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2 years ago

6

An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius

of the circle?

1 answer:

Gnom [1K]2 years ago

7 0

Answer:

r = mv^2/ Fc

Explanation:

Got it right

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Which option is part of designing a set of experimental procedures

-BARSIC- [3]

determining how data will he gathered

Explanation: Apex

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3 years ago

The tension in the horizontal towrope pulling a water-skier is 250 N while the skier moves due west a distance of 50 m. How much

icang [17]

Answer:

W = 250(50) = 12500 J

Explanation:

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3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface

Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is 0.950999 m/s

Explanation:

Given data

legs of length r = 0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr) ...............1

here r is length of leg

we know g = GM/r² from universal gravitational law

so G we know 6.67 × $10^{-11}$ N-m²/kg²

g = 6.67 × $10^{-11}$ ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is 0.950999 m/s

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3 years ago

What is the most likely evidence of an energy transformation taking place when a skater glides over ice?

Triss [41]

Small tracks of water are left on the ice as kinetic energy is transformed into thermal energy

5 0

3 years ago

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