Answer:
The rate at which the container is losing water is 0.0006418 g/s.
Explanation:
- Under the assumption that the can is a closed system, the conservation law applied to the system would be:
, where 
is all energy entering the system, 
is the total energy leaving the system and, 
is the change of energy of the system. - As the purpose is to kept the beverage can at constant temperature, the change of energy () would be 0.
- The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by:
where 
is the emissivity of the surface, 
known as the Stefan–Boltzmann constant, 
is the total area of the exposed surface, 
is the temperature of the surface in Kelvin, 
is the environment temperature in Kelvin. - For the can the surface area would be ta sum of the top and the sides. The area of the top would be
, the area of the sides would be 
. Then the total area would be 
- Then the radiation heat transferred to the can would be
. - The can would lost heat evaporating water, in this case would be
, where 
is the rate of mass of water evaporated and, 
is the heat of vaporization of the water (
). - Then in the conservation balance:
, it would be
. - Recall that
, then solving for :
Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, 
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
M is the mass of the sun
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
Answer:
Options d and e
Explanation:
The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.
We can get the length of the pendulums likely to oscillate with the formula;
where g=9.8m/s
ω= 2rad/s to 4rad/sec
when ω= 2rad/sec
L = 2.45m
when ω= 4rad/sec
L = 9.8/16
L=0.6125m
L is between 0.6125m and 2.45m.
This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.
Have a great day ahead
Answer:
(b) EAST
Explanation:
you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:
You use the Lorentz formula to known which is the direction of the magnetic force over the electron:
which implies the cross product between the unitary vecors j and i, that is
(WEST)
However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST
