The acceleration of the box as it slides down the ramp is 4.9 m/s².
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What is the acceleration of the box down the incline?</h3>
The acceleration of the box down the incline is determined by applying Newton's second law of motion as shown below;
F - Ff = ma
where;
- F is the parallel force on the box
- Ff is the frictional force on box = 0
- m is the mass of the box
- a is the acceleration of the box
F - 0 = ma
F = ma
mg sinθ = ma
g sinθ = a
where;
- g is acceleration due to gravity
- θ is the angle of inclination of the incline
a = 9.8 x sin(30)
a = 4.9 m/s²
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Answer:
Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height
Explanation:
As we know that initially ball is held at height h = 40 cm
So here we can say that kinetic energy of the ball is zero and potential energy is given as
now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies
So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height
While if we assume that there is no energy loss during collision then in that case ball will reach to same height again
Answer:
129.96
At a Gravitational acceleration of 32.17405 ( which is the normal rate for a freefall) you will geta velocity of 129.96 and the time of fall will be 4.039 seconds from 80 meters.
Why is the weight of a free falling body zero? It is not, an object in free fall will still have a weight, governed by the equation W = mg, where W is the object's weight, m is the object's mass, and g is acceleration due to gravity. Weight, however, has no effect on an objects free falling speed, two identically shaped objects weighing a different amount will hit the ground at the same time.
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the more pressure put on the string, the more frequency and higher pitch.
Answer:
(a) 1.054 m/s²
(b) 1.404 m/s²
Explanation:
0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ)) - μk·m·g·(1 - sin(θ)) = m·a
Which gives;
0.5·g·cos(θ) - μ·g·(1 - sin(θ) = a
Where:
m = Mass of the of the block
μ = Coefficient of friction
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block
θ = Angle of elevation of the block = 20°
Therefore;
0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°) - μk×9.81×(1 - sin(20°) = a
(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;
0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°) - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²
(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;
0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°) - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².
