Uk=2m*the normal force
the distance d is needed only if you were asked about work
Depends on if the samples are different kinds of sand
Answer:
The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.
Explanation:
Hope this helps Plz mark brainliest
The relationship between the impulse and the momentum allows to find the result for the question about the contact time is:
-
The contact time on the floor is t = 0.60 s
The impulse is the push force on a body and is related to the variation of the amount of momentum.
I = ∫ F. dt = Δp
p = mv
where I is the momentum, F the force, t the time, p the amount of momentum and m the mass .
They indicate the average force F = 330 N, the jump height is y = 12.0 cm = 0.120 m.
F. t = m v_f - m v₀
Let's search with kinematics for the initial speed of the jump.
v² = v₀² - 2 g y
At the highest point the velocity is zero.
0 = v₀² - 2ay
Let's calculate.
v₀ = 1.53 m / s
Suppose that the jumps are elastic, that is, that the speed with which it reaches the ground is equal to the speed with which it leaves, but in the opposite direction.
Let's calculate.
t = 0.60 s
In conclusion using the relationship between impulse and momentum we can find the result for the question about the contact time is:
The contact time on the floor is t = 0.60 s
Learn more about impulse here: brainly.com/question/904448
Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=