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alexdok [17]
3 years ago
10

The gravitational force between the Sun (mass -1.99 1030 kg) and Mercury (mass 3.30 x 1023 kg) is 899 1021 N. How far is Mercury

from the Sun?
Physics
2 answers:
Daniel [21]3 years ago
7 0
The gravitational force between the two objects is given by:
F=G  \frac{m_1 m_2}{d^2}
where
G=8.99 \cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant
m_1 = 1.99 \cdot 10^{30} kg is the Sun mass
m_2 = 3.30 \cdot 10^{23} kg is the mass of Mercury
and d is the distance between Sun and Mercury. Since we know the force:
F=8.99 \cdot 10^{21} N
we can re-arrange the formula to find d:
d= \sqrt{ G \frac{m_1 m_2}{F} }=6.98 \cdot 10^{10}m
balandron [24]3 years ago
5 0

Answer:

B - 6.98 x 10^7 km

Explanation:

took it on egde

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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.
Bond [772]
calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
density\ of\ water=\ 999.9720 \frac{kg}{m^3} \\\\mass=0,15*999.9720=149,9958kg=149995,8grams\\\\ Answer\ is\ :\\water\ which\ weight\ is\ equal\ to\ 149995,8grams.

4 0
3 years ago
5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi
lakkis [162]

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

3 0
3 years ago
The actual mechanical advantage of any machine is its__________divided by its________________
zaharov [31]

Answer: Load divided by it effort

Explanation:

Mechanical advantage of any machine is its load divided by its effort

4 0
3 years ago
If a star is 720 , 000 , 000 , 000 , 000 meters from Earth, how many seconds does it take light to travel from the Earth to the
Minchanka [31]

Answer:

anywhere between 100000 to about 400000 human years .

Explanation:

5 0
2 years ago
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