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Naya [18.7K]
3 years ago
10

Approximate the field by treating the disk as a +6.1 C point charge at a distance of 21 cm Answer in units of N/C.

Physics
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

Electric field, E=1.24\times 10^{12}\ N/C

Explanation:

It is given that,

Charge, Q = +6.1 C

Distance, r = 21 cm = 0.21 m

We need to find the electric field. It is given by :

E=k\dfrac{Q}{r^2}

E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}

E=1.24\times 10^{12}\ N/C

So, the electric field at this distance is 1.24\times 10^{12}\ N/C. Hence, this is the required solution.

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The reason being nuclear reactions are slow in less massive stars than in massive stars hence they age slowly.
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4 years ago
Two objects exert a gravitational force on 8N on one another what would that force be if the mass of both objects were doubled?
jolli1 [7]
Gravitational force of attraction between two objects can be calculated from the following formula:
F = \frac{m_1 \cdot m_2}{r^{2}}
Where m with subscript stands for mass of the object and r is the distance between them.
When we double the mass of those two objects, distance between them stays the same, while in the numerator we have:
2\cdot m_1 \cdot 2 \cdot m_2 = 4\cdot m_1 \cdot m_2
When numerator in the second case 4 times greater than the 'original' numerator and denominator stays the same, force becomes 4 times greater.

3 0
3 years ago
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0
4 years ago
. A car with a mass of 700-kg changes its speed from 10.0 m/s to 30.0 m/s in a displacement of 50.0 meters. Calculate the net fo
Nezavi [6.7K]

Answer:

5600N

Explanation:

Given parameters:

Mass of car  = 700kg

Initial velocity  = 10m/s

Final velocity  = 30m/s

Displacement  = 50m

Unknown:

Net force acting on the car  = ?

Solution:

To find the force acting on a body, it is pertinent we know the mass and acceleration.

 Force  = mass x acceleration

Now;

 Let us find the acceleration from the kinematics equations:

 v²  = u²  + 2aS

 v is the final velocity

 u is the initial velocity

 a is the acceleration

 S is the distance

  30²  = 10²  + (2 x a x 50)

 900 = 100 + 100a

      100a  = 800

           a  = 8m/s²

Therefore;

     Force  = 700 x 8  = 5600N

6 0
3 years ago
A coil of 10 turns is in the shape of an ellipse having a major axis of 13.0 cm and a minor axis of 6.00 cm. The coil rotates at
bearhunter [10]

Answer:

the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V

Explanation:

Given the data in the question;

Number of turns N = 10

major axis Ma = 13 cm = 0.13 m

a = 0.13/2 = 0.065 m

Minor axis Mi = 6 cm = 0.06 m

b = 0.06/2 = 0.03 m

we know that; 1 RPM = 0.10472 rad/s

rate of rotation R = 73rpm = 7.64 rad/s

Magnetic field = 55 uT

we know that, Area of ellipse = π × a × b

we substitute

A = π × 0.065 m × 0.03 m

A = 0.006126 m²

so

Maximum Voltage = N × Area × Magnetic field × rate of reaction

we substitute

Maximum Voltage = 10 × 0.006126 × ( 55 × 10⁻⁶ ) × 7.64

Maximum Voltage = 2.574 × 10⁻⁵ V

Therefore, the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V

6 0
3 years ago
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