Answer:
a. 25000 J
b. 2500 J/s
Explanation:
Given,
Distance ( s ) = 50 m
Force ( f ) = 500 N
a.
To find : -
Work done ( W ) = ?
Formula : -
W = fs
W
= 500 x 50
= 25000 J
Therefore,
the work done by the force the horse exerts is
25000 J.
b.
To find : -
Power ( P ) = ?
Formula : -
W = Pt
P = W / t
P
= 25000 / 10
= 2500 J/s
Therefore,
the power produced if the movement took 10 s
is 2500 J/s.
<span>It bounces back toward a single spot. This is how the focusing mirrors on headlights work</span>
Explanation:
We have,
Spring constant of the spring, k = 165 N/m
Mass, m = 2 kg
It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

The frequency of vibration is reciprocal of its time period. So,

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²