The speed will be 0.2 m/s and the velocity will be 0 m/s.
Speed = Total Distance / Total time
We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.
Speed = 145/ 720 m/s
Speed = 0.2 m/s
Velocity = Total Displacement / Total time
As the initial and final is the home, hence the net displacement is 0 in that case.
In this case also the total time we have given is 12 minutes or we can say 720 seconds.
Velocity = 0 / 720 m/s
Velocity = 0 m/s
So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.
Learn more about velocity here:
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The (more) speed an object has, the (lower) the potential energy and the (higher) the kinetic energy. (I believe that is correct but it’s been a while since I’ve done this)
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
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