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3 years ago

7Which of the following terms describes how glaciers move?

Physics

2 answers:

tatuchka [14]3 years ago

7 0

Answer:

D is the answer I think (0 w 0 )

Explanation:

ollegr [7]3 years ago

5 0

Your answer should me D

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As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

rusak2 [61]

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; $\mathbf{W_{bottom} = F_{adhesion}}$

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = $F_{adhesion}$

Weight of the top block = $W_{top}$

Weight of the bottom block = $W_{bottom}$

Given that with the exertion of the tension, the two blocks remain at rest, we have;

T = $W_{top}$ + $W_{bottom}$

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

$W_{bottom}$ = $F_{adhesion}$

Learn more here:

brainly.com/question/18907970

8 0

3 years ago

As a person squeezes and applies pressure to the outside of a balloon air particles inside

MAVERICK [17]

They will occupy less volume which will cause an increased density of the particles.

3 0

4 years ago

A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav

uysha [10]

In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0

3 years ago

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0

3 years ago

Read 2 more answers

If a car was moving at a speed of 25mph for 6 hours, how far would the car travel?

Anestetic [448]

Answer:

150 miles

Explanation:

25mph x 6hrs = 150miles

7 0

3 years ago