In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T|=0.55N, and the ang
1 answer:
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Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
Answer:
Explanation:
The equivalent of Newton's second law for rotational motions is:
where
is the net torque applied to the object
I is the moment of inertia
is the angular acceleration
In this problem we have:
(net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
is the moment of inertia
Solving for , we find the angular acceleration:
Answer:
P = I²r
Explanation:
ε= IR + Ir
where r is the internal resistance
