Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
Answer:
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Explanation:
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Answer:
0.06 N
1.08 m/s
Explanation:
m = mass of the fan cart = 0.250 kg
a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²
F = Net force on the cart
Net force on the cart is given as
F = ma
F = (0.250) (0.24)
F = 0.06 N
v₀ = initial velocity of the cart = 0 m/s
v = final velocity of the cart
t = time interval = 4.5 s
Using the equation
v = v₀ + a t
v = 0 + (0.24) (4.5)
v = 1.08 m/s
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316