A particle in uniform circular motion requires a net force acting in what direction?

A ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum he

Andrew [12]

Answer:

10.2 metres

Explanation:

Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?

Solution

To get the maximum height, let us use the formula

V^2 = U^2 sin^2ø - 2gH

At maximum height V = 0

U^2 sin^2ø = 2gH

Substitute all the parameters into the formula

20^2 ( sin 45 )^2 = 2 × 9.8 × H

400 × 0.5 = 19.6 H

Make H the subject of formula

H = 200 / 19.6

H = 10.204 metres.

Therefore, the maximum height reached by the projected ball is 10.2 metres.

5 0

3 years ago

If the moon moved closer to the earth, how would the gravitational force between them change?.

hichkok12 [17]

If the moon moved closer to the earth, The gravitational force between them will increase.

This can be understood by Newton's law of universal gravitation. This is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

$F_{gravitation}$ = G $\frac{M_{1} M_{2}}{r^{2} }$

If the moon moved closer to earth, the distance between them will be reduced, and as the force of gravitation is inversely proportional to the distance between the masses, the gravitation force increases.

Learn more about gravitation force here

brainly.com/question/12528243

#SPJ4

7 0

2 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

(c) In the first part we have:

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$