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Ghella [55]
3 years ago
12

9.The force of gravity between two asteroids is 10,000 newtons (N).

Physics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

F1 = G m1 m2 / R^2        force of attraction

F2 = G m1 m2 / (R/2)^2

F2 / F1 = 4       the force of gravity will be quadrupled

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The first is that the object moves. The second is that a force must act on the object in the direction the object moves.
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Which of the following represents a covalent compound?<br> a. BeF2<br> b. KF<br> C. PCIE<br> d. Ti
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KF is the covalent compound
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3 years ago
A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 29.0 m. After what time interval does it
Y_Kistochka [10]

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}

Then the solutions are:

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4 0
4 years ago
Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)ı^−(4.0m/s)ȷ^(20.0m/s)ı^−(4.0m/s
Art [367]

Answer:

(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^

Explanation:

Given

W = 0.56N = mg

m = 0.56/g = 0.56/9.8 = 0.057kg

t = 3.00ms = 3.00×10-³s

Impulse is a vector quantity so we would treat it as such

We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that

u = (20.0m/s)ı^−(4.0m/s)ȷ^

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Fx = –380N

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Impulse also equals the change in momentum of the body. So

J = m(v–u)

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Vx = -20 + 20 = 0m/s

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Vy= Jy/m + uy

Vy= 0.33/0.057 + (-4.0)

Vy= 5.79 + (-4.0) = 1.79m/s

V = (0m/s)ı^−(1.79m/s)ȷ^

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