A motorcycle moving at a constant velcoity suddenly accelerates at a rate of 4.0 m/s/s to a speed of 35 m/s in 5.0 s. What was t
1 answer:
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Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
The average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds = 63.84 ft/s
(ii) 0.001 seconds = 63.984 ft/s
Given that a ball is thrown with an initial velocity = 80 ft/s
Let 'y' be the height in feet after 't' seconds.
Given, gives the height in 't' seconds.
Average velocity = Rate of change of distance
= Change in distance/Change in time.
The initial time can be taken as 0 s.
When t =1 s, y = 80 - 16 = 64 ft
(1) t = 0.01 s
y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft
Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s
(2) t = 0.001 s
y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft
Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s
The question is incomplete. Find out the complete question below:
If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by .Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds
(ii) 0.001 seconds
Learn more about average velocity at brainly.com/question/6504879
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