Three students derive the following equations in which x refers to distance traveled, v the speed, a the acceleration (m/s^2), a
1 answer:
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Answer:
The average acceleration of the bearings is
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy
Put the value into the formula
Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy
Put the value into the formula
We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration
Put the value into the formula
Hence,The average acceleration of the bearings is
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
Answer:
20 hertz of frequency produced.
Explanation:
Here we will find frequency and period should be in second, here given: 0.05 seconds
using the formula: