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larisa [96]
3 years ago
11

If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

d a diameter of 25.0 km?Take the mass of the sun to be ms= 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 .
Physics
1 answer:
lana [24]3 years ago
5 0

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

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A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
For an object moving with constant negative acceleration, draw the following:
Tcecarenko [31]

Answer:

Kindly find the graphs attached

Explanation:

For figure 1: There is a steady increase in the position of the object as time increases. This is because despite the negative acceleration (deceleration), the object continues to move and cover more ground as time goes by.

<em>The straight line graph is observed because the acceleration is constant  and not varying.</em>

For Figure 2: The graph of velocity vs time will have an inverted nature. This is because since the object is decelerating, it is reducing in its velocity as time goes by (increases). <em>This is also in a straight line since the deceleration is constant.</em>

3 0
3 years ago
Where does fusion regularly occur and what kind of energy is produced
AlexFokin [52]
Sun is the place where fusion continuously occurs.. heat energy is produce plus raditions are by product <span />
4 0
3 years ago
Read 2 more answers
Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady
Bumek [7]

We define acceleration as the rate of change of the velocity

Thus, if you have positive velocity and positive acceleration, your <u>speed increases.</u>

If you have positive velocity and negative acceleration, your speed decreases.

Now you get the idea, we will see that the correct option is graph 1.

We know that the car moves towards the right (let's define this as "the car has positive velocity") and we also know that te car is slowing down constantly (thus the acceleration needs to be negative and constant).

By looking at the graphs, the only one with these properties is graph 1.

If you want to learn more, you can read:

brainly.com/question/12550364

5 0
2 years ago
Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ic
Setler [38]

Answer:

final temperature will be 0 degree C

Total amount of ice will be

m_{ice} = 182 g

total amount of water

m_{water} = 17 g

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

Q_1 = m_1s_1 \Delta T_1

Q_1 = 0.030(4186)(50 - 0)

Q_1 = 6279 J

heat absorbed by ice cubes to reach 0 degree

Q_2 = m_2 s_2 \Delta T_2

Q_2 = (0.169)(2100)(30)

Q_2 = 10647 J

so we will have

Q_2 > Q_1

so here we can say that few amount of water will freeze here to balance the heat

10647 - 6279 = mL

m = \frac{10647 - 6279}{335000}

m = 13 g

so final temperature will be 0 degree C

Total amount of ice will be

m_{ice} = 84.5 + 84.5 + 13

m_{ice} = 182 g

total amount of water

m_{water} = 30 - 13

m_{water} = 17 g

4 0
3 years ago
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