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
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,

Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,

Therefore, the resistance of the bulb is 484 Ω
Distance travelled in south direction= 1.5hr*0.75km/hr= 1.125km
Distance travlled in north direction= 0.90*2.5=2.25
Net displacement = 2.25-1.125= 1.125 to the north
Answer:
the rate of change in volume with time is 280πr² cm³/min
Explanation:
Data provided in the question:
Radius of the sphere as 'r'
= 70 cm/min
Volume of the sphere, V =
Surface area of the sphere as 4πr²
Now,
Rate of change in volume with time,
=
=
Substituting the value of 
=
= 280πr² cm³/min
Hence, the rate of change in volume with time is 280πr² cm³/min