Answer:
6692J
Explanation:
Power is defined as the rate at which work is being done.
So,
Power =
Work done = Power x time
Given parameters:
Power = 478watts
Time = 14s
So;
Work done = 478 x 14 = 6692J
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
Answer:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

Solving for the force F, we obtain that:

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

In this case, the force from the rocket's thrusters is equal to 118,000N.
For this case, the first thing you should know is that the length of a football field is around 100 meters.
We must then look for a measure close to this value.
We have the following unit conversion:
1 meter = 10 decimeters
Applying the conversion we have:

Therefore, the measure closest to a soccer field is:
1000 dm
Answer:
The length of a football field is closest to:
(2) 1000 dm