Answer:
44.3 m/s
Explanation:
a) Draw a free body diagram of the mass M. There are three forces:
Weight force mg pulling down,
Normal force N pushing perpendicular to the ramp,
and tension force T pulling parallel up the ramp.
Sum of forces in the parallel direction:
∑F = ma
T − Mg sin 30° = 0
T = Mg sin 30°
T = Mg / 2
Draw a free body diagram of the hanging mass m. There are two forces:
Weight force mg pulling down,
and tension force T pulling up.
Sum of forces in the vertical direction:
∑F = ma
T − mg = 0
T = mg
Substitute:
mg = Mg / 2
m = M / 2
M = 2m
b) Velocity of a standing wave in a string is:
v = √(T / μ)
T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:
v = √(49 N / 0.025 kg/m)
v = 44.3 m/s
Answer:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Explanation:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Mass.
Because mass doesn't depend on weight but weight depends on mass.
Answer:
Amplitude = 0.058m
Frequency = 6.25Hz
Explanation:
Given
Amplitude (A) = 8.26 x 10-2 m
Frequency (f) = 4.42Hz
Conversation of energy before split
½mv² = ½KA²
Make A the subject of formula
A =
Conversation of energy after split
½(m/2)V'² = ½(m/2)V² = ½KA'²
½(m/2)V² = ½KA'²
Make A the subject of formula
First divide both sides by ½
(m/2)V² = KA'²
Divide both sides by K
V² = A'²
= A'
Substitute
for A in the above equation
A' = A/√2
A' = 8.26 x 10^-2/√2
A' = 0.05840702012600882
Amplitude after split = 0.058 (Approximated)
Frequency (f') = f√2
f' = 4.42√2
f' = 6.25082394568908011
Frequency after split = 6.25Hz (approximated)
Answer:
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