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2 years ago

If a circling object is released, centrifugal force will make it travel away from the center of its original path.

1 answer:

5 0

The answer is True. The object is thrown away from the center of its original path due its inertia. Centrifugal force acts in the opposite direction as centripetal force. Centripetal force applies towards the center of the curvature in a spinning object. Centrifugal force is considered an apparent force while centripetal force is an actual force.

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How to find initial velocity in projectile motion problems when you are not given the vlocity?

Papessa [141]

Vi= square root vi^2 -2ad

7 0

3 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

2 years ago

A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

sp2606 [1]

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

6 0

2 years ago

The volume of a gas is 605 liters at 27.0°C. The new temperature is -3.0°C. What is the new volume?

nlexa [21]

From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545

5 0

2 years ago

Read 2 more answers

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0

2 years ago

Read 2 more answers