A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
2 answers:
Answer:
Kinetic energy, KE = 3.43 Joules
Explanation:
It is given that,
Mass of the ball, m = 0.2 kg
Initially, it is placed at a height of 2.25 meters. We need to find the kinetic energy of the ball when it reaches 1.5 meters above the ground. It can be calculated using the conservation of energy.
To find,
Kinetic energy of the ball when it reaches 1.5 meters above the ground
KE = 3.43 Joules
So, the kinetic energy of the ball is 3.43 joules.
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Answer:
L = 1.11 x m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x m
Length = ?
So,
we know that,
U = 1/2 C Δ
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ
52000 J = (0.5) x (C) x ()
C = 0.28 F
And we also know that,
C =
E = 8.85 x
K = 3.7
A = 0.20 x L
d = 2.6 x m
Plugging in the values into the formula, we get:
0.28 =
Solving for L, we get:
L = 1.11 x m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.