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2 years ago

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A student sits motionless on a stool that can turn friction-free about its vertical axis (total rotational inertia I). The stude

nt is handed a spinning bicycle wheel, with rotational inertia I1, that is spinning about a vertical axis with a counterclockwise rotational velocity ω0. The student then turns the bicycle wheel over (that is, through 180∘). Determine the final rotational velocity acquired by the student. Express your answer in terms of the variables

1 answer:

vampirchik [111]2 years ago

6 0

Answer:

Explanation:

The problem can be solved with the help of conservation of angular momentum.

Initial angular momentum

= I₁ω₀

When wheel is turned by 180 degree , its angular momentum becomes

- I₁ω₀ .

So total angular momentum

= - I₁ω₀ . + I W where W is angular velocity of student .

Applying conservation of angular momentum

=I₁ω₀= - I₁ω₀ +I W

2 I₁ω₀ = I W

W = 2 I₁ω₀ / I

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A man pushes a drum of oil 10kg up an incline OA. If OA is 5m and AB is 3m, what is the potential energy of the drum at A.

Paladinen [302]

Answer:

294 Joules

Explanation:

From the question;

Mass of the drum is 10 kg
The length of inclined surface, OA is 5 m
The final height of the drum on the plane AB is 3 m

We are required to determine the potential energy of the drum at A

We know that potential energy is given by the formula;

Potential energy = mgh

where m is the mass, g is the gravitational pull and h is the height

Taking g as 9.8 N/kg

Then;

Potential energy = 10 kg × 9.8 N/kg × 3 m

= 294 Joules

Thus, the potential energy of the drum at A is 294 Joules

6 0

2 years ago

Read 2 more answers

Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capac

SVETLANKA909090 [29]

Answer:

Explanation:

Before the dialectic was inserted the capacitor is Co

When the slab is inserted,

The capacitor becomes

C=kCo

The charge Q is given as

Q=CV

Then, when C=Co

Qo=CoV

Then, when C=kCo

Q=kCoV

Then, the change in charges is given as

Q-Qo= kCoV - CoV

∆Q= kCoV - CoV

Current is given as

I=dQ/dt

I= (kCoV - CoV) / dt

I=Co(kV-V)/dt

Note Co is the value capacitor

So, Capacitance of parallel plates capacitor is given as

Co=εoA/d

Then,

I=εoA(kV-V)/d•dt

I=VεoA(k-1)/d•dt

Where A=πr²

I = V•εo•πr²•(k-1) / d•dt

This is the required expression for current is in the required term

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2 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

8 0

2 years ago

A cylindrical tank of methanol has a mass of 40 kgand a volume of 51 L. Determine the methanol’s weight, density,and specific gr

mezya [45]

Answer:

Weight W = 392.4 N

Density $\rho$ = 784.31 $\frac{kg}{m^{3} }$

Specific gravity S = 0.78431

Force required F = 10 N

Explanation:

Given data

Mass (m) = 40 kg

Volume (V) = 0.051 $m^{3}$

Weight W = m × g

⇒ W = 40 × 9.81

⇒ W = 392.4 N

This is the weight of the methanol.

Density $\rho$ = $\frac{mass }{volume}$

⇒ $\rho$ = $\frac{40}{0.051}$

⇒ $\rho$ = 784.31 $\frac{kg}{m^{3} }$

This is the density of the methanol.

Specific gravity (S) = $\frac{\rho}{\rho_{water} }$

⇒ $S = \frac{784.31}{1000}$

⇒ S = 0.78431

This is the specific gravity of the methanol.

Force needed to accelerate this tank F = ma

⇒ F = 40 × 0.25

⇒ F = 10 N

This is the force required to accelerate the tank.

4 0

2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago