The answer would be no. Both will have same charge, in both caes charge will reside on surface. In case of solid sphere it will be distributed evenly throughout. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Answer:
λ = 2.7608 x 10⁻⁷ m = 276.08 nm
Explanation:
The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:
Work Function = hc/λ
where,
Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = longest wavelength capable of releasing electron.
Therefore,
7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)
<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>
Answer:
a) v= 2.1 m/s
b) ω = 0.807 rad/s
Explanation
Conceptual analysis :
The dog and the merry-go- round describes a circular motion, then, the following formulas apply :
Formula (1)
v = ω *r Formula (2)
Where:
: Centripetal acceleration(m/s²)
v: linear speed or tangential (m/s)
r : radius of the circle (m)
ω : angular speed ( rad/s)
Data
r= 2.6 m
= 1.7 m/s²
Problem develpment
a) We replace data in the formula 1 to calculate the dog's linear speed(v):
v= 2.1 m/s
b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).
v = ω *r
2.1 = ω *2.6
ω = 2.1/2.6
ω = 0.807 rad/s