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EleoNora [17]
3 years ago
14

What is the gravity force between two stars with mass of 5,000,000 kg and 1,000,000 kg if the distance between them is 100 m

Physics
1 answer:
Luda [366]3 years ago
8 0

Answer:

0.0334N

Explanation:

Given parameters:

M1  = 5 x 10⁶kg

M2  = 1 x 10⁶kg

Distance  = 100m

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

     Fg  = \frac{G m1 m2}{r^{2} }  

G is the universal gravitation constant

m is the mass

r is the distance

         Fg  = \frac{6.67 x 10^{-11} x 5 x 10^{6}  x 1 x 10^{6} }{100^{2} }   = 0.0334N

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A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is  Q = 5.866 * 10^9 J

The power is  P = 5866\  MW

Explanation:

From the question we are told that

      Mass of the water per second is m = 1917 \ kg

      The initial temperature of the water is T_i  = 35^oC

      The boiling point of water is  T_b = 100^oC

      The final temperature T_f = 450^oC

      The latent heat of vapourization of water is  c__{L}} = 2256*10^3 J/kg

      The specific heat of water c_w = 4184 J/kg^oC

      The specific heat of stem is C_s =1520 \ J/kg ^oC

Generally the heat needed each second is mathematically represented as

         Q = m[c_w (T_i - T_b) + m* c__{L}}  + m* c__{S}} (T_f - T_b)]

Then substituting the value

        Q = m[c_w [T_i - T_b] + c__{L}}  + C__{S}} [T_f - T_b]]

         Q = 1917 [(4184) [100 - 35] + [2256 * 10^3]  +[1520]  [450 - 100]]

         Q = 1917 * [3.05996 * 10^6]

         Q = 5.866 * 10^9 J

The power required is mathematically represented as

         P = \frac{Q}{t}

From the question t = 1\ s

So  

        P = \frac{5.866 *10^9}{1}

        P = 5866*10^6 \ W

        P = 5866\  MW

6 0
3 years ago
A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat
emmasim [6.3K]

Answer:

sorry I need points

Explanation:

IM SO SORRY

5 0
3 years ago
The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is
stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

7 0
3 years ago
Please help me to do this problem
Novosadov [1.4K]

Answer:

we got time and velocity over time.

so the distance is again the area underneath the graph

for a triangle with known base and height it's

4*10 / 2

distance traveled is 20

deceleration occurs when velocity decreases. that happens from t=2 till t=4

in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time

a (from t=2 to t=4) = -5v/t

7 0
3 years ago
an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m
Ksenya-84 [330]

Answer:

7.87x10^5m/s^2

Explanation:

3 0
3 years ago
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