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Dahasolnce [82]
3 years ago
13

A scientist observes debris added to a landform from a melting glacier. This is evidence for which type of natural process?

Physics
2 answers:
wlad13 [49]3 years ago
6 0

Answer:

Deposition

Explanation:

deposition is a physical process in which debris in the form of soil, rocks etc are added to some landform . This process is carried out by wind, water( river ) or ice- river (glacier). When their movement is stopped , they become unable

to bear their weight and deposit them on the bed.

Phantasy [73]3 years ago
5 0
It may be deposition
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This could be wrong but I am giving a decent suggestion: All ions are atoms with a negative charge.

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classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per
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Answer:

t_1 = 0.28 s

t_2 = 7.72 s

Explanation:

Given that height of the projectile as a function of time is

h = -16 t^2 + 128 t + 112

here we know that

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so from above equation

147 = -16 t^2 + 128 t + 112

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now by solving above quadratic equation we know that

t_1 = 0.28 s

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2 years ago
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
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...do

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24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

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3 years ago
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

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In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

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Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

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Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

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