A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball
1 answer:
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Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Answer:
F = 4.48N
Explanation:
In order to calculate the net gravitational force on the rocket, you take into account the formula for the gravitational force between two objects, which is given by:
(1)
G: Cavendish's constant = 6.674*10^-11 m^3kg^-1s^-2
r: distance between the objects
You have a rocket at the middle of the distance between Earth and Moon, then, you have opposite forces on the rocket.
If you assume the origin of a system of coordinates at the rocket position, with the Moon to the left and the Earth to the right, you have:
(2)
Me: mass of the Earth = 5.98*10^24 kg
Mm: mass of the Moon = 7.35*10^22 kg
m: mass of the rocket = 1200kg
r1: distance from the rocket to the Earth = 3.0*10^8m
r: distance between rocket and Moon = 3.84*10^8m - 3.0*10^8m = 8.4*10^7m
You replace the values of the parameters in the equation (2):
The net force exerted over the rocket is 4.48N