Answer:
72.53 mi/hr
Explanation:
From the question given above, the following data were obtained:
Vertical distance i.e Height (h) = 8.26 m
Horizontal distance (s) = 42.1 m
Horizontal velocity (u) =?
Next, we shall determine the time taken for the car to get to the ground.
This can be obtained as follow:
Height (h) = 8.26 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
8.26 = ½ × 9.8 × t²
8.26 = 4.9 × t²
Divide both side by 4.9
t² = 8.26 / 4.9
Take the square root of both side by
t = √(8.26 / 4.9)
t = 1.3 s
Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:
Horizontal distance (s) = 42.1 m
Time (t) = 1.3 s
Horizontal velocity (u) =?
s = ut
42.1 = u × 1.3
Divide both side by 1.3
u = 42.1 / 1.3
u = 32.38 m/s
Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:
1 m/s = 2.24 mi/hr
Therefore,
32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s
32.38 m/s = 72.53 mi/hr
Thus, the car was moving at a speed of
72.53 mi/hr.
Answer:
7. free fall -- h. 9.8m/s^2
3. Velocity -- x. 60 km/hr west
6. Acceleration -- d. change in velocity/time
8. Centrifugal -- s. towards the centre
13. Work done --w. Force * displacement
5. Uniform circular motion --j. spin cycle in washer
18. Power -- r. kW an hour
7. g -- a. 10N
hope this helps
<span>The size of the ocean, along with the large heat capacity of water,
allows the ocean to store large amounts of heat energy.</span>
Answer:
Net force, F = 7 N
Explanation:
It is given that,
Initial kinetic energy of the car, 
Final kinetic energy of the car, 
Distance, d = 1 m
We need to find the average net force on the car during this interval. It is given by using the work energy theorem as :
Also, W = F.d d = distance and F = net force
F = 7 Newton
So, the average net force on the car during this interval is 7 newton. Hence, this is the required solution.
Answer:
period in case 2 is 
times the period in case 1
Explanation:
The period of oscillation of a spring is given by:
where
m is the mass hanging on the spring
k is the spring constant
Therefore, in order to compare the period of the two springs, we need to find their m/k ratio.
We know that when a mass hang on a spring, the weight of the mass corresponds to the elastic force that stretches the spring by a certain amplitude A:
So we find
The problem tells us that the amplitude of case 1 is d, while the amplitude in case 2 is 2d. So we can write:
- for case 1:
- for case 2:
And by comparing the two periods, we find:
So, the period of oscillation in case 2 is times the period of oscillation in case 1.
