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EleoNora [17]
3 years ago
14

What is the gravity force between two stars with mass of 5,000,000 kg and 1,000,000 kg if the distance between them is 100 m

Physics
1 answer:
Luda [366]3 years ago
8 0

Answer:

0.0334N

Explanation:

Given parameters:

M1  = 5 x 10⁶kg

M2  = 1 x 10⁶kg

Distance  = 100m

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

     Fg  = \frac{G m1 m2}{r^{2} }  

G is the universal gravitation constant

m is the mass

r is the distance

         Fg  = \frac{6.67 x 10^{-11} x 5 x 10^{6}  x 1 x 10^{6} }{100^{2} }   = 0.0334N

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Answer:

F = m x a

Explanation:

F = force

m = mass of an object

a = acceleration

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Formed through longshore drift<br> a. sea stack<br> b. sandbar<br> c. spit<br> d. headland
ANTONII [103]
<h3>Answer;</h3>

<em><u>Sand Spit or Spit </u></em>

<h3><u>Explanation;</u></h3>
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7 0
3 years ago
Read 2 more answers
Based on the graph, which data point is most likely to have experimental
Len [333]

Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

8 0
2 years ago
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi
Snowcat [4.5K]

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ V=20\ \Omega,\ V_1=30\ \Omega

⇒ G=1680\ \Omega,\  G_1=2930\ \Omega

Concept to be used:

Conversion of galvanometer into voltmeter.

Let G be the resistance of the galvanometer and I_g the maximum deflection in the galvanometer.

To measure maximum voltage resistance R is connected in series .

So,

⇒ V=I_g(R+G)

We have to find the value of R we know that in series circuit current are same.

For G=1680                                    For G_1=2930

⇒ I_g=\frac{V}{R+G}   equation (i)                ⇒ I_g=\frac{V_1}{R+G_1} equation (ii)

Equating both the above equations:

⇒ \frac{V}{R+G} = \frac{V_1}{R+G_1}

⇒ V(R+ G_1) = V_1 (R+G)

⇒ VR+VG_1 = V_1R+V_1G

⇒ VR-V_1R = V_1G-VG_1

⇒ R(V-V_1) = V_1G-VG_1

⇒ R =\frac{V_1G-VG_1}{(V-V_1)}

⇒ Plugging the values.

⇒ R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}

⇒ R =\frac{(50400 - 58600)}{(-10)}

⇒ R=\frac{-8200}{-10}

⇒ R=820\ \Omega

The coil resistance of the circuit is 820 Ω .

4 0
3 years ago
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