Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that force on the passenger while moving in circle is given as

now variation in force is given as

here speed is constant
Part b)
Now if the variation in force is required such that r is constant then we will have

so we have

Part c)
As we know that time period of the circular motion is given as

so here if radius is constant then variation in time period is given as

Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer:
the correct answer is C
Explanation:
When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.
Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.
Applying this to our case if there is
10 boulder on the chart
in reality there are #_boulder = 10 30
#_boulder = 300 boulder
so the correct answer is C
Weight. Because there is less gravity on the moon.