Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
Answer:
solution given:
acceleration (a)=?
initial velocity (u)=3m/s
final velocity (v)=6m/s
distance (s)=90m
we have
v²=u²+2as
substituting value
6²=3²+2*a*90
36=9+180a
36-9=180a
a=25/180
<u>a=0.1388m/s²</u>
Answer:
Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples.
<span>A. three</span><span>
Oil spill can be very harmful to marine life because the chemical make up of oils can be poisonous to marine life. The oil can also affect the natural body temperature of marine animals especially to the small fish. Sea otters and sea birds are the most commonly affected by oil spills and those other marine animals that can be found in the shoreline. Heavy oils like the bunker oils used to fuel ships are the most harmful oil because when this oil stick to birds feathers, they may have an inability to warm themselves that could lead them to die.</span>
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
brainly.com/question/13754413
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