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DerKrebs [107]
4 years ago
7

Two tiny conducting spheres are identical and carry charges of -18.4 µC and +53.0 µC. They are separated by a distance of 2.73 c

m. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
Physics
1 answer:
melomori [17]4 years ago
4 0

Answer:

-11776.36 N

This force is attractive since both charges are of opposite sign

Explanation:

Given that

q_1=-18.4\mu C=-18.4\times 10^{-6}C\ and\\\ q_2=53\mu C=53\times 10^{-6}C

Distance between the spheres = 2.73 cm =0.0273 m

where K is Coulomb's constant = 9.10^ 9 [N.m^2 /C^2]

According to coulombs law we know that force between two charges is given by

F = \frac{kq_1q_2}{r^2}

F = \frac{(9\times10^9)\times (-18.4 \times 10^-^6)(53  \times 10^-^6}{0.0273^2)} \\= -11776.36N

This force is attractive since both charges are of opposite sign

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A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos
brilliants [131]

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic  collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25                   --------1

v₂-v₁=u₁-u₂            ( e= 1)

v₂-v₁ = -35        --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

4 0
3 years ago
Please help on this one?
diamong [38]

Answer:

ITS C

Explanation:

7 0
3 years ago
If the frequency of a periodic wave is doubled, the period of the wave will be
Lunna [17]
The period of the wave would be halved 
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3 years ago
There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture
Pani-rosa [81]

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

You can Put it inside a container of water and see the bubble and rotate the inner tube to pass all of it through the water

7 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
3 years ago
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