Answer:
The angular speed of the object is 0.0281 rad/s
The linear speed of the object is 0.169 ft/s
Explanation:
Given;
radius of the circle, r = 6 ft
time of motion of the object around the circle, t = 80 s
central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad
The angular speed of the object is calculated as;

The linear speed of the object is calculated as;
v = ωr
v = 0.0281 rad/s x 6ft
v = 0.169 ft/s
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Answer:
Explanation:
This is a circular motion questions
Where the oscillation is 27.3days
Given radius (r)=3.84×10^8m
Circular motion formulas
V=wr
a=v^2/r
w=θ/t
Now, the moon makes one complete oscillation for 27.3days
Then, one complete oscillation is 2πrad
Therefore, θ=2πrad
Then 27.3 days to secs
1day=24hrs
1hrs=3600sec
Therefore, 1day=24×3600secs
Now, 27.3days= 27.3×24×3600=2358720secs
t=2358720secs
Now,
w=θ/t
w=2π/2358720 rad/secs
Now,
V=wr
V=2π/2358720 ×3.84×10^8
V=1022.9m/s
Then,
a=v^2/r
a=1022.9^2/×3.84×10^8
a=0.0027m/s^2
Answer:
B.The box experiences less friction on the marble floor
Explanation:
Answer:
b. The current stays the same.
Explanation:
In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .
Now, when an identical bulb is connected in parallel to the original bulb .
Therefore, both the resistance( bulb) are in parallel.
We know, when two resistance are in parallel , current through them is same and voltage is divided between them.
Therefore, in this case current stays same in the original bulb.
Hence, this is the required solution.