Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
mediums only, mechanical waves always propagate through medium
2 oxygen atoms are used to bond with sulfer
Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that


As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N


We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W

Now solve for v2


= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,

Now solve for Us


= 0.00494
work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.
T = tension force in the rope = 115 N
d = displacement of the crate = 7.0 m
θ = angle between the direction of tension force and displacement = 37 deg
work done on the crate is given as
W = F d Cosθ
inserting the values given above
W = (115) (7.0) Cos37
W = 643 J