The first law of Newton’s law state an object in motion will stay in motion and an object at rest will stay at rest unless acted upon with an outside force. Other know as the law of inertia so yes it does
Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
Answer:
<u>Frequency</u>- number of wave cycles that occur in a given amount of time.
<u>Pitch</u>- number of wavelengths in a given amount of time.
<u>Amplitude</u>- fluctuation or displacement of a wave from its mean value. That means how high or low they are away from the center line.
<u>Volume</u>- The perception of loudness from the intensity of a sound wave. The higher the intensity of a sound, the louder it is perceived in our ears, and the higher volume it has.
<u>Wavelength</u>- the distance between the tops of the "waves".
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;

Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;

substituting;

Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s
Answer:
The critical stress required for the propagation of an initial crack
= 21.84 M pa
Explanation:
Given data
Modulus of elasticity E = 225 ×

Specific surface energy for magnesium oxide is
= 1 
Crack length (a) = 0.3 mm = 0.0003 m
Critical stress is given by
=
-------- (1)
⇒ 2 E
= 2 × 225 ×
× 1 = 450 ×
⇒
a = 3.14 × 0.0003 = 0.000942
⇒ Put these values in equation 1 we get
⇒
=
⇒
= 4.77 × 
⇒
= 2.184 ×

⇒
= 21.84 
⇒
= 21.84 M pa
This is the critical stress required for the propagation of an initial crack.