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VikaD [51]
3 years ago
7

QUICK WILL MARK BRAINLIEST

Physics
1 answer:
Tju [1.3M]3 years ago
4 0

Answer:

37.1°

Explanation:

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A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s

In free fall

v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s

The time the rocket is accelerating is 6.75 seconds

5 0
4 years ago
A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of the car?
rodikova [14]

Answer:

16 m/s.

Explanation:

The following data were obtained from the question:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Mass of car = 2500 kg

Velocity of car =..?

Next, we shall determine the momentum of the truck. This can be obtained as follow:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Momentum of truck =.?

Momentum = mass × velocity

Momentum = 5000 × 8

Momentum of the truck = 40000 Kg.m/s

Finally, we shall determine the velocity of the car as follow:

From the question given above, we were told that the car and truck has the same momentum.

This implies that:

Momentum of the truck = momentum of car = 40000 Kg.m/s

Thus, the velocity of the car can be obtained as shown below:

Mass of car = 2500 kg

Momentum of the car = 40000 Kg.m/s

Velocity of car =..?

Momentum = mass × velocity

40000 = 2500 × velocity

Divide both side by 2500

Velocity = 40000/2500

Velocity = 16 m/s

Therefore, the velocity of the car is 16 m/s.

5 0
4 years ago
Which surface exhibits the highest albedo measurement
k0ka [10]
Fresh snow is the surface that exhibits the highest albedo measurement. 
8 0
3 years ago
When a pendulum is swinging, the velocity is highest at which point?
Tems11 [23]
At the center, when the bob is hanging straight down

3 0
3 years ago
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