Answer:
You should use this for work related questions :/
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
Answer:
44 N
Explanation:
The electrostatic forces between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is their separation
We notice that the force is directly proportional to the charges.
In this problem, initially we have a force of
F = 22 N
on a q2 = 4.0 C, exerted by a charge q1.
If the charge is doubled,
q2 = 8.0 C
This means that the force will also double, so it will be
E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m
