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Brut [27]
4 years ago
5

When a glacier moves, rocks and sediments in the bottom of the glacier scrape against the ground. This creates grooves known as

Physics
2 answers:
Leya [2.2K]4 years ago
8 0

Answer: D. glacial striations

Explanation:

Glaciers striations are the features formed over the bedrock due to glacier erosion. This occurs due to glacier abrasion that happens during glacier erosion or retreating glacier. This results in the formation of striations into multiple straight parallel lines and scrape marks over the bedrock or ground. These striations appear as grooves.

nikdorinn [45]4 years ago
5 0

Answer:

D. Glacial striations

Explanation:

Glacial striations are marks and grooves carved on a rock surface as the materials carried by it abraded the surfaces of the earth.

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7) You think you have found a diamond. Its mass is 5.28 g and its volume is 2 cm3. Calculate the density
lesya [120]

Answer:

\boxed {\tt d=2.64 \ g/cm^3}

\boxed {\tt Not \ a \ diamond}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass is 5.28 grams and the volume is 2 cubic centimeters.

m=5.28 \ g\\v= 2 \ cm^3

Substitute the values into the formula.

d=\frac{5.28 \ g }{2 \ cm^3}

Divide.

d=2.64 \ g/c^3

The density of the unknown substance is 2.64 grams per cubic centimeter.

The density of a diamond is about 3.5 grams per cubic centimeter. Since 2.64 is not equal to 3.5, the unknown substance is not a diamond.

6 0
4 years ago
What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?
Wewaii [24]

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

3 0
4 years ago
The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m
lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

8 0
3 years ago
Which instrument is used in submarine to see the objects above sea level ?
svetoff [14.1K]
Periscope i believe but i may be wrong.
8 0
3 years ago
Read 2 more answers
A golfer, driving a golf ball off the tee, gives the ball a velocity of 38 m/sec. The mass of the ball is 0.045 kg, and the dura
atroni [7]

Answer:

\Delta p=1.71\frac{kg\cdot m}{s}

Explanation:

The momentum of a body is defined as the product of its mass and its velocity at a given time. Therefore the change in the momentum of the ball is given by the difference between the final momentum and the initial momentum:

\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)\\\Delta p=0.045kg(38\frac{m}{s}-0\frac{m}{s})\\\Delta p=1.71\frac{kg\cdot m}{s}

6 0
4 years ago
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