Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Answer:
False
Explanation:
A white dwarf star that is easy to locate and see with small telescopes.
An energy bar contains of 20grams of carbohydrates and 1gram
is equal to 17 KJ. If the energy bar was his only fuel the total energy
available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal
to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and
his Distance is equal to (5,000 x .2209) = 1.1045km.
Explanation:
a) The rope obeys Hooke's law, so:
F = k Δx
The elastic energy in the rope is:
EE = ½ k Δx²
Or, in terms of F:
EE = ½ F Δx
Use trigonometry to find the stretched length.
cos 20° = 35 / x
x = 37.25
So the displacement is:
Δx = 37.25 − 24
Δx = 13.25
The elastic energy per rope is:
EE = ½ (3.7×10⁴ N) (13.25 m)
EE = 245,000 J
There's two ropes, so the total energy is:
2EE = 490,000 J
Rounded to one significant figure, the elastic energy is 5×10⁵ J.
b) The elastic energy in the ropes is converted to gravitational energy.
EE = PE = mgh
5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h
h = 42 m
Rounded to one significant figure, the height is 40 m. So the claim is not justified.
The paper clip did because it is a conductor. A pencil is made with wood and that is an insulator.
