Answer:
1) The magnetic field outside the loop is zero.
In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V
The diagram is attached
Good work on solving part a).
b) may look complicated, but it's not too bad.
It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do. (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)
You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.
-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.
-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules
-- Portion of a kilogram he needs to use = (9.6 x 10⁶) / (3.8 x 10⁷)
Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone. It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)
Answer:
a) W = - 318.26 J, b) W = 0
, c) W = 318.275 J
, d) W = 318.275 J
, e) W = 0
Explanation:
The work is defined by
W = F .ds = F ds cos θ
Bold indicate vectors
We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces
Let's use trigonometry to break down weight
sin θ = Wₓ / W
Wₓ = W sin 60
cos θ = Wy / W
Wy = W cos 60
X axis
How the body is going at constant speed
fr - Wₓ = 0
fr = mg sin 60
fr = 15 9.8 sin 60
fr = 127.31 N
Y Axis
N - Wy = 0
N = mg cos 60
N = 15 9.8 cos 60
N = 73.5 N
Let's calculate the different jobs
a) The work of the force of gravity is
W = mg L cos θ
Where the angles are between the weight and the displacement is
θ = 60 + 90 = 150
W = 15 9.8 2.50 cos 150
W = - 318.26 J
b) The work of the normal force
From Newton's equations
N = Wy = W cos 60
N = mg cos 60
W = N L cos 90
W = 0
c) The work of the friction force
W = fr L cos 0
W = 127.31 2.50
W = 318.275 J
d) as the body is going at constant speed the force of the tape is equal to the force of friction
W = F L cos 0
W = 127.31 2.50
W = 318.275 J
e) the net force
F ’= fr - Wx = 0
W = F ’L cos 0
W = 0
Answer:
As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.
Explanation:
As block 2 goes down , work is done by gravity on block 2 . This is converted
into kinetic energy of block 1 and block 2 . Work done by gravity is mgh which can be measured easily . kinetic energy of both the blocks can also be measured.
Answer:
C) The pressure reading stays the same.
Explanation:
