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klemol [59]
3 years ago
14

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

a diameter of 19.0 km ?]
Physics
1 answer:
NeTakaya3 years ago
5 0
-- If you weigh 690N on the Earth, then your mass is about

           (690N) / (9.8 m/s²) = 70 kg .

-- If your mass is 70 kg then your weight on that neutron star would be

              G  M₁  M₂  /  R²

         =  (6.67 x 10⁻¹¹ N m²/kg²) (70 kg) (1.989 x 10³⁰ kg) / (19,000 m)²

         =  (6.67 x 10⁻¹¹ x 70 x 1.989 x 10³⁰) / (3.61 x 10⁸)        N

         =  (6.67 x 70 x 1.989 x 10¹⁹)  /  (3.61 x 10⁸)          N

         =    2.57 x 10¹³    Newtons .

  (about  2,890,000,000 tons !)
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I think the answer would be A.
8 0
2 years ago
All organisms in Kingdom Animalia are multicellular, meaning their bodies are composed of more than one cell. Which other charac
k0ka [10]
Well,

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4 0
3 years ago
Read 2 more answers
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
2 years ago
A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram
Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0
1 year ago
A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t
zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0
2 years ago
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