For each load, Work = (mass) x (gravity) x (distance .
Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .
Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.
The work required is equal in both cases.
The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .
Answer: The trip takes 
Explanation:
Velocity
is the variation of the position of a body (distance traveled
) with time
:
In this case, the car travels a distance
at a velocity
and we need to find the time it takes the trip.
Isolating
:

Finally:

Answer:
0.4
Explanation:
Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance
Using the power formula
P = IV
Substitute all the parameters
P = 0.8 × 2
P = 1.6 W
But P = I^2 R
Substitute power and current
1.6 = 0.8^2 R
R = 1.6 / 0.64
R = 2.5 ohms
Inductance = reciprocal of resistance
Inductance = 1 / 2.5
Inductance = 0.4
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ
............1
put here value
= 1.75×
× 
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 ×
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 ×
v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 ×
v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds