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Blizzard [7]
2 years ago
15

The formula $F = \frac{9}{5} C + 32$ can be used to convert temperatures between degrees Fahrenheit ($F$) and degrees Celsius ($

C$). How many degrees are in the Celsius equivalent of $-22^\circ F$?
Physics
1 answer:
slamgirl [31]2 years ago
7 0

Answer:

-30° C

Explanation:

Data provided in the problem:

The formula for conversion as:

F  = (9/5)C + 32

Now,

for the values of F = -22 , C = ?

Substituting the value of F in the above formula, we get

-22 = (9/5)C + 32

or

-22 - 32 = (9/5)C

or

(9/5)C = - 54

or

C = - 54 × (5/9)

or

C = - 30 °

Hence, -22 Fahrenheit equals to -30°C

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What causes a satellite to move tangentially to a circular path?
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Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

So, we can say gravity is responsible for the motion of the satellite in a circular path.

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Define the unit of time and unit of length​
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Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo
Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

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What do scientists use to explain an atom or the universe?
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The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

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\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
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