Ask question

Ask question

All categories

3 years ago

11

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

t: The centripetal acceleration of the ball can be increased by a factor of 4 by Entry field with incorrect answer keeping the radius fixed and increasing the period by a factor of 4. keeping the radius fixed and increasing the speed by a factor of 4. keeping the radius fixed and decreasing the period by a factor of 4. keeping the speed fixed and decreasing the radius by a factor of 4. keeping the speed fixed and increasing the radius by a factor of 4.

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

You might be interested in

A centrifuge rotor (hollow disk) is rotating at 10,000 rpm is shut off and brought to rest by a constant frictional torque of 1.

Y_Kistochka [10]

Answer:

The no of revolutions rotor turn before coming to rest is 1,601.1943 and time taken is equal to 19.21 seconds

Explanation:

here we know that torque = I×α

α= angular acceleration

I = moment of inertia of hollow disc = m× $k^{2}$

given that m=4.37kg

k=0.0710m

torque=1.2Nm

$w_{o}=10000\times \frac{\pi}{30} rad /s$

$\alpha =\frac{torque}{I}$

from the above equation we can calculate the angular acceleration of the hollow disc .

since $w^{2}-w_o^{2} =2\alpha$ $\theta$

from this above equation $\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }$

no of revolutions = $\frac{\theta}{2\pi}$ = 1,601.1943.

Now to calculate time we know that time = $\frac{2\theta}{w+w_{o}}$

so upon calculating we will be getting t=19.21 seconds

5 0

3 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

3 years ago

The distance from one Crest to the next Crest is the blank

Luda [366]

The distance from one crest to the next crest in a set of waves is called the wavelength. The distance from the crest of one wave to the equilibrium point is called frequency.
Hope that helped! =)

8 0

3 years ago

Read 2 more answers

3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance

Digiron [165]

Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 0

3 years ago

A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg

Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0

3 years ago