A hanging mass of 23 kg will exert a force of:
23 * 9.81 = 225.6 Newtons on the string
Thus, during rotation, the force exerted should remain less than this. Force on an object in circular motion is given by:
F = (m * v^2) / r
225.6 = (2.95 * v^2) / 0.791
v = sqrt(60.491)
v = 7.78 m/s
Thus, the magnitude of the velocity should not exceed 7.78 meters per second.
Answer:
Conduction, radiation and convection all play a role in moving heat between Earth's surface and the atmosphere. Since air is a poor conductor, most energy transfer by conduction occurs right near Earth's surface
Answer:
9. (B) ¼ Mv²
10. (A) √(3gL)
11. 20 N
12. 5 m/s²
Explanation:
9. The rotational kinetic energy is:
RE = ½ Iω²
RE = ½ (½ MR²) (v/R)²
RE = ¼ Mv²
10. Energy is conserved.
Initial potential energy = rotational energy
mgh = ½ Iω²
Mg(L/2) = ½ (⅓ ML²) ω²
g(L/2) = ½ (⅓ L²) ω²
gL = ⅓ L² ω²
g = ⅓ L ω²
ω² = 3g / L
ω = √(3g / L)
The velocity of the top end is:
v = ωL
v = √(3gL)
11. Sum of torques about the hinge:
∑τ = Iα
-(Mg) (L/2) + (T) (r) = 0
T = MgL / (2r)
T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)
T = 20 N
12. Sum of forces on the block in the -y direction:
∑F = ma
mg − T = ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a / R)
T = ½ Ma
Substitute:
mg − ½ Ma = ma
mg = (m + ½ M) a
a = mg / (m + ½ M)
Plug in values:
a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))
a = 5 m/s²
The answer is d) 20 m.
The depth of pond is 20 m if it seems to be 10 m.
The formula relating real depth and app. depth :
In this case, refractive index has to be an whole number for real depth to give a whole number for app. depth.
- Then, real depth must be a multiple of 10, but refractive index cannot be 1
- The option is 20 m
F = ma
double f and m gives
(2f) = (2m)a
notice a is unchanged = 2.2
m/s^2