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Ne4ueva [31]
3 years ago
14

Two uncharged, conducting spheres are separated by a distance d. When charge −Q is moved from sphere A to sphere B, the Coulomb

force between them has magnitude F0.
(a)Is the Coulomb force attractive or repulsive?



attractive repulsive



(b) If an additional charge −Q is moved from A to B, what is the ratio of the new Coulomb force to the original Coulomb force,(c) If sphere B is neutralized so it has no net charge, what is the ratio of the new to the original Coulomb force,
Physics
1 answer:
rosijanka [135]3 years ago
3 0

Answer: a) the force will be repulsive

b) the ratio of the new force to the old force will be 2

c) O

Explanation:

a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.

b) initial force will be -q(-Q)/d2

Adding extra charge -Q will cause change on B to become -2Q

The new force will be - 2Q(-q)/d2

Dividing new force by old force will give 2

C) if B is neutralized, the net charge becomes 0 and there will be no force on it.

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Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

6 0
3 years ago
A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela
larisa86 [58]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

<span>Now we will write the equation for a Change in Potential energy ΔU,</span>

ΔU = Uf - Ui

= 0-m*g*h

<span>  ΔU = -m*g*h --Equation 2</span>

Now compare the both equation

<span>Wint = -ΔU</span>

we can rewrite the above equation

ΔU = -W.

<span>So our Answer is ΔU = -W. .</span>

<span> </span>

5 0
3 years ago
Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr
Aleksandr [31]
<span>c. faster, less

is the answer
</span>
6 0
3 years ago
Read 2 more answers
A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
pashok25 [27]

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

5 0
3 years ago
Help m-e-e people -_-!​
xxTIMURxx [149]

Answer:

the answer is

Explanation:For equilibrium

Weight = Tension

mg=T

∴T=4×3.1π=12.4πN (as can be inferred from the question)

Y=

△l/l

T/A

​

=

1000

0.031

​

/20

12.4π/π(

1000

2

​

)

2

​

=

4×0.031

12.4×20×1000×(1000)

2

​

=2×10

12

N/m

2

6 0
3 years ago
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