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3 years ago

What is the most common feature on the far side of the Moon?

Physics

2 answers:

Bond [772]3 years ago

6 0

Answer:

The far side of the moon is the hemisphere of the moon that always faces away from Earth. The far sides terrain is rugged with a multitude of impact craters and relativity few flat lunar Maria compared to the nera side. It is one of the largest craters in the solar system, the south pole- Aitken basin

Sati [7]3 years ago

3 0

The far side of the Moon is the hemisphere of the Moon that always faces away from Earth. The far side's terrain is rugged with a multitude of impact craters and relatively few flat lunar maria compared to the near side. It has one of the largest craters in the Solar System, the South Pole–Aitken basin.

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Please help me quickly!!!

vichka [17]

Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.

Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.

Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.

6 0

3 years ago

A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

Explanation:

The centripetal acceleration is given by:

$a_{c} = \frac{v^{2}}{r}$

Where:

$v^{2}$ : is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope

Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir to the work input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0

3 years ago

The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid

timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0

3 years ago

When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm .

vlada-n [284]

Answer:

m = 56.5 kg

Explanation:

Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m

Δm*g / Area = p * g * Δh ..... Eq1

$m = \frac{p*h*A}{g}$

$m = \frac{850 * 0.4 * pi*0.46^2}{4*9.81}\\\\m= 56.5 kg$

8 0

2 years ago