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crimeas [40]
3 years ago
11

Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

^- 8 N how far apart are centers of mass?

Physics
1 answer:
nydimaria [60]3 years ago
8 0

Answer:

Explanation:

The formula for this is

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2} and moving things around to solve for r:

r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} } Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

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When talking about variables in a scientific experiment, describe how you know what the independent variable, dependent variable
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An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.

4 0
3 years ago
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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
3 years ago
Radon is the heaviest naturally radioactive ________________ gas. A) noble B) halogen C) group 1 D) diatomic
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(A) noble gases / group 18
6 0
3 years ago
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When applying a horizontal force of 30N, an object of mass 6 kg accelerates at 4m/s2. The force of friction on the surface must
Rus_ich [418]
Using Newton's Second Law, F = ma, where F is the net force

So the net force is:

F = (6kg)(4m/s^2) = 24N

Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.

30N-24N = 6N

F_{f} = 6N
4 0
3 years ago
A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the
Nesterboy [21]

Answer:

A.) 1372 N

B.) 1316 N

C.) 1428 N

Explanation:

Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:

a. The load moves downward at a constant velocity

At constant velocity, acceleration = 0

T - mg = ma

T - mg = 0

T = mg

T = 140 × 9.8

T = 1372N

b. The load accelerates downward at a rate 0.4 m/s??

Mg - T = ma

140 × 9.8 - T = 140 × 0.4

1372 - T = 56

-T = 56 - 1372

- T = - 1316

T = 1316N

C. The load accelerates upward at a rate 0.4 m/s??

T - mg = ma

T - 140 × 9.8 = 140 × 0.4

T - 1372 = 56

T = 56 + 1372

T = 1428N

8 0
3 years ago
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