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Viefleur [7K]
3 years ago
12

What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =

8.99 × 109 N • ___ V
What is the electric potential at a distance of 99 m from the charge?___ V
Physics
2 answers:
Orlov [11]3 years ago
6 0
According to Edge, the answers is 

3100 V

and 

200v
iren2701 [21]3 years ago
3 0
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
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An object with 274 J of GPE is 140cm above the ground. What is its mass?
Amanda [17]
E=274J
h=140cm=1,4m
g≈9,8m/s²

m=?

E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg



"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."



Regards M.Y.
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3 years ago
A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the
SCORPION-xisa [38]

Answer:

Magnitude of the net force acting on the kayak = 39.61 N

Explanation:

Considering motion of kayak:-

Initial velocity, u =  0 m/s

Distance , s = 0.40 m

Final velocity, v = 0.65 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0.65² = 0² + 2 x a x 0.4

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    F  = ma = 75 x 0.53 = 39.61 N

Magnitude of the net force acting on the kayak = 39.61 N

6 0
3 years ago
How many times did john glenn orbit the earth
galben [10]

Answer:

2 times

Explanation:

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navik [9.2K]

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A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.
Zielflug [23.3K]

<u>Answer</u>

8. 2 Hz

9. 0.5 seconds

10. 20 cm


<u>Explanation</u>

<u>Q 8</u>

Frequency is the number of oscillation in a unit time. It is the rate at which something repeats itself in a second.

In this case, the spring bob up and down 2 times per second.

∴ Frequency = 2 Hz

<u>Q 9</u>

Period is the time taken to complete one oscillation.

2 oscillations takes 1 second

1 oscillation = 1/2 seconds.

∴ Period = 0.5 seconds


<u>Q 10</u>

Amplitude is the the maximum displacement of the spring.

In this case the spring bob up 20 cm. This is it's displacement.

∴ Amplitude = 20 cm

5 0
2 years ago
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