The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Answer:
300000.01008 Pa
123.76237 m/s²
Explanation:
= Density of liquid nitrogen = 808 kg/m³
h = Depth
g = Acceleration due to gravity
P = Atmospheric pressure
Absolute Pressure is given by
Below 2 m from surface
Below 5 m from surface
Subtracting the above equations we get
The acceleration due to gravity on the planet is 123.76237 m/s²
Equating the value of g in the first equation
The atmospheric pressure on the planet is 300000.01008 Pa
One liter is equivalent to around 34 ounces (33.81 more exactly).
Fun fact: there are different ounces. So if you meant the imperial ounce then one liter would be equivalent to 35.2 ounces
OPTIONS :
A.) the force that the ball exerts on the wall
B.) the frictional force between the wall and the ball
C.) the acceleration of the ball as it approaches the wall
D.) the normal force that the wall exerts on the ball
Answer: D.) the normal force that the wall exerts on the ball
Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.
In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.
The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.
