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alukav5142 [94]
3 years ago
13

From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same poi

nt on the earth’s surface?​
Physics
1 answer:
yulyashka [42]3 years ago
3 0

In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.

But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.

The Moon has nothing to do with any of this.

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In a chemical reaction, molecules of hydrogen gas (H2) react with molecules of oxygen gas (O2) in a sealed reaction chamber to p
11Alexandr11 [23.1K]

Answer:

Option (B) is correct.

Explanation:

Given that the molecules of hydrogen gas (H_2) react with molecules of oxygen gas (O_2) in a sealed reaction chamber to produce water (H_2O).

The governing equation for the reaction is

2H_2 +O_2 \rightarrow 2H_2O

From the given, the only fact that can be observed that 2 moles of H_2 and 1 mole of O_2 reacts to produce 2 moles of H_2O.

As the mass of 1 mole of H_2 = 2 grams ... (i)

The mass of 1 mole of O_2 = 32 grams ...(ii)

The mass of 1 mole of H_2O = 18 grams (iii)

Now, the mass of the reactant = Mass of 2 moles of H_2 + mass 1 mole of  O_2

= 2 \times 2 + 32  [ using equations (i) and (ii)]

=4+32 = 36 grams.

Mass of the product = Mass of 2 moles of H_2O

=2\times 18 [ using equations (iii)]

=36 grams

As the mass of reactants = mass of the product.

So, mass is conserved.

Hence, option (B) is correct.

8 0
3 years ago
What is the quantity of motion ??
horsena [70]

Answer:

The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. In other words, rather than defining the quantity of motion of a given object as simply the kinematic velocity v of the object, he defined it as the product mv, where m is the mass of the object.

Explanation:

3 0
3 years ago
How do I solve this step by step? I’m really confused
LekaFEV [45]

Step-#1:

Ignore the wire on the right.

Find the strength and direction of the magnetic field at P,

caused by the wire on the left, 0.04m away, carrying 5.0A

of current upward.

Write it down.


Step #2:

Now, ignore the wire on the left.

Find the strength and direction of the magnetic field at P,

caused by the wire on the right, 0.04m away, carrying 8.0A

of current downward.

Write it down.


Step #3:

Take the two sets of magnitude and direction that you wrote down

and ADD them.


The total magnetic field at P is the SUM of (the field due to the left wire)

PLUS (the field due to the right wire).


So just calculate them separately, then addum up.

4 0
3 years ago
The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0
3 years ago
Why does a plastic comb attract small bits of paper after rubbing with wool?
navik [9.2K]
Static electricity. Like the balloon against hair
5 0
3 years ago
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