Speed of wave is given as
Wavelength of the wave is given as
now from the formula of wave time period we can say
so it will have time period of T = 4 s
What distance does a police car travel if it is going 3.0 m/s for 20 seconds
1 answer:
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Answer:
The distance is 0.53 m.
Explanation:
Given that,
Target distance = 100.0 m
Speed of bullet = 300 m/s
We need to calculate the total time
Using formula of time
Put the value into the formula
Now, consider vertical motion of bullet.
Initial velocity of bullet in vertical direction = 0 m/s
We need to calculate the vertically distance
Using equation of motion
Put the value in the equation
Hence, The distance is 0.53 m.
Refer to the diagram shown below.
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).