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AysviL [449]
3 years ago
9

How does Earth's surface and the structures on the surface change as a result of an earthquake? Help me pls and I will give Brai

nliest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Physics
1 answer:
EastWind [94]3 years ago
6 0
<span>Earthquakes often cause dramatic changes at Earth's surface. In addition to the ground movements, other surface effects include changes in the flow of groundwater, landslides, and mudflows. Earthquakes can do significant damage to buildings, bridges, pipelines, railways, embankments, dams, and other <span>structures</span></span>
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A 521-kg meteor is subject to a force of 2520 N. What is its acceleration?
Vladimir79 [104]
Put it in the equation a=f/m
3 0
3 years ago
Read 2 more answers
A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati
kirill115 [55]

Answer:1.039 N

Explanation:

Given

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta is parallel to the tray =1.039 N

5\cos \theta is perpendicular to the tray =4.890 N

5 0
3 years ago
Is it possible to determine from the data in the graphs whether the sensor attached to cart a is actually plugged into ch a and
ladessa [460]
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4 0
3 years ago
Using Planck’s constant as h = 6.63 E-34 J*s, what is the wavelength of a proton with a speed of 5.00 E6 m/s? The mass of a prot
Marrrta [24]
De Broglie's identity gives the relationship between the momentum and the wavelength of a particle:
p=mv= \frac{h}{\lambda}
where
p is the particle momentum
m is its mass
v its velocity
h is the Planck constant
\lambda is the wavelength

By re-arranging the equation, we get
\lambda=  \frac{h}{mv}
and by using the data about the proton, given in the text, we can find the proton's wavelength:
\lambda= \frac{h}{mv} = \frac{6.63 \cdot 10^{-34} Js}{(1.66 \cdot 10^{-27} kg)(5.00 \cdot 10^6 m/s)} =7.99 \cdot 10^{-14} m
6 0
3 years ago
The flaming gorge bridge, in wyoming rises above a dry gulch. If you throw a rock straight out from the bridge, horizontally, an
Novosadov [1.4K]

Answer:

12.495m/s

Explanation:

Horizontal displacement is the range of the projectile motion.

The range is expressed as;

R = 2U/g

U is the speed at which the rock is thrown (initial speed)

g is the acceleration due to gravity.

Given

R = 255cm = 2.55m

g = 9.8m/s²

Required

Speed U

Substitute the given parameters into the formula as shown;

2.55 = 2U/9.8

Cross multiply

2U = 2.55×9.8

2U = 24.99

U = 24.99/2

U = 12.495m/s

Hence the speed that you thew the rock is 12.495m/s

7 0
2 years ago
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