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slamgirl [31]
1 year ago
5

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

press your answer in terms of the given quantities and appropriate constants.1. What is the electric field at the surface of the sphere?E=2.What is the electric potential at the surface of the sphere?V=
Physics
1 answer:
amid [387]1 year ago
3 0

Answer:

E =   k*Q₁/R₁² V/m

V =  k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                                        F = k*Q₁/R₁²

- Then the electric field at that point is

                                        E =  F/1

                                        E =  k*Q₁/R₁²  V/m

- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                                        V =  k*Q₁/R₁  Volt

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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
1 year ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk
HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

          P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of ​​the disk, T the absolute temperature and e the emissivity that for a black body is  1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

           I = P / A

Writing this expression for both discs

          I₁ A₁ = I₂ A₂

          I₂ = I₁ A₁ / A₂

The area of ​​a sphere is

          A = 4π r²

           I₂ = I₁ (r₁ / r₂)²

          r₂ = r₁ ± 5

          I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

        Let's write the Stefan equation

         P / A = σ e T⁴

          I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

         σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values ​​of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

        e₁ T₁⁴ = T₂⁴   (r1 + 5)²/r₁²

       T₁ = T₂  {(e₂/e₁)}^{1/4}  √(1 ± 1/ r₁)  

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

           (1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

          T₁ = T₂ {(e₂/e₁)}^{1/4}  (1 ± ½   5/r1)

           T₁ = T₂ {(e₂)}^{1/4}   (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

              T₁ = T₂  {(e₂)}^{1/4} ¼

              T<u>₁</u> = T₂  0.90.9^{1/4}

               5500 = T₂  0.974

               T₂ = 5646 K

3 0
2 years ago
abin is doing work by lifting a bowling ball. Which statement could be made about the energy in this situation?
PtichkaEL [24]
The statement that could be made about the energy in this situation would be :
It being transferred from his arms muscles to the ball.

The muscle contraction from his arms created a force that could be used to lift the ball up.<span />
8 0
2 years ago
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