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amid [387]
3 years ago
9

how did the ice age affect the landscape of snowdonia? explain whether this process happened over short or a long period of time

Physics
2 answers:
nignag [31]3 years ago
6 0

Answer:

It took over a long period of time to form the ice age. During the last Ice age, a thick sheet of the ice covered the landscape of Snowdonia.

The ice sheets shifted downwards and the rock from the back of the valley deposited the material again because the glaciers melted.

The main influence has been the climate and human actions in Snowdonia. The ice age formed around 450 million years ago over the Ordovician period.

PSYCHO15rus [73]3 years ago
5 0

Answer:

Glaciers formed during the ice age eroded Earth’s surface as they moved. This process resulted in the formation of a U-shaped valley. The process took place over a long time. The valleys were revealed when the glaciers thawed after the ice age ended.

Explanation:

PLATO ANSWER

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A CAR ACCElerates FROM REST To
Ilia_Sergeevich [38]

Answer:

<u>666.6 kW</u>

Explanation:

<u>Power Formula</u>

  • Power = Force × Velocity

<u>Calculating Force</u>

  • F = ma
  • F = m(v - u / t) [From the equation v = u + at]
  • F = 2,000 (50/7.5)
  • F = 2,000 (20/3)
  • F = 40,000/3
  • F = 13333.3 N

<u>Solving for Power</u>

  • P = 13333.3 × 50
  • P = 666666.6 W
  • P = <u>666.6 kW</u>
5 0
2 years ago
A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .
swat32

From the calculations, the power expended is 43650 W.

<h3>What is the power expended?</h3>

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v =  95 km/h or 26.4 m/s

t =  6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work =  3298 N * 90 m = 296820 J

Power =  296820 J/ 6.8 s

= 43650 W

Learn more about power expended:brainly.com/question/11579192

#SPJ1

5 0
2 years ago
A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present
Varvara68 [4.7K]

Answer:

a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.

b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

Explanation:

4 0
3 years ago
Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say
True [87]

Answer:

a. Ssystem  > 40 J/K

Explanation:

Given that

The entropy of first block = 10 J/K

The entropy of second block = 30 J/K

When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40  J/K.

Therefore the answer is a.

Ssystem  > 40 J/K

6 0
3 years ago
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio
lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

W = F\times d

KE = 0.5\times m\times v^2

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

W = F \times d = 2.62 N \times 100 m

W = 261.6 N\times m

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

0.5 m\times v2^2 = 0.5 m\ v1^2 - W

Now solve for v2

v2 = \sqrt{v1^2 - {\frac{2W}{M}}}

= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Ff = Us\times N

Now solve for Us

= \frac{Ff}{N}

= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}

= 0.00494

4 0
3 years ago
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