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Dafna1 [17]
3 years ago
12

4.0 kg objects mive a distance of 7.8 m under the action of a cinstant gorce of 5.6 N. how much work is dine on object?

Physics
1 answer:
alexdok [17]3 years ago
6 0

Answer:

43.68 J

Explanation:

Distance moved= 7.8 m

Force = 5.6 N

Work Done= Distance moved * Force

                   = 7.8 *5.6

                      =43.68 Joules

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Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
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Answer:

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Explanation

1. <u>Data</u>:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. <u>Formulae</u>

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3. <u>Solution</u>:

a) Abbie:

  • Ac₁ = ω² r₁  =  ω² (1m)

b) Zack:

  • Ac₂ = ω² r₂  = ω² (2m)

c) Divide Ac₁ / Ac₂

  • Ac₁ / Ac₂ =  ω² (1m) / [ω² (2m) ] = 1/2

⇒      Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

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Special relativity can be used to study an object in which frame of reference? A. A frame of reference that has no change in vel
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A 1,680 kg satellite is in a circular orbit around Earth with a tangential speed of 6,578 m/s. What is the height of the satelli
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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
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Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

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Finally, we replace (5) in (4) and solving for q:

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