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Snezhnost [94]
3 years ago
8

find the gravitational force between the two bodies of unit mass each and separated by unit distance​

Physics
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

Here gravitational force is G or gravity constant.

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Roger drives his car at a constant speed of 80 km/hr. How far can he travel in 2 hrs. and 30 minutes?
bonufazy [111]

Answer:

200 km/hr

Explanation:

Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.

80 x 2.5 = 200 km/hr.

6 0
2 years ago
Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
a major difference radio waves, visible light, and gamma rays is the _________ of the photons, which results in different photon
Aleks04 [339]

There is a spectrum of electromagnetic radiation with variable wavelengths and frequency, which in turn imparts different characteristics. ... X-rays and gamma rays have the same nature as visible light, radiant heat, and radio waves; however, they have shorter wavelengths and consequently a larger photon energy.

3 0
3 years ago
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What distance should be used to pattern a shotgun hunter ed?
Murljashka [212]
I think the distance that should be used is the distance that one expects to be from the game you are hunting. Before taking a shotgun for a gobbler or even for ducks or other animals, you need to see how your gun performs by patterning it at various ranges with the load you want to use.
7 0
3 years ago
The light coming out of a concave lens:
torisob [31]
The light coming out of a concave lens will never meet.

So, the answer is A. will never meet.

Happy Studying! ^^
4 0
3 years ago
Read 2 more answers
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