Answer:
313.92w
Explanation:
Formula for power:
P=W/∆t = Fv
Givens:
m=20kg
∆y=4.0m
∆t=2.5s
a=9.81m/s²
In order to find power, we first need to solve for work.
W=Fd (force*displacement), f=mg
W=mg∆y
W=(20kg)(9.81m/s²)(4.0m)
W=784.8J
P=W/∆t
P=784.8J/2.5s
P=313.92 watts
<em>Answer: </em>tellurium (Te)
<em>atomic number = 52 ,</em>
<em>Number of energy levels = 5;</em>
First energy level = 2
Second energy level = 8
Third energy level = 18
Fourth energy level = 18
Fifth energy level = 6
<em>In this electron configuration, 0uter most electrons are 6.</em>
If the applied force is in the same direction as the object's displacement, the work done on the object is:
W = Fd
W = work, F = force, d = displacement
Given values:
F = 45N
d = 12m
Plug in and solve for W:
W = 45(12)
W = 540J
Answer:
θ= 5 radian
Explanation:
Given data:
Radius r = 0.70 m
Initial angular speed ω_i = 2rev/s
Time t = 5 s
Final angular speed ω_f =0
so we have angular displacement
putting values
= 5 rad
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