Answer:
Option C: Third Class
Explanation:
This is third class because the effort or the input force is in the middle between the fulcrum and the load.
Answer:
48.26 m
Explanation:
time to goes up (till stop for a while in the air - maximum height)
vt = vo + a t
0 = 15 + g . t
0 = 15 + (-9.8) . t
9.8t = 15
t = 1.531 s
so the time left to goes down is
4.0 - 1.531 = 2.469 s
height from the top of building can find it by using
vo =√(2gh)
15 = √(2)(9.8).h
15² = 19.6h
h = 225/19.6 = 11.48 m
so the distance of maximum height to the ground is
t = √(2H/g)
2.469 = √(2H/9.8)
2.469² = 2H/9.8
6.096 = 2H/9.8
2H = 6.096 x 9.8 = 59.74 m
so the vertical distance of the building (or the building height's is)
H - h = 59.74 - 11.48 = 48.26 m
Answer:
a) 6.26 m/s
b) 7.67 m/s
Explanation:
The potential energy at height h0 is initially ...
PE0 = mgh0
At height h1, the potential energy is ...
PE1 = mgh1
The difference in potential energy is converted to kinetic energy:
PE0 -PE1 = KE1 = (1/2)m(v1)^2
Solving for v1, we have ...
mg(h0 -h1) = (1/2)m(v1)^2
2g(h0 -h1) = (v1)^2
v1 = √(2g(h0 -h1))
__
a) When the body is 1 m high, its speed is ...
v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high
__
b) When the body is 0 m high, its speed is ...
v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground
Since the equation for frequency is:
frequency = velocity/wavelength, then the correct answer is B. velocity.
In order to calculate the velocity, we just need to multiply frequency and wavelength.
Answer:
The height of the wave is determined by the wind strength and fetch.
Explanation:
The height of the wave is determined by the wind strength and fetch.
The more the strength and the more the fetch size the more will be the height of the wave.
Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.
