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Viefleur [7K]
3 years ago
12

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

meter. If you weigh 675 N on the earth, what would you weigh at the surface of a neutron star that has the same mass as our sun and a diameter of 20 km
Physics
1 answer:
Crank3 years ago
3 0

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

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g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the sa
Olegator [25]

Answer:

The correct answer is "\frac{4E}{3}".

Explanation:

According to the question,

Energy of satellite,

⇒ E_s=-\frac{GM_sM_E}{2r}

For the very 1st case:

r = R_E+R_E

  =2R_E

or,

⇒ E=-\frac{GM_sM_E}{4R_E}...(1)

For the new case:

r = R_E+\frac{R_E}{2}

  =\frac{3R_E}{2}

then,

⇒ E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }

        =-\frac{GM_sM_E}{3R_E}...(2)

From equation (1) and (2), we get

⇒ E'=\frac{1}{3}(4E)

        =\frac{4E}{3}  

7 0
3 years ago
A steam power plant produces 50MW of net work while burning fuel to produce 150MW of heat energy at the high temperature. Determ
Mazyrski [523]

Nth =\frac{W net, out}{Qh}

50MW is our net output.

Then, we plug it in.

= \frac{50MW}{150MW} = 0.333 or 33.33%

W net, out = Qh - Ql

Ql = Qh - W net,out

Plug the values in.

Then, it becomes:

= 150MW - 50 MW

= 100MW

Thus, the cycle thermal efficiency is 33.33% and the heat rejected is 100MW.

3 0
3 years ago
A cycle travels along a circular track of diameter 42 m. Calculate the distance travelled and the displacement of the cycle in (
DENIUS [597]

Answer:

(a) i) The distance travelled by the cycle in half round is approximately 65.97 m

ii) The displacement is 42 m

(b) (i) The distance travelled in one round is approximately 131.95 m

(ii) The displacement of the cycle in one round is 0

Explanation:

The diameter of the track through which the cycle travels, D = 42 m

(a) i) Half round is the motion of half the length of the circular path

The distance travelled by the cycle in half round = The length of half the circular track = (1/2) × π × D

∴ The distance travelled by the cycle in half round = (1/2) × π × 42 m = 21·π m ≈ 65.97 m

ii) The displacement half round = The change in the location of the cycle = The difference between the start and stop locations of the cycle on a straight line after half round

The angle at the center of the circular path the cycle turns in half round  = 180°

Therefore, the path between the start and stop location of the cycle in half round = The diameter of the circular track

The displacement of the cycle in half round = The diameter of the circular track = 42 m

(b) (i) The distance travelled in one round = The perimeter of the circular track = π·D

∴ The distance travelled in one round = π × 42 m ≈ 131.95 m

(ii) The displacement of the cycle in one round = The change in the location of the cycle

The start and stop location of the cycle after moving one round is the same, therefore, there is no change in the location of the cycle.

Therefore we have;

The displacement of the cycle in one round = 0 (no change in location of the cycle)

7 0
3 years ago
Force = mass times acceleration
Alona [7]

Yes indeed ! Very good !


3 0
3 years ago
Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad
Anettt [7]

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, d=1.9\times 10^7\ m

Mass of Earth, M=5.97\times 10^{24}\ kg

Radius of Earth, r=6.37\times 10^6\ m

We need to find the force of gravity above the surface of Earth. It is given by :

F=G\dfrac{mM}{R^2}

R = r + d

R = 25370000 m

F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

5 0
4 years ago
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