Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force
by the worker must be equal to the friction force
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2

b. The work is done on the crate by this force is the product of its force
and the distance traveled s = 4.35

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

Answer:
Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.
Answer:
Energy, E = 178.36 J
Explanation:
It is given that,
Mass 1, 
Mass 2, 
Mass 3, 
Height from which they are dropped, h = 1.3 m
Let m is the energy used by the clock in a week. The energy is equal to the gravitational potential energy. It is given by :


E = 178.36 J
So, the energy used by the clock in a week is 178.36 Joules. Hence, this is the required solution.
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector

.
The positive y-axis = the northern direction, with unit vector

.
The airplane flies at 340 km/h at 12° east of north. Its velocity vector is

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
![\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B2%7D%20%3D40%28cos%2834%5E%7Bo%7D%29%5Chat%7Bi%7D%20-%20sin%2824%5E%7Bo%7D%29%5D%5Chat%7Bj%7D%29%20%3D%2033.1615%5Chat%7Bi%7D%20-22.3677%5Chat%7Bj%7D%29)
The plane's actual velocity is the vector sum of the two velocities. It is

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h
The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°
Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.