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erica [24]
3 years ago
8

What are the five events that can trigger a mass movement

Physics
1 answer:
den301095 [7]3 years ago
8 0
Wind, water, air, rain, hot water 

hope it helpes!!!


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A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size
Olenka [21]

Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10  cm

Mass of ball\left ( m\right )=0.05kg

time of contact\left ( t\right )=34.3\times 10^{-3} s

Now we know impulse =Force\times time\ of\ contact=Change in momentum

F_{average}\times t=m\left ( v-(-v)\right )

and velocity at the bottom is given by

v=\sqrt{2gh}

F_{average}\times 34.3\times 10^{-3}=0.05\left ( 1.4-(-1.4)\right )

F_{average}=4.081N

5 0
3 years ago
Please help ASAP I have to turn this in soon
Allisa [31]

Answer:

See in explanation

Explanation:

Scientific use: The Einstein's THEORY of relativity states that "Time Is Absolute".

Everyday use: Einstein's LAW of relativity says that time is not the same at all places and events.

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3 years ago
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An expert archer tried to shoot an arrow into a target lying at the bottom of a shallow pool. Even though the water did not chan
frutty [35]
The interaction that caused the archer to miss was probably refraction.
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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. the two protons thenmov
Kay [80]

Because the two paths are perpendicular, therefore the target proton's new path must be at 30 degrees from the original direction. 

Using the law of conservation of momentum about the original direction: 
m (400 m/s) = m (v1) cos(60) + m (v2) cos(30) 
Cancelling m since the two protons have similar mass.
(v1)cos(60) + (v2)cos(30) = 500 m/s                         ---> 1

Now by using the law conservation of momentum perpendicular to the original direction: 
m (0 m/s) = m (v1) sin(60) – m (v2) sin(30) 
Which simplifies to:
(v1)sin(60) - (v2)sin(30) = 0 m/s                                
v2 = v1 * sin(60) / sin(30) = v1 * sqrt(3)                  ---> 2

Plugging equation 2 to equation 1: 
(v1) (1/2) + (v1 * sqrt(3)) sqrt(3)/2 = 500 m/s 
(1/2) (v1) + (3/2) (v1) = 500 m/s 
2 (v1) = 500 m/s 
v1 = 250 m/s 

Thus, from equation 2:

v2 = v1*sqrt(3) = (250 m/s) sqrt(3) = 433.01 m/s 


So,
A. The target proton's speed is about 433 m/s 
B. The projectile proton's speed is 250 m/s

8 0
4 years ago
PLEASE HELP!!! THANKS I GIVE BRAINLIEST !!A student examines the effect of the number of D batteries in a closed circuit on the
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Answer:

The batteries

Explanation:

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