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Firdavs [7]
2 years ago
15

The objects shown in the following diagrams have different masses and are different distances apart. Which diagram shows the two

objects that have the greatest force of gravity acting between them?
Physics
1 answer:
sergiy2304 [10]2 years ago
5 0

Answer:

C. 10kg to 10kg

Explanation:

You have to picture to it I think

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Ede4ka [16]
Its to capture light or to focus. don't forget to like. :D

6 0
3 years ago
A force of 99 N causes a box to accelerate at a rate of 12 m/s2. What is the mass of the box?
ira [324]


m = F/a

m = 99/11 = 9

The answer is 9 Kg

5 0
3 years ago
Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force
melamori03 [73]

Answer:

C

Explanation:

gravity is a pulling force according to Newton

4 0
2 years ago
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0
1 year ago
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the dr
bulgar [2K]

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

     

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

6 0
3 years ago
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