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: 16N
Explanation:
Given that:
mass of box M= 2 kg
Initial speed V1 = 4 m/s
Final speed V2 = 8 m/s
Time taken T= 0.5 s
Average strength of this force F = ?
Now, recall that Force is the rate of change of momentum per unit time
i.e Force = momentum / time
Hence, F = M x (V2 - V1)/T
F = 2kg x (8 m/s - 4 m/s) / 0.5s
F = 2kg x (4 m/s / 0.5s)
F = 2kg x 8 m/s/s)
F = 16N
Thus, the average strength of this
force is 16 newton.
Answer:
The correct option is C
Explanation:
According to third equation of motion, v
2
=u
2
+2ax
Here, u=0 m/s
a=−g and x=−h
Negative sign indicates downward direction. Displacement and acceleration both are downwards.
So,v=±
2(−g)(−h)
We take minus sign because it is downwards.
v=−
2gh
After bouncing. velocity becomes 80% of v, i.e.,
v
′
=+0.8
2gh
(positive sign because the direction of ball has reversed after bouncing and is upwards.
Applying third equation of motion again, for u=v
′
, v=0 and a=−g
v
2
=u
2
+2×a×x
Thus,
0=0.64(2gh)+2(−g)x
or
x=0.64h