An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
1 answer:
You might be interested in
Answer:
The height of the water column = 1.62405 × 10⁻¹ m
Explanation:
The air cavity in the Coke bottle = 0.220 m deep
The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz
The given speed of sound in air, v = 343 m/s
We note that the air cavity in the coke bottle is equivalent to a tube closed at one end
The fundamental frequency for a tube closed at one end, 'f', is given as follows;
f = v/(4·L) = v/λ
Where;
L = The height of the water column
λ = The wavelength of the wave
∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496 m
∴ L = 0.6496 m/4 = 0.162405
m
The height of the water column = 1.62405 × 10⁻¹ m.
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
