Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
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
Because it’s made of metal. And metal is a good conductor
You see, during the day the ocean collects heat from the sun. So the air above the ocean get warm at night, but the rest of the air on the land gets cooler because water has the ability to collect energy from the Sun.
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,
E = V/d
However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts
Therefore,
E = 499.32 volts /2.5 m
E = 199.73 N/C
The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.