Answer:
D Electromagnetic and gravitational
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
- Pressure = 5000 Pa
- Contact Area = 0.04 m^2
- Acceleration due to gravity = 9.8 m/s^2
- Let the force be F.
- We know, Force = Pressure × Contact Area
- Therefore, Force = 5000 Pa × 0.04 m/s^2
- or, Force = 200 N
- We know, force = mass × acceleration
- Therefore, mass = force ÷ acceleration
- or, mass = 200 N ÷ 9.8 m/s^2 = 20.4 Kg
<u>Answer</u><u>:</u>
<u>2</u><u>0</u><u>.</u><u>4</u><u> </u><u>Kg</u>
Hope you could understand.
If you have any query, feel free to ask.
Here is the full question:
The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Answer:
a) 0.85 m
b) 0.98 m
c) 0.76 m
Explanation:
Given that: the radius of gyration
So, moment of rotational inertia (I) of a cylinder about it axis = 





k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = 




k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = 




k = 0.7560
k ≅ 0.76 m