Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
Exothermich cools and endo heats so it heats
<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>
The force needed to keep the space shuttle moving at constant speed is 0.
The given parameters;
- <em>weight of the space shuttle, F = 750,000 N</em>
- <em>constant speed of the space shuttle, v = 28,000 km/h</em>
The mass of the space shuttle is calculated as follows;

The force needed to keep the space shuttle moving at constant speed is calculated as follows;


where;
a is the acceleration of the space shuttle
At a constant speed, acceleration is zero.
F = 76,530.61 x 0
F = 0
Thus, the force needed to keep the space shuttle moving at constant speed is 0.
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Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= 
So the lines terminating at - 2 micro coulomb
= 
So, the number of filed lines terminates at - 2 micro Coulomb are 6.