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OLEGan [10]
2 years ago
10

A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How far did the bulle

t travel horizontally when it hit the ground?
595.2 m
478.4 m
364.0 m
247.2 m
Physics
2 answers:
Vesnalui [34]2 years ago
5 0
It’s most likely A , i’m not so sure though
USPshnik [31]2 years ago
4 0

Answer:

im pretty sure its a

Explanation:

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A solid ball of radius rb has a uniform charge density rho.
Oksana_A [137]

Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a.  What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?  Express your answer in terms of ρ, rb, r, and ϵ0.

b.   What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?  Express your answer in terms of ρ, r, rb, and ϵ0.

c.   Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ ,  ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V=  ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E=  ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b)  The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E=  ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

(c)  E(0)= 0

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

4 0
3 years ago
1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)
kotegsom [21]

As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.

Then   friction force = 80 Newtons              but in the opposite direction.

Friction force =  Mu  * Normal force exerted by ground  =  Mu * weight of box

So we find Mu.

Mu = coefficient of friction between box and horizontal surface

          = Force of friction / weight  =  80 / 50 * 9.81 = 0.163

When an identical box is placed on top, the force of friction is

      = Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons

7 0
3 years ago
PLZ HELP ME FAST A relationship between two variables is called:
Irina18 [472]

Answer:

B- Correlation

Explanation:

6 0
2 years ago
Read 2 more answers
A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h
katen-ka-za [31]

Answer:

v_r=5.89\ m.s^{-1}

Explanation:

Given:

  • mass of rocket, m_r=50\ g
  • time of observation, t=2\ s
  • mass lost by the rocket by expulsion of air, m_a=10\%\ of m_r=5\ g
  • velocity of air, v_a=53\ m.s^{-1}

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

m_a.v_a=(m_r-m_a)\times v_r

5\times 53=(50-5)\times v_r

v_r=5.89\ m.s^{-1}

7 0
3 years ago
Distance = 6 km south<br><br> 60 minutes<br><br> What was the average velocity
blsea [12.9K]

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

3 0
3 years ago
Read 2 more answers
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