Answer:
At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.
Explanation:
Answer: E) A) salt water.
Explanation:
E) In equilibrium, pressure exerts equally in all directions, so for a given depth, the pressure is the same for all points located at the same depth, and it can be written as follows:
p = p₀ + ρ.g.h, where p₀ = atmospheric pressure, ρ=fluid density, h=depth from the surface.
A) The buoyant force, as discovered by Archimedes, is an upward force, that opposes to the weight of an object (as it is always downward), and is equal to the weight of the volume of the liquid that the object removes, which means that is proportional to the density of the liquid.
As salt water is denser than fresh water, the buoyant force exerted by the salt water is always greater than the one produced by the fresh water, so objects will float more easily in salt water than in fresh water.
In the limit, it is possible that one object float in salt water and sink in fresh water.
When Emmett is lifting a box
vertically, the forces that must be added to calculate the total force are: the
gravitational force, tension force(the force exerted by Emmett to the box and
the force exerted by the box to Emmett), and air resistance force.
The sun orbits the eth at 2kilogram per sec
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.