Answer:
Solar eclipses result from the Moon blocking the Sun relative to the Earth; thus Earth, Moon and Sun all lie on a line. Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.
So, the time needed before you hear the splash is approximately <u>2.06 s</u>.
<h3>Introduction</h3>
Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:
With the following condition :
- t = interval of the time (s)
- h = height or any other displacement at vertical line (m)
- g = acceleration of the gravity (m/s²)
Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :
With the following condition :
- t = interval of the time (s)
- s = shift or displacement (m)
- v = velocity (m/s)
<h3>Problem Solving</h3>
We know that :
- h = height or any other displacement at vertical line = 19.6 m
- g = acceleration of the gravity = 9.8 m/s²
- v = velocity = 343 m/s
What was asked :
- = ... s
Step by step :
- Find the time when the object falls freely until it hits the water. Save value as
- Find the time when the sound propagate through air. Save value as
- Find the total time
<h3>Conclusion</h3>
So, the time needed before you hear the splash is approximately 2.06 s.
The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?
Answer:
Answer is 20
Explanation:
As we know
\Large\boxed{ \tt{}F =ma}
F=ma
Here Given
★ F = 100
★ m = 25 kg
Putting in the equation
\leadsto\tt{100=25\times\:a}⇝100=25×a
On solving
\sf{\dashrightarrow\:a = 4 \:ms^{-2}}⇢a=4ms
−2
Now the body initially is stationary therefore initial velocity (u) must be zero
u = 0
Applying Newton 1st Law of Motion :
\sf\Large\boxed{ \tt{}v =u + at}
v=u+at
Substituting the values to find final velocity at t= 5 sec
\leadsto \tt{v = 0 + 4 \times \:5}⇝v=0+4×5
\tt{ \pink{\dashrightarrow\:v = 20 \:m/s}}⇢v=20m/s
☞ Hence, the final velocity is 20 m/s