Answer:
The biggest factor affecting coastal erosion is the strength of the waves breaking along the coastline. A wave's strength is controlled by its fetch and the wind speed. Longer fetches & stronger winds create bigger, more powerful waves that have more erosive power.
Explanation:
hope it helps !
Answer:
The acceleration of the earth is 7.05 * 10^-25 m/s²
Explanation:
<u>Step 1:</u> Data given
mass of the apple = 0.43 kg
acceleration = 9.8 m/s²
mass of earth = 5.98 * 10 ^24 kg
<u>Step 2:</u> Calculate the acceleration of the earth
Following the third law of Newton F = m*a
F(apple) = F(earth) = m(apple)*a(apple)
F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N
a(earth) = F(apple/earth)/m(earth)
a(earth) = 4.214N /5.98 * 10 ^24 kg
a(earth) = 7.05 * 10^-25 m/s²
The acceleration of the earth is 7.05 * 10^-25 m/s²
The two will fall at the same speed and reach the surface at the same time. This is because the two will experience the same gravitational acceleration on the moon. However, on the earth surface the two will land on the surface of the earth at the same time due to air resistance such that the egg will experience a higher air resistance than the hammer. On, the moon, where there is no noticeable atmosphere there is no air resistance on either object and both fall at the same speed. It is also important to note that their mass doesn't affect their speed.
Answer:
B. 47-54 miles/hour
Explanation:
Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.
On the basis of force of wind gale is divided in four groups:
- Near gale - 32-38 mph
- Gale - 39-46 mph
- Strong gale - 47-54 mph
- Storm - 55-63 mph
Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

Substituting the numerical values, we find
