Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
Answer:
(a) 
(b) 
(c) K.E. = 21.168 J
(d) 
Explanation:
Given:
- mass of a block, M = 3.6 kg
- initial velocity of the block,

- constant downward acceleration,

That a constant upward acceleration of
is applied in the presence of gravity.
∴
- height through which the block falls, d = 4.2 m
(a)
Force by the cord on the block,



∴Work by the cord on the block,


We take -ve sign because the direction of force and the displacement are opposite to each other.

(b)
Force on the block due to gravity:

∵the gravity is naturally a constant and we cannot change it


∴Work by the gravity on the block,



(c)
Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.
mathematically:


K.E. = 21.168 J
(d)
From the equation of motion:

putting the respective values:

is the speed when the block has fallen 4.2 meters.
Answer: W =
J
Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.
The work to transport an ion from a lower potential side to a higher potential side is calculated by

q is charge;
ΔV is the potential difference;
Potassium ion has +1 charge, which means:
p =
C
To determine work in joules, potential has to be in Volts, so:

Then, work is


To move a potassium ion from the exterior to the interior of the cell, it is required
J of energy.
Answer:
Explanation:
To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.
The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>