3 years ago

When a constant force acts on an object, what does the object's change in momentum depend upon?

Physics

1 answer:

Eva8 [605]3 years ago

8 0

You don’t need time (1) or the force (4) either so just 3,5

You might be interested in

3. Looking at the image below label the areas of highest and lowest kinetic energy of both gravitational potential

MArishka [77]

Explanation:

max potential your welcome i love helping outhers :)

8 0

3 years ago

How long will it take this wave to travel 3000 m in the x-direction?

REY [17]

2.4 secs I think

?????????????????????????

5 0

3 years ago

Which component of health-related fitness is developed by performing a wall sit?

baherus [9]

Answer:

The answer is Muscular endurance

Explanation:

It is this because your seeing how long your muscles can with stand. I answered by guessing lol.

4 0

3 years ago

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a

dem82 [27]

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m$

The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

$v_{y}^{2}=u_{y}^{2}+2gh'$

here, uy = 0

So,

$v_{y}^{2}=2\times 9.8 \times 6.32$

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{11.87^{2}+11.13^{2}}$

v = 16.27 m/s

7 0

4 years ago

You are a doctor and Jack is your patient. Jack wants to know about his overall health. What is the best thing you can do to hel

Novay_Z [31]

Answer:

determine his body composition

Explanation:

5 0

3 years ago