n an oscillating LC circuit, L = 3.76 mH and C = 3.13 μF. At t = 0 the charge on the capacitor is zero and the current is 2.95 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?

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:

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:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

[tex]mgx - \frac{1}{2}kx^2 = 0[/tex]

[tex](2.07)(9.8)(0.131) - \frac{1}{2}k(0.131)^2 = 0[/tex]

now we will have

[tex] k = 310 N/m[/tex]

Part B)

Time period of oscillation is given as

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

[tex]T = 2\pi\sqrt{\frac{2.07}{310}}[/tex]

[tex]T = 0.51 s[/tex]

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:

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:

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:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

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:

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:

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:

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:

(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:

3.9 atm

Explanation:

Ideal gas law states:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The volume of the tank is constant, so we can say:

V₁ = V₂

Using ideal gas law to write in terms of P, n, R, and T:

n₁ R T₁ / P₁ = n₂ R T₂ / P₂

n₁ T₁ / P₁ = n₂ T₂ / P₂

Initially:

P₁ = 9.8 atm

T₁ = 29.0°C = 302.15 K

n₁ = n

Afterwards:

P₂ = P

T₂ = 83.0°C = 356.15 K

n₂ = n/3

Substituting:

n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P

(302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P

P = 3.85 atm

Rounding to 2 significant figures, P = 3.9 atm.

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:

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

Final answer:

The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

Explanation:

To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:

Impulse = change in momentum

Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).

Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:

Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s

So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.

To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:

Average force = impulse / time

Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:

Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N

Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

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:

[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:

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.

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:

Original speed of the bullet = 305.21 m/s

Explanation:

Here momentum is conserved.

Mass of bullet = 9.6 g = 0.0096 kg

Mass of wood = 4.95 kg

Let velocity of bullet be v.

Initial momentum = 0.0096 v

Final mass = 0.0096 + 4.95 = 4.9596 kg

Final velocity = 0.591 m/s

Final momentum = 4.9596 x 0.591 = 2.93 kgm/s

Equating both momentum

           0.0096 v = 2.93

                      v = 305.21 m/s

Original speed of the bullet = 305.21 m/s

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:

f = 1.354*10^{20} Hz

Explanation:

By conservation of linear momentum, wavelength shift due to collision of photon to electron is given by following formula

[tex]\lambda ^{'}-\lambda =\frac{h}{m_{o}c}(1-cos\theta )[/tex]

where h is plank constant = 6.626*10^{-34}

c = speed of light = 3*10^{8} m/s

scattered angle  =  60 degree

m = rest mass of electron  = 9*10^{-31}

[tex]\lambda ^{'}=10^{-12} +\frac{6.626*10^{-34}}{9*10^{-31}*3*10^{8}}(1-cos60^{o} )[/tex]

[tex]\lambda ^{'}= 2.215 pm[/tex]

we know that 1 pm = 10^{-12}m

[tex]f = \frac{c}{\lambda ^{'}}[/tex]

[tex]f = \frac{3*10^{8}}{2.215 *10^{-12}} = 1.354*10^{20} Hz[/tex]

f = 1.354*10^{20} Hz

Answer:

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

Explanation:

The forces acting on a rock thrown up are force due to gravity and air resistance. Air resistance is directly proportional to velocity of rock, when velocity is zero air resistance is zero. When it is at the top of its trajectory its velocity is zero. So air resistance is also zero. Hence only gravitational force acts on the rock.

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

At the peak of its trajectory, the only significant force acting on a rock is the force of gravity. Thus, the net external force on the system is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity. At this point, the rock begins its descent due to the gravitational pull.

The question is asking about the net external force acting on a rock when it is at the top of its trajectory. At the peak of its trajectory, the only significant force acting on the rock is the force of gravity (weight of the rock), assuming air resistance is negligible. Thus, the net external force on the system, if air resistance is neglected, is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity.

For example, if a rock is thrown straight upwards with an initial velocity, as it reaches the topmost point of its trajectory, the vertical velocity momentarily becomes zero before the rock starts descending again. At this point, the only major force acting on the rock is the gravitational pull exerted by the Earth, also known as the rock's weight.

It is important to note that this net external force (-mg) is what determines the acceleration of the object at the top of its trajectory. Simply put, even at the top of its path, the rock is still experiencing the force of gravity which pulls it back towards Earth, leading to its eventual fall back to the ground.

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Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water  x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

Further Explanation:

The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.

Given:  

The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].  

The mass m of Max is [tex]15\text{ kg}[/tex].

Concept:

The stress developed in the bungee cord is given by:  

[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex]                                                                               ... (1)         

Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.  

The radius of the bungee cord will be the half of its diameter.  

[tex]r=\dfrac{d}{2}[/tex]

Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].  

[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex] 

Convert the radius of the bungee cord in meter.  

[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex] 

The area of cross-section of the bungee cord is given by:  

[tex]A=\pi r^{2}[/tex]  

Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].  

[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]  

The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:  

[tex]F=mg[/tex]  

Here, m is the mass of Max's body and g is acceleration due to gravity.  

Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].  

Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.  

[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]  

Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).  

[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]

Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Stress and Strain

Keywords:

Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.

The stress in the bungee cord due to Max’s weight is 294,000 N/m².

The given parameters;

The area of the bungee cord is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]

The weight of Max is calculated as follows;

[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]

The stress in the bungee cord due to Max’s weight is calculated as;

[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]

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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