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3 years ago

2) What are the directions of the velocity and acceleration of an object in

Physics

2 answers:

zaharov [31]3 years ago

7 0

Motion because the motion is the range

r-ruslan [8.4K]3 years ago

3 0

Uniform circular motion is when an object travels in a circular path at a constant speed.

Direction of the velocity is constant
Direction of the acceleration is inward.

You might be interested in

What is the centripetal force

Lelechka [254]

Answer:90N

Explanation:

Mass=30kg

Centripetal acceleration=3m/s^2

centripetal force=mass x centripetal acceleration

Centripetal force=30 x 3

Centripetal force =90

Centripetal force =90N

4 0

3 years ago

[TRUE/FALSE] If an object is moving at constant velocity it must be in equilibrium.

DENIUS [597]

Answer:

yes

Explanation:

objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium

4 0

3 years ago

1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7

xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0

1 year ago

2.)

Oduvanchick [21]

Answer:

$-22.7 m/s^2$

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2$

4 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago