How fast is a 5 kg object traveling that has 20 kg m/s of momentum?

Write down the factors on which moment depends upon ?

timurjin [86]

Answer:

The size of a force
The perpendicular distance from the pivot the line of action of force

Explanation:

Factors that affect the moment of a force are;

The size of a force
The perpendicular distance from the pivot the line of action of force

The magnitude of force applied is directly proportional to the moment of force in that for a perpendicular distance d, increased in force applied will result to a higher moment of force. When the perpendicular distance from the pivot is decreased while the force applied remains constant, the moment of force decreases.

5 0

3 years ago

A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free f

FrozenT [24]

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s$

Velocity at the end of its upward acceleration is 138.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m$

Maximum height the rocket reaches is 762.3 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s$

The velocity with which the rocket crashes to the Earth is 122.3 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s$

Total time from launch to crash is 12.47+11 = 24.47 seconds

6 0

3 years ago

ASAP please!<br> need physics help on test

givi [52]

Answer:

I think its the second one sorry if im wrong

Explanation:

6 0

2 years ago

The mass of the Earth is 5.98 × 1024 kg . A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceler

jekas [21]

Answer:

v = 7.18_m/s

Explanation:

Velocity of the earth towards the ball is = velocity of the ball moving towards earth

For object in free fall, we have

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

S = height of ball above ground

v^2 = u^2 - 2×g×(-S)

= 0 + 2×9.8×2.63 = 51.55_m^2/s^2

Velocity of the ball just before it hits the ground is

v = 7.18_m/s

3 0

3 years ago

You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed

BlackZzzverrR [31]

Answer:

$6.5\; {\rm N}$ .

Explanation:

When an object travel at a speed of $v$ in a circle of radius $r$ , the (centripetal) acceleration of that object would be $a = (v^{2} / r)$ .

In this question, the ball is travelling at $v = 3\; {\rm m\cdot s^{-1}}$ in a circle of radius $r = 1.8\; {\rm m}$ . The (centripetal) acceleration of this ball would be:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}$ .

By Newton's Laws of Motion, for an object of mass $m$ , if the acceleration of that object is $a$ , the net force on that object would be $m\, a$ . Since the acceleration of this ball is $a = 5\; {\rm m\cdot s^{-2}}$ , the net force on this ball would be:

$\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}$ .

7 0

2 years ago