The environment between the outer and inner nuclear membranes is most accurately referred to as the:
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:Given mass = 4.1kg
Radius = 0.0117m
Velocity V = 8.4m/s
Coefficient of friction = 0.25
Explanation: Below is an attached solution to the problem stated above.
1. The angular acceleration is equal to 524rad/s^2
2. The linear acceleration is equal to 2.45m/s^2
3.the time it takes the ball to begin rolling = 0.98s
4. The distance the ball slides before it begins to roll = 7.05m
