Answer:
10.8 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Acceleration (a) = 5 m/s/s
Distance travelled (s) = 291 m
Time (t) taken =?
We can calculate the time taken for the car to cover the distance as follow:
s = ut + ½at²
291 = 0 × t + ½ × 5 × t²
291 = 0 + 2.5 × t²
291 = 2.5 × t²
Divide both side by 2.5
t² = 291 / 2.5
t² = 116.4
Take the square root of both side
t = √116.4
t = 10.8 s
Thus, it will take the car 10.8 s to cover the distance.
Answer: North
Explanation: I believe the friction will go the opposite way of the object being pushed.
The equal and opposite reaction to the gravitational force of the Earth pulling down on a man standing on the beach is the gravitational force of the man pulling up on the Earth. <em>(A)</em>
To put it in other words ... which should begin to blow your mind if you just take a moment to think about them ...
Your weight on Earth is exactly equal to the Earth's weight on YOU.
Answer:
1. v = 30 m/s
2. v = 5 m/s
3. f = 40 Hz
4. f = 400 Hz
5. f = 300 Hz
6. λ = 0.772 m
7. λ = 0.386 m
8. λ = 0.625 m
9. v = 100 m/s
10. v = 50 m/s
Explanation:
The relationship between frequency, wavelength, and speed of a wave is given by the following formula:

where,
v = speed of wave
f = frequency of wave
λ = wavelength
1.
f = 100 Hz
λ = 0.3 m
Therefore,
v = (100 Hz)(0.3 m)
<u>v = 30 m/s</u>
<u></u>
2.
f = 50 Hz
λ = 0.1 m
v = (50 Hz)(0.1 m)
<u>v = 5 m/s</u>
<u></u>
3.
v = 20 m/s
λ = 0.5 m

<u>f = 40 Hz</u>
<u></u>
4.
v = 80 m/s
λ = 0.2 m

<u>f = 400 Hz</u>
<u></u>
5.
v = 120 m/s
λ = 0.4 m

<u>f = 300 Hz</u>
<u></u>
6.
v = 340 m/s
f = 440 Hz

<u>λ = 0.772 m</u>
<u></u>
7.
v = 340 m/s
f = 880 Hz

<u>λ = 0.386 m</u>
<u></u>
<u></u>
8.
v = 250 m/s
f = 400 Hz

<u>λ = 0.625 m</u>
<u></u>
9.
f = 50 Hz
λ = 2 m
v = (50 Hz)(2 m)
<u>v = 100 m/s</u>
<u></u>
10.
f = 100 Hz
λ = 0.5 m
v = (100 Hz)(0.5 m)
<u>v = 50 m/s</u>
Physics can be used in a variety of ways, a specific example would be rollercoaster. Physics can be used to determine the speed at which a rollercoaster can travel without going off the rails or when it needs to begin to slow down to make a sharp turn.