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3 years ago

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A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

gravity on this satellite will double but the satellite's orbit will not be affected. A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. True False

2 answers:

adelina 88 [10]3 years ago

7 0

Answer:

it is True

Explanation:

I hope this helps

Fofino [41]3 years ago

5 0

Answer:

Explanation:

True, provided that the extra mass is also already in the same orbit itself.

If the extra mass is NOT in the same orbit, it's called a collision and the orbits of both will be altered due to the changes in momentum and kinetic energy.

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Find the equivalent resistance in the circuit between A and B

Monica [59]

Answer:

6Ω

Explanation:

The 6Ω resistor and the 3Ω resistor are in parallel.

R = 1 / (1/6 + 1/3)

R = 1 / (1/6 + 2/6)

R = 1 / (3/6)

R = 6/3

R = 2

The 8Ω resistor and the other 8Ω resistor are also in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

This 4Ω resistance is in series with the 4Ω resistor.

R = 4 + 4

R = 8

The 2Ω resistor and the 6Ω resistor are also in series.

R = 2 + 6

R = 8

These two 8Ω resistances are in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

Finally, this 4Ω resistance is in series with the 2Ω resistance we found earlier.

R = 4 + 2

R = 6

The total equivalent resistance is 6Ω.

5 0

3 years ago

Choose the correct statement below about the interaction between magnets, magnetic materials and charges.

bekas [8.4K]

Answer: B. The south magnetic pole attracts the north pole of another magnet.

Explanation:

The correct statement below about the interaction between magnets, magnetic materials and charges is that the south magnetic pole attracts the north pole of another magnet.

We should note that a magnet typically consists of two poles which are the North Pole and the South pole. The opposite poles doesn't repel each other but attracts each other while same pole repels one another.

In this case, the North Pole will attract the South Pole while same magnetic piles repel e.g the North Pole will repel the North Pole.

5 0

3 years ago

Please help. Explain why it is correct. I will give brainliest

blsea [12.9K]

Answer: Hey

B an encyclopedia entry about an experiment

A primary resource is information that was documented during an event that was written about.

Please wait tell you get another answer because im not sure about this

7 0

3 years ago

A rock has a mass of 25kg. The acceleration due to gravity is 10m/s and the height of the rock is 42 m. Find the potential energ

alisha [4.7K]

Answer:

The potential energy of the rock = 10.5 kN

Explanation:

Mass of rock = 25 kg

Acceleration due to gravity = 10 m/s²

Height = 42 m

Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.

PE = 25 x 10 x 42 = 10500 N = 10.5 kN

The potential energy of the rock = 10.5 kN

7 0

3 years ago

Read 2 more answers

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago