Answer:
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Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
Velocity = displacement/time
1.6 = 253/t
t = 158.125
It takes them about 158 seconds
Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector 
y component of the vector 
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector 
y component of the vector 
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j