That is

          Q' = C x 18 = 9C x 2 = 2Q

Option C is the correct answer.      

When the voltage on a capacitor is doubled, the charge on the capacitor doubles as well. Hence, if a capacitor initially has a charge Q at 9V, it will have a charge of 2Q at 18V.

The subject of the question is the behavior of a capacitor when it is connected to various voltage sources. The charge Q on a capacitor is given by the equation Q = CV, where C is the capacitance of the capacitor and V is the voltage across it. So if a capacitor is charged by a 9V battery (Q = C*9V), and then connected to an 18V battery, its new charge (lets call it Q1) would be C*18V. Therefore, Q1 = 2Q, since the voltage is doubled, hence, the charge on the capacitor also doubles.

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Answer:

2069.76 N

Explanation:

density of wood = 824 kg/m^3

density of water = 1000 kg/m^3

Volume of wood = 1.2 m^3

True weight of wood = volume of wood x density of wood x gravity

True weight of wood = 1.2 x 824 x 9.8 = 9690.24 N

Buoyant force acting on wood = volume of wood x density of water x gravity

Buoyant force acting on wood = 1.2 x 1000 x 9.8 = 11760 N

Tension in the rope = Buoyant force - True weight

tension in rope = 11760 - 9690.24 = 2069.76 N

Answer:0.642

Explanation:

Given

initially car is at rest i.e u=0

When car hits the truck it's velocity is 4.5 m/s

it hits the truck after 7 sec

Now average acceleration of car is =[tex]\frac{change\ in\ velocity}{time\ taken}[/tex]

average acceleration of car =[tex]\frac{\left ( final velocity\right )-\left ( initial velocity\right )}{time}[/tex]

average acceleration of car=[tex]\frac{\left ( 4.5\right )-\left ( 0\right )}{7}[/tex]

average acceleration of car=[tex]0.642 m/s^{2}[/tex]

Taking Down hill as positive x axis

average acceleration of car=[tex]0.642\hat{i}[/tex]

Answer:

0.01663 Nm

Explanation:

N = 1000, r = 12 cm = 0.12 m, i = 15 A, B = 5.8 x 10^-5 T, θ = 25

Torque = N i A B Sinθ

Torque = 1000 x 15 x 3.14 x 0.12 x 0.12 x 5.8 x 10^-5 x Sin 25

Torque = 0.01663 Nm

Answer:

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

Explanation:

Initial total mechanical energy is given as

[tex]ME = U + KE[/tex]

here we will have

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.05) = 0.68 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(15.1)^2[/tex]

[tex]KE = 7.52 J[/tex]

[tex]ME_i = 0.68 + 7.52 = 8.2 J[/tex]

Now similarly final mechanical energy is given as

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.59) = 1.03 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(10.9)^2[/tex]

[tex]KE = 3.92 J[/tex]

[tex]ME_f = 1.03 + 3.92 = 4.95 J[/tex]

Now change in mechanical energy is given as

[tex]\Delta E = ME_i - ME_f[/tex]

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

Answer:

(a) 0.204 Weber

(b) 0.22 Volt

Explanation:

N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.

(a) Magnetic flux, Ф = N x B x A

Ф = 100 x 0.0650 x 3.14 x 0.1 0.1

Ф = 0.204 Weber

(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s

dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s

induced emf, e = N dФ/dt

e = N x A x dB/dt

e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V

(a) The magnetic flux through a coil 0.204 Weber

(b) The emf induced in the coil during the time interval 0.22 Volt

It is given that

Number of the turns N = 100,  

Radius of the coil r = 10 cm = 0.1 m,

The magnetic field B = 0.0650 T

Since the angle is 90 degrees with the plane of the coil, so  [tex]\Theta[/tex]= 0 degrees with the normal coil.

(a) The Magnetic flux, will be given as

[tex]\rm\phi =N\times B\times A[/tex]

By putting the values

[tex]\phi =100\times0.0650\times3.14\times 0.1\times0.1[/tex]

[tex]\phi=0.204 \rm \ weber[/tex]

(b) The emf induced will to be given as

[tex]B_1=0.0650T\ \ B_2=0.1T\ \ dt=0.5s[/tex]  

[tex]\dfrac{dB}{dt} =\dfrac{B_2-B_1}{dt} =\dfrac{0.1-0.650}{0.5} =0.07\dfrac{T}{s}[/tex]

induced emf,

[tex]e=N\dfrac{d\phi}{dt}[/tex]

[tex]e=N\times A \times \dfrac{dB}{dt}[/tex]

[tex]e=100\times 3.14\times0.1\times0.1\times0.07=0.22V[/tex]

Thus

(a) The magnetic flux through a coil 0.204 Weber

(b) The emf induced in the coil during the time interval 0.22 Volt

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Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

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

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

[tex]i[/tex] = magnitude of current

magnitude of current is given as

[tex]i = \frac{Ane^{2}\tau E}{m}[/tex]

[tex]i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}[/tex]

[tex]i[/tex]  = 1.87 A

Answer:

213.18 volt

Explanation:

N = 500, A = 0.02 m^2, B = 0.30 T, f = 960 rpm = 960 / 60 = 16 rps

The maximum value of induced emf is given by

e0 = N x B x A x ω

e0 = N x B x A x 2 x π x f

e0 = 500 x 0.3 x 0.02 x 2 x 3.14 x 16

e0 = 301.44 Volt

The rms value of induced emf is given by

[tex]e_{rms} = \frac{e_{0}}{\sqrt{2}}[/tex]

[tex]e_{rms} = \frac{301.44}{\sqrt{2}}[/tex]

e rms = 213.18 volt

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

[tex]EMF = \frac{\phi_2 - \phi_1}{\Delta t}[/tex]

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

[tex]\phi_1 = NBAcos 0 = BA[/tex]

[tex]\phi_2 = NBAcos 90 = 0[/tex]

now we will have

[tex]EMF = \frac{NBA}{t}[/tex]

[tex]10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}[/tex]

[tex]B = 0.126 T[/tex]

A cool experiment with a testable hypothesis is testing an apple and if it will be able to oxidize in different temperatures with the inside exposed to Oxygen.  

Hypothesis: If the cut piece of an apple is exposed to warm air then the apple will turn brown and oxidize faster than a cut piece of apple in a cooler temperature.

Supplies: Three thermometers, An apple, an apple cutter, a plate, etc.

Directions place the three thermometers in a cool cold area,

Answer:

Temperature change in the wire is 625°C

Explanation:

Let the initial current in the wire be I₀

Thus,

the final current in the wire will be, I = (1 - 0.20) × I₀ = 0.80I₀

Given α = 0.0004 1/°C

Now,

Resistance = (potential difference (E))/current  (I)

also

R = R₀ [1 + αΔT]

Where ΔT is the increase in temperature , thus

E/I = E/I₀ [1 + α × ΔT)

on substituting the values in the above equation, we get

1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]

or

1/(0.80) = [1 + 0.0004 × ΔT)]

or

ΔT = 625°C

The temperature change in the Nichrome wire, based on a 20% drop in current, is calculated to be approximately 625°C.

A Nichrome wire's resistance increases with temperature, primarily determined by its temperature coefficient.

Step-by-Step Solution :

      E/I = E/I₀ [1 + α × ΔT)

on substituting the values in the above equation, we get

= 1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]

= 1/(0.80) = [1 + 0.0004 × ΔT)]

= ΔT = 625°C

The temperature change in the Nichrome wire is  approximately 625°C.

Answer:

All the balls hit the ground with same velocity.  

Explanation:

Let the speed of throwing be u and height of building be h.

Ball A is launched 45 degrees above the horizontal

Initial vertical speed = usin45

Vertical acceleration = -g

Height = -h

Substituting in v² = u² +2as

                       v² = (usin45)² -2 x g x (-h)

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]  

Ball C is launched horizontally

Initial vertical speed = 0

Vertical acceleration = -g

Height = -h

Substituting in v² = 0² +2as

                       v² = 0² -2 x g x (-h)

                       v² = 2 x g x h

Final vertical speed² = 2 x g x h

Initial horizontal speed = final horizontal speed =  u

Final horizontal speed² =u²

Magnitude of final velocity

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

Ball B is launched 45 degrees below the horizontal

Initial vertical speed = usin45

Vertical acceleration = g

Height = h

Substituting in v² = u² +2as

                       v² = (usin45)² +2 x g x h

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]    

All the magnitudes are same so all the balls hit the ground with same velocity.  

The ball C which was launched horizontally will hit the ground fastest.

With respect to angle of projection above the horizontal, the time of motion each ball is calculated as follows;

[tex]T = \frac{u sin\theta }{g}[/tex]

where;

When θ is 45 degrees above the horizontal;

[tex]T = \frac{u \times sin(45)}{9.8}\\\\ T = 0.072u[/tex]

When θ is 45 degrees below the horizontal = 135 degrees above the horizontal

[tex]T = \frac{ u \times sin(135)}{9.8} \\\\T = 0.072 \ s[/tex]

When launched horizontally, θ = 90 degrees above the horizontal

[tex]T = \frac{u \times sin(90)}{9.8} \\\\T = 0.1 u[/tex]

Thus, the ball C which was launched horizontally will hit the ground fastest.

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Answer:

Very fast!

Explanation:

The waves are traveling very fast. I know this because they're not traveling slow for sure.

If this is incorrect, they're probably traveling moderately fast.

Answer:

Explanation:

The half-life of a radioisotope, in this case carbon-14, is the time that a sample requires to reduce its amount to half, and it is a constant for every radioisotope (it does not change with the amount of sample).

Then, the formula for the remaining amount of a radioisotope is:

Where:

The number of half-lives for carbon-14 elapsed for the dinosaur fossil is:

Then, A / A₀ = (1/2)ⁿ = (1/2)¹¹⁸⁶⁷ ≈ 0.00000 .

The number is too small, and when you round to five decimal places the result is zero. That is why carbon-14 cannot be used to date dinosaur fossils, given that they are too old.

C-14 dating is not applicable to a 68 million years old dinosaur fossil, because the half-life of C-14 (5730 years) only allows for accurate dating up to about 50,000-57,000 years. After such time period, the remaining C-14 would be too small to measure accurately. Therefore, for a 68 million year old fossil, the remaining C-14 would be effectively zero.

The question is asking about the amount of Carbon-14 (C-14) remaining in a dinosaur fossil that is approximately 68 million years old. However, C-14 dating is not applicable to such old samples. This is because C-14 decays back to Nitrogen-14 (N-14) with a half-life of approximately 5,730 years. Because of this half-life, the best-known method for determining the absolute age of fossils with C-14 allows for reliable dating of objects only up to about 50,000 years old. After 10 half-lives, which would be about 57,000 years, any original C-14 existing in a sample would become too small to measure accurately

For very old samples like a 68 million year old dinosaur fossil, isotopes with much longer half-lives are used for dating, such as uranium-235 and potassium-40. Therefore, for a 68 million year old dinosaur fossil, the remaining C-14 would effectively be zero.

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Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance [tex]C=260\ pF[/tex]

Charge [tex]q=0.155\ \mu\ C[/tex]

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

[tex]V= \dfrac{Q}{C}[/tex]

Where, Q = charge

C = capacitance

Put the value into the formula

[tex]V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}[/tex]

[tex]V=596.2\ volts[/tex]

Hence,The potential difference between the plates is 596.2 volts.

The potential energy stored in the spring is calculated using the formula U = 1/2kx^2. With a force of 20 N and a spring constant of 10 N/m, the displacement is found to be 2 m, and thus the potential energy is 20 J.

The question is related to the calculation of the potential energy stored in an ideal spring. To find this, we use the formula for elastic potential energy, which is U = 1/2kx2, where U is the potential energy, k is the spring constant, and x is the displacement (or compression) of the spring from its rest position.

In this case, since the force holding the spring in compression is 20 N and the spring constant is 10 N/m, we first need to find the displacement x. The force F applied to a spring is also given by F=kx, so rearranging for x gives us x = F/k = 20 N / 10 N/m = 2 m. Plugging this into the potential energy formula, we get U = 1/2 * 10 N/m * (2 m)2 = 1/2 * 10 * 4 = 20 J.

Therefore, the potential energy stored in the spring is 20 J.

Answer:

option (d)

Explanation:

There are two types of oscillations.

1. Damped oscillations: The oscillations in which the amplitude of a wave goes on decreasing continuously are called damped oscillations.

For example, the oscillations of a seismic wave.

2. Undamped oscillations: The oscillations in which the amplitude o a wave remains constant are called undamped oscillations.

for example, oscillations of light waves.

Answer:

[tex]a = -16 m/s^2[/tex]

Explanation:

As we know that the initial speed of the particle when it enters the nozzle is given as

[tex]v_i = 10 m/s[/tex]

after travelling the distance d = 3 m it will have its final speed as

[tex]v_f = 2 m/s[/tex]

now we know that when acceleration is constant the equation of kinematics is applicable

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]2^2 - 10^2 = 2(a)(3)[/tex]

[tex]a = -16 m/s^2[/tex]

Final answer:

To find the average acceleration from a fluid dynamics perspective, one can use the Eulerian viewpoint to observe velocity changes at a fixed point, or follow the particle through its motion using the Lagrangian viewpoint. However, the provided information is insufficient for precise calculation without the time variable or assumption of constant deceleration.

Explanation:

To find the average acceleration of a particle that slows down from 10 m/s to 2 m/s over a distance of 3 meters, we can use the concepts of Eulerian and Lagrangian viewpoints in fluid dynamics, though it is more common to apply these in the context of fluid flow rather than individual particles in motion.

In the Eulerian viewpoint, we observe the change of velocity at a fixed point in space as the particle passes through. Writing the kinematic equation δv = aδt, where δv is the change in velocity and a is the average acceleration, we can rearrange to find a = δv / δt.

The Lagrangian viewpoint, on the other hand, follows the particle along its path. Given the initial and final velocities (v₀=10 m/s and vf=2 m/s) and the distance traveled (3 m), we can use the kinematic equation vf^2 = v₀^2 + 2aδx to solve for a. In this case, δx is 3m, and vf and v₀ are given. From this equation, a = (vf^2 - v₀^2) / (2δx).

If actual calculations are performed, please note that the given information in the problem statement is insufficient to solve for time, as neither the time taken to travel through the nozzle nor the deceleration time is provided. Assuming the particle moves at a constant deceleration, one can calculate the time through δv = aδt and then average acceleration using that time duration. However, without additional data or time measurement, we cannot find a precise numerical value for acceleration.

Answer:

64.08 rad

Explanation:

f = 78 rpm = 76 / 60 rps

θ = 0.45 rad, t = 8 s

(θ2 - θ1) / t = ω

θ2 - θ1 = 2 x 3.14 x 76 x 8 / 60 = 63.63

Angle turns from initial position = 0.45 + 63.63 = 64.08 rad

Answer:

The final speed of the train car is 8.57 m/s.

Explanation:

Given that,

Mass of train car = 1000 kg

Mass of sand = 400 kg

Speed of train car = 12 m/s

We need to calculate the final speed of the train car after the sand is dumped

Using conservation of momentum

[tex]m_{1}u_{1}=(m_{1}+m_{2})v[/tex]...(I)

Put the value in the equation (I)

[tex]1000\times12=(1000+400)v[/tex]

[tex]v=\dfrac{1000\times12}{1400}[/tex]

[tex]v=8.57\ m/s[/tex]

Hence, The final speed of the train car is 8.57 m/s.

Explanation:

It is given that,

Mass of lithium, [tex]m=1.16\times 10^{-26}\ kg[/tex]

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV[/tex]

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

q is the charge on an electron

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}[/tex]

v = 78608.58 m/s

[tex]v=7.86\times 10^4\ m/s[/tex]

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

[tex]qvB=\dfrac{mv^2}{r}[/tex]

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}[/tex]

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

Answer:

33.33 m/s

Explanation:

m = 450 kg. T = 5000 N, t = 3 seconds,

let the net acceleration is a.

T = m a

a = 5000 / 450 = 11.11 m/s^2

u = 0 , v = ?

Let v be the velocity after 3 seconds.

Use first equation of motion

v = u + a t

v = 0 + 11.11 x 3 = 33.33 m/s

Answer:

Option (B)

Explanation:

L = 0.5 H, I = 2 A, R = 10 Ohm

V = I R

In case of dc the inductor cannot works.

V = 2 x 10 = 20 V

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

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

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

[tex]0-(17.7)^2=2a\times 33.1[/tex]

[tex]a=\dfrac{-(17.7)^2}{33.1}[/tex]

[tex]a=-9.46\ m/s^2[/tex]

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

[tex]F = ma[/tex]

[tex]F =1400\times(-9.46)[/tex]

[tex]F=-13244\ N[/tex]

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

[tex]|F| =13244\ N[/tex]

Hence,  The magnitude of the horizontal net force is 13244 N.

Answer:

3 x 10^-4 N/m

Attractive

Explanation:

r = 0.1 m, i1 = 10 A, i2 = 15 A

The force per unit length between two parallel wires is given by

[tex]F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}[/tex]

[tex]F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}[/tex]

F = 3 x 10^-4 N/m

Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.

Answer:

B.Exhaust Value

Explanation:

The waste products of combustion leave the internal combustion engine through the Exhuast Value.

Answer:

Exhaust Value

Explanation:

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

[tex]F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N[/tex]

[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\[/tex]

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\[/tex]

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

Answer:

His velocity when t= 3 sec is V= 1.56 m/s

Explanation:

a= 0.5 m/s²

ω= 0.2 rad/s

t= 3 sec

Vr= a*t

Vr= 1.5 m/s

r= a*t²/2

r= 2.25m

Vt= w*r

Vt= 0.45 m/s

V= √(Vr²+Vt²)

V= 1.56 m/s

The boy's tangential velocity, after accelerating radially for 3 seconds at 0.5 m/s^2 while the platform rotates at 0.2 rad/s, is 1.5 m/s.

To determine the boy's velocity (magnitude) at t = 3 sec while he runs outward radially from the center of a rotating platform, we must consider both his radial (tangential) velocity due to his acceleration and the rotational movement of the platform.

The boy starts from rest with a constant radial acceleration of 0.5 m/s2. Using the kinematic equation v = u + at (where u is the initial velocity, a is the acceleration, and t is the time), we can calculate his radial velocity after 3 seconds as:

vradial = 0 + (0.5 m/s2)(3 s) = 1.5 m/s

The platform's constant rotation rate is given as 0.2 rad/s. This rotational movement does not change the boy's radial (tangential) velocity directly, as the question only asks for his velocity in the radial direction. Therefore, the total magnitude of the boy's velocity after 3 seconds is 1.5 m/s tangentially.