The momentum of the ball when it hits the ground is 4.89 kg.m/s.
The given parameters;
- <em>mass of the baseball, m = 0.145 kg</em>
- <em>height of fall of the ball, h = 58 m</em>
The final velocity of the ball when it hits the ground is calculated as follows;
The momentum of the ball when it hits the ground is calculated as follows;
P = mv
P = 0.145 x 33.72
P = 4.89 kg.m/s
Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.
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Answer:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Explanation:
From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration (
), measured in meters per square second, is entirely centripetal and is expressed as:
(1)
Where:
- Period of rotation, measured in seconds.
- Radius of rotation, measured in meters.
If we know that 
and 
, then the time taken by the object to complete one revolution is:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".
Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.
Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
Answer:
B. changing shape and changing volume
Explanation:
*no definite shape (takes the shape of its container)
*no definite volume
