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VLD [36.1K]
3 years ago
15

Two large boulders are 8.5 m apart.one boulder has a mass of 3600 kg and the other rock has a mss of 2850 kg find the gravitatio

nal force betweeen them
Physics
1 answer:
Inessa [10]3 years ago
8 0

9.47 \times 10^{-6}\; \text{N}.

<h3>Explanation</h3>

Constant of universal gravitation:

G = 6.67 \times 10^{-11} \;\text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}.

Apply Newton's law of universal gravitation:

F = \dfrac{G \cdot m_1 \cdot m_2}{r^{2}} =\dfrac{6.67 \times 10^{-11}\times 3600 \times 2850}{8.5^{2}} = 9.47 \times 10^{-6} \;\text{N}.

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A spring has a natural length of 14 cm. If a 23-N force is required to keep it stretched to a length of 20 cm, how much work W i
lubasha [3.4K]

Answer:

1.78 J

Explanation:

Find the spring coefficient using Hooke's law:

F = k Δx

23 N = k (0.20 m − 0.14 m)

k = 383.33 N/m

The work is the change in energy:

W = PE₂ − PE₁

W = ½ kx₂² − ½ kx₁²

W = ½ k (x₂² − x₁²)

W = ½ (383.33 N/m) ((0.17 m)² − (0.14 m)²)

W = 1.78 J

4 0
3 years ago
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignor
pogonyaev

Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

y =- \frac{1}{2} gt^2 + v_0 t + y_0

where:

acceleration g = 9.8 m/s²

time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0

This leaves only one unknown: v₀

v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2

and how fast will it go:

v' = -gt + v_0

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2

Now let's use this result in the equation for the first ball:

0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m

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3 years ago
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Acceleration= change in velocity/time=27.8-0/4.8=5.792 ms⁻²
Force= Mass*acceleration=66*5.792=382.25 N
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