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ratelena [41]
3 years ago
7

What are glacial grooves?

Physics
2 answers:
KonstantinChe [14]3 years ago
5 0
<span>Glacial striations are scratches or gouges cut into bedrock by glacial abrasion.</span>
Lina20 [59]3 years ago
3 0
Glacial striations are scratches or gouges cut into bedrock by glacial abrasion. These scratches and gouges were first recognized as the result of a moving glacier in the late 18th century when Swiss alpinists first associated them with moving glaciers. They also noted that if they were visible today that the glaciers must also be receding.[1]

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Entific findings are useful
Kaylis [27]

That's true.  The give-away is "observation ... in the real world".  That's what "field reports" are.

The part I'm not so sure about is those "entific findings".

4 0
3 years ago
A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m ​ 2
asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s

4 0
3 years ago
4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the
vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

t = 4.0s

u = 0m/s

v = 0 + (-9.8)(4)

v = 0 + (-9.8)(4)

v = -39.2m/s

Hence the speed of the ball before release is -39.2m/s

6 0
3 years ago
A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.
aleksley [76]
Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
                            d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
                          d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters. 
3 0
3 years ago
Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto
kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

  • The force of tugboat A, F_A = 3000 N, acting in a certain direction
  • The force of tugboat B, F_B = 5000 N, also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

F=F_A+F_B = 3000 + 5000 = 8000 N

and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

#LearnwithBrainly

5 0
3 years ago
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