The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
<h3>
What is the energy in a capacitor?</h3>
The energy stored in a capacitor is an electrostatic potential energy.
It is related to the charge(Q) and voltage (V) between the capacitor plates.
It is represented as 'U'.
<h3>
How to determine the potential difference</h3>
Formula:
Potential difference, V is the ratio of the charge to the capacitance of a capacitor.
It is calculated using:
V = Q ÷ C
Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9
Substitute the values into the equation
Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts
<h3>
How to determine the energy stored</h3>
Formula:
Energy, U = 1 ÷ 2 (QV)
Where Q= charge and V = potential difference across the capacitor
Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)
= 0.5 × 25 × 10∧-1
= 0.5 × 2.5
= 1. 25 Joules
Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
Learn more about capacitance here:
brainly.com/question/14883923
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From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.
<h3>What is weight?</h3>
The weight of an object is obtained as the product of the mass of the body and the acceleration due to gravity.
Thus;
When;
mass = 120 kg
weight = 46 N
acceleration due to gravity = 46 N/120 kg
=0.38 m/s^2
Learn more about acceleration due to gravity :brainly.com/question/13860566
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It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.
Hopr it helps :)
D garden salad : )
A heterogenous mixture can be easily taken apart visually/physically
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
