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gladu [14]
3 years ago
7

Rocks are elastic too! Explain how elasticity and Harry Reid's elastic rebound theory lead to a better understanding of earthqua

kes.
Physics
1 answer:
ANEK [815]3 years ago
7 0
After the great 1906 San Francisco earthquake, geolophysicistHarry Fielding Reid examined the displacement of the ground surface along the San Andreas Fault. He concluded that the quake must have been the result of the elastic reboundof the strain energy in the rocks on either side of the fault.
strain energy is 0. 5x force x (compression) X (compression)
There is a lot of force and a bit of compression when rocks squash up against other rocks causing earthquakes
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The work that a force does by acting on an object is equal to what?
maw [93]
2nd and only 2nd option is right
3 0
3 years ago
An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa
Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

f'=f\frac{v\pm v_o}{v\pm v_x}

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

v_x=\frac{f}{f'}(v+v_o)-v

on substituting the values, we get

v_x=\frac{1790}{1780}(343+2.36)-343

or

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0
3 years ago
A roller coaster car speed is 4 m/s, but 3 seconds later its speed is 22 m/s. What is its average acceleration?
Juli2301 [7.4K]
The velocity increased from 4 m/s to 22 m/s in 3 seconds. 18 m/s in 3 seconds so the average acceleration is change in velocity divided by time. 18 m/s divided by 3 seconds = 6 m/s^2
6 0
3 years ago
Which organelle is the powerhouse of the cell, the site of cellular respiration? A) 2 - nucleus B) 5 - endoplasmic reticulum C)
Paladinen [302]
D) 9- mitochondria
Hope this helps :D
7 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
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