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earnstyle [38]
3 years ago
13

the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force bet

ween them change? Also, assume the distance between the centers of the objects stayed constant. A. It would increase. B. It would decrease. C. It would remain the same.
Physics
2 answers:
slamgirl [31]3 years ago
7 0

Answer:

a

Explanation:

kherson [118]3 years ago
6 0

The gravitational force between the objects A. It would increase.

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m_1,m_2 are the masses of the two objects

r is the separation between the objects

In this problem, we are told that one of the object (the one on the right) gains mass: this means that, for instance, the value of m_2 increases. We can see from the equation that the gravitational force is directly proportional to the masses: therefore, if one of the masses increases (while the distance between the two objects remains constant), it means that the force also increases.

Therefore, the correct answer is

A. It would increase.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
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Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

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We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

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Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

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Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

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Answer:

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Explanation:

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4 years ago
Read 2 more answers
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