Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
Answer:
IMA = 2.5 metres
EFFICIENCY = 80%
Explanation:
The AMA of a machine is referred to as the Actual Mechanical Advantage of a machine, calculated as the ratio of the output to the input force.
The Ideal Mechanical Advantage is the ratio of the input distance to the output distance.
From the diagram, the input distance which is also the distance moved by effort = 5metres
The load distance (output distance) = 2 metres
IMA = INPUT DISTANCE / OUTPUT DISTANCE
IMA = 5metres / 2 metres = 2.5 meters
Efficiency is the ratio of AMA TO IMA
AMA = 2, IMA = 2.5
EFFICIENCY = AMA / IMA
EFFICIENCY = (2 / 2.5) × 100%= 0.8 × 100%
EFFICIENCY = 80%
Answer:
22 N applied force
Explanation: Since they are both pushing the wagon in the same direction the force adds up.
Answer:
e. all of these
Explanation:
The fatigue strength is improved by then high alloy steels , high yield steels , high hardened steel , high ultimate steel .
Due to the formation of the improved materials in alloy steels will increase the fatigue strength . Similarly for a high yield steels and hardened steels these cycles to failure will improve .