I NEED HELPP!!!!!!!!!!!!!!

The impulse experienced by a body is equivalent to the body’s change in?

vovangra [49]

<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

#brainlyzkool</span>

5 0

3 years ago

Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X

Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

$F=k\frac{q_x q_y}{r^2}$ (1)

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_x, q_y$ are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago

5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-

Airida [17]

D. 980, this is the best answer because 35 x 7 is 980 :)

3 0

2 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$