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vredina [299]
3 years ago
9

Where will the spacecraft be when the gravitational forces acting on it are equal?

Physics
1 answer:
klio [65]3 years ago
6 0
It would be not be able to move yet it would be in the air

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Can someone help me solve this its a simple series circuit
daser333 [38]

Answer:

See the explanation below.

Explanation:

Indeed this is a serial circuit, in order to calculate the different questions, we must calculate the current of the circuit. The total resistance of the circuit can be calculated by means of an algebraic summation.

Then using ohm's law,we can calculate the current.

Rt = 2 + 4 + 6 = 12 [ohm]

The current

I = V / Rt

I = 12 / 12

I = 1 [amp]

Therefore the voltage in each resistance will be:

V1 = 2 * 1 = 2 [V]

V2 = 4 * 1 = 4 [V]

V3 = 6 * 1 = 6 [V]

The total voltage will be:

V = V1 + V2 + V3 = 12 [V]

The power can be calculated by multiplying the voltage by the total current.

P = V * I

P = 12 * 1

P = 12 [W]

8 0
3 years ago
what is the source of centripetal force that planet requires revolving around the sun on what basis the factor does that was the
Taya2010 [7]

Answer:

The source of centripetal force is the gravitational force between the planet and Sun. Gravitational force depends on the mass of the planet, mass of the Sun and distance between them.

5 0
3 years ago
Do anyone answer this question​
bezimeni [28]

Answer:

B) 10^-2 cm/s

in term of meter. it is 10^-4 m/s

Explanation:

7 0
3 years ago
Read 2 more answers
Please help? physics question...
ratelena [41]

I believe the answer is A.

Hope This Helps!   Have A Nice Day!!

7 0
3 years ago
A particle has a de Broglie wavelength of 2.1 × 10-10m. Then its kinetic energy increases by a factor of 3. What is the particle
svlad2 [7]

Answer:

\lambda' =1.21\times 10^{-10}\ m

Explanation:

given,

de broglie wavelength of the particle = 2.1 × 10⁻¹⁰m

Kinetic energy increase factor = 3

de broglie  wavelength of new particle = ?

we know,

P = \dfrac{h}{\lambda}

Kinetic energy

K =\dfrac{p^2}{2m}

K =\dfrac{h^2}{2m\lambda^2}...........(1)

Kinetic energy of other particle

3K =\dfrac{h^2}{2m\lambda'^2}............(2)

dividing equation (1)/(2)

\dfrac{1}{3} =\dfrac{\lambda'^2}{\lambda^2}

\lambda' = \dfrac{\lambda'}{\sqrt{3}}

\lambda' = \dfrac{2.1\times 10^{-10}}{\sqrt{3}}

\lambda' =1.21\times 10^{-10}\ m

6 0
3 years ago
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