On which planet would your weight be the most and the least?
1 answer:
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Answer:
a) v = 18.86 m / s, b) h = 8.85 m
Explanation:
a) For this exercise we can use the conservation of energy relations.
Starting point. Like the compressed spring
Em₀ = K_e + U = ½ k x² + m g x
the zero of the datum is placed at the point of the uncompressed spring
Final point. With the spring if compress
Em_f = K = ½ m v²
how energy is conserved
Em₀ = Em_f
½ k x² + m g x = ½ m v²
v² = x² + 2gx
let's reduce the magnitudes to the SI system
m = 500 g = 0.500 kg
x = -45 cm = -0.45 m
the negative sign is because the distance in below zero of the reference frame
let's calculate
v² = 0.45² + 2 9.8 (- 0.45)
v = √355.68
v = 18.86 m / s
b) For this part we use the conservation of energy with the same initial point and as an end point at the point where the rock stops
Em_f = U = m g h
Em₀ = Em_f
½ k x²2 + m g x = m g h
h = ½ x² + x
let's calculate
h = - 0.45
h = 8.85 m
measured from the point where the spring is uncompressed
(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ()
(b) The plates can be close together to 1.7 mm with this applied voltage
<u>Explanation:</u>
Given data:
Dielectric strength of air =
Distance between the plates = 2.00 mm =
Potential difference, V =
We need to find
a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not
b) the minimum distance at which the plates can be close together with this applied voltage.
The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,
From the above, concluding that The electric field strength between two parallel conducting plates () does not exceed the breakdown strength for air (
)
b) To find how close together can the plates be with this applied voltage:
The formula would be,
Apply all known values, we get