Answer:
a 10.6 m/s²
Explanation:
Since F = ma (Force = mass * acceleration), acceleration would be...
a = F/m
a = 302 N/28.6
a 10.6 m/s²
Two objects with different masses accelerate at the same rate according to newton’s 2nd law,what must be true about the forces p
1 answer:
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Answer:
112 N
Explanation:
going through the question you would notice that some detail is missing, using search engines u was able to find a similar question on "https://socratic.org/questions/a-2-3-kg-steel-ball-strikes-a-wall-with-a-speed-of-8-5-m-s-at-an-angle-of-64-wit"
and here is the question i found
"A 2.3 kg steel ball strikes a wall with a speed of 8.5 m/s at an angle of 64⁰ with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.448 s. What is the average force exerted by the ball?"
you would notice that there is a change in the values from the question posted, hence we would only take the following part to complete our question, " If the ball is in contact with the wall for 0.448 s what is the average force exerted by the ball?" while retaining all original detail.
solution
mass of ball (m) = 2.1 kg
speed of ball (v) = 13.2 m/s
angle of contact (p) = 64.8°
time of contact (t) = 0.448 s
What is the average force exerted by the ball?
The average force exerted by the ball =
where
- The momentum changes only the direction perpendicular to the wall, hence the component of momentum perpendicular to the wall = m x v x sin (p) = 2.1 x 13.2 x sin 64.8 = 25.1 kg.m/s
since the ball strikes the wall and bounces off it with the same speed
and at the same angle, the component of momentum acting
perpendicular to the wall remains the same while hitting and leaving
the wall but in opposite directions.
Hence the component of momentum acting perpendicular to the wall while hitting and leaving the wall will be 25.1 kg.m/s and -25.1 kg.m/s respectively.
change in momentum = 25.1 - (-25.1) = 25.1 + 25.1 = 50.2 kg.m/s
- change in time = 0.448 s
- now substituting the above into the equation we have
The average force exerted by the ball = = 112 N
The frictional force required is 9000 N
Explanation:
In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.
Therefore, we can write:
where the term on the left is the frictional force while the term on the right is the centripetal force, and where:
m is the mass of the car
v is its speed
r is the radius of the curve
For the car in this turn, we have
m = 1000 kg
v = 30 m/s
(since the diameter is 0.20 km, the radius is half that value)
And substituting, we find
Learn more about friction:
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