Answer:
let us use the expression v = ./2gS in both the cases.
a) In the first case g = 9.8 m/s^2 and S = 10.3 m.
So v = 14.21 m/s
b) in the second case, g = 1.64 m/s^2 and S = 4.2 m = 3.71 m/s
00
Explanation:
D
Because if an object is moving at a constant speed the force of friction must equal the applied (horizontal) force, and for it to be accelerating or decelerating, the force of friction and the applied force must be unequal
Answer:
191.36 N/m
Explanation:
From the question,
The Potential Energy of the safe = Energy of the spring when it was compressed.
mgh = 1/2ke²............... Equation 1
Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression
Making k the subject of the equation,
k =2mgh/e²................ Equation 2
Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m
Constant: g = 9.8 m/s²
Substitute into equation 2
k = 2(1100)(9.8)(0.0024)/0.52²
k = 51.744/0.2704
k = 191.36 N/m
Hence the spring constant of the heavy-duty spring = 191.36 N/m
The statements of both students are incorrect.
-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms' means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).
There are no 'following statements'.
All in all, a very disappointing question.
