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Ilia_Sergeevich [38]
3 years ago
7

A rubbit gets down from a rump which its /\x=0.85m in 0.5s, The rubbit's mass is 2kg, what is the net Force?

Physics
1 answer:
mestny [16]3 years ago
8 0

Answer: 13.6 N

Explanation:

The equation of motion for the rabbit is:

\Delta x=V_{ox}t+\frac{1}{2}a_{x}t^{2} (1)

Where:

\Delta x=0.85 m is the distance traveled by the rabbit

V_{ox}=0 m/s is the rabbit's initial velocity, assuming it started from rest

t=0.5 s is the time

a_{x} is the acceleration

Isolating a_{x}:

a_{x}=\frac{2 \Delta x}{t^{2}} (2)

a_{x}=\frac{2 (0.85 m)}{(0.5)^{2}} (3)

a_{x}=6.8 m/s^{2} (4)

On the other hand, the force F_{x} is given by:

F_{x}=m.a_{x} (5)

Where m=2 kg is the mass of the rabbit

Substituting (4) in (5):

F_{x}=(2 kg)(6.8 m/s^{2}) (6)

Finally:

F_{x}=13.6 N

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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.

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