Answer:
The electric potential in volts is 1.618 x 10⁻¹⁷ V
Explanation:
The electric potential, in volts, at point P, can be calculated as follow;
Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.
Thus, the work done in this process in moving the charge to point p is 101eV.
Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V
= 1.618 x 10⁻¹⁷ V
Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
The power that heat pump draws when running will be 6.55 kj/kg
A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).
Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8
We have to determine the power the heat pump draws when running.
To solve this question we have to assume that the heat pump is at steady state
Let,
Q₁ = 22000 kj/kg
COP = 2.8
Since heat pump used to heat a house runs about one-third of the time.
So,
Q₁ = 3(22000) = 66000 kj/kg
We known the formula for cop of heat pump which is as follow:
COP = Q₁/ω
2.8 = 66000 / ω
ω = 66000 / 2.8
ω = 6.66 kj/kg
Hence the power that heat pump draws when running will be 6.55 kj/kg
Learn more about heat pump here :
brainly.com/question/1042914
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Answer:
Explanation:
mass of airplane, m = 12000 kg
altitude, h = 10 km = 10,000 m
velocity, v = 175 m/s
(a) Angular momentum
L = m x v x h x Sin 90
L = 12000 x 10,000 x 175
L = 2.1 x 10^10 kg m^2/s
(b) As there is no external torque, so the angular momentum remains constant.
