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uranmaximum [27]
2 years ago
12

What processes were involved in the formation of the model? How is the model similar to the real sedimentary sample? How is the

model different from the real sedimentary sample?

Physics
1 answer:
Stels [109]2 years ago
8 0

Answer:

1. sediment layering and compacting on top of each other and solidifying

2. sediment layering and compacting (not as much) on top of each other and solidifying, just not as much

Explanation:

hope this helps! :))

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A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten
lions [1.4K]

Answer:

 h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

 100 = m g h

5 h = 100

 h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0
3 years ago
The color that an object appears to be depends on the select one:
ddd [48]
B is the right answer
3 0
3 years ago
Importance of liquid pressure
trapecia [35]

Answer:

hope it helps...

Explanation:

Both the water in the ocean and the air in the atmosphere exert pressure because of their moving particles. ... This causes greater pressure. Denser fluids such as water exert more pressure than less dense fluids such as air. The particles of denser fluids are closer together, so there are more collisions in a given area.

7 0
2 years ago
A business letter is informal correspondence and is sent to people within
jarptica [38.1K]

Answer:

FALSE!!!

Explanation:

Business letters are formal and normally given to someone of higher importance.

5 0
3 years ago
You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?
Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
4 0
3 years ago
Read 2 more answers
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