Answer:
Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
At its maximum height, the ball will have zero vertical velocity, so the ball's velocity at this point is exactly equal to its horizontal velocity.
At any time <em>t</em>, the horizontal component of its velocity is
<em>v</em> = (15 m/s) cos(40°) ≈ 11.49 m/s
so at the highest point of its trajectory, the ball has a velocity of about 11.49 m/s pointed in the positive horizontal direction.
Answer:
The induced voltage in the Secondary is 18 volt.
Explanation:
Given that,
Voltage = 120 volt
Number of turns in primary = 500
Number of turns in secondary = 75
We need to calculate the induced voltage in the Secondary
Using relation number of turns and voltage in primary and secondary
Where, 
= Number of primary coil
= Number of secondary coil
= Voltage of primary coil
= Voltage of primary coil
Put the value into the formula
Hence, The induced voltage in the Secondary is 18 volt.
Answer:
6.86 meters
Explanation:
Let the compression of the string be represented by x, and the height of projection of the toy rocket be represented by h.
So that;
x = 9 cm = 0.09 m
In its rest position (i.e before the launch), the spring has a stored potential energy which is given as;
PE = 
K
= x 830 x 
= 415 x 0.0081
= 3.3615
The potential energy in the string = 3.36 Joules
Also,
PE = mgh
where: m is the mass, g is the gravitational force and h the height.
m = 50 g = 0.05 kg, g = 9.8 m
Thus,
PE = 0.05 x 9.8 x h
3.3615 = 0.05 x 9.8 x h
3.3615 = 0.49h
⇒ h = 
= 6.8602
The height of the toy rocket would be 6.86 meters.
