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Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C
Answer:
When the temperature increases
Explanation:
Answer:
a.) L = 2.64 kgm^2/s
b.) V = 4.4 m/s
Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.
So,
Radius r = 0.6 m
Mass M = 2 kg
Velocity V = 1.1 m/s
Angular momentum L can be expressed as;
L = MVr
Substitute all the parameters into the formula
L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1
the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1
b. If she pulls her arms into 0.15 m,
New radius = 0.15 m
Using the same formula again
L = 2( MVr)
2.64 = 2( 2 × V × 0.15 )
1.32 = 0.3 V
V = 1.32/0.3
V = 4.4 m/s
Her new linear speed will be 4.4 m/s
Answer:
L = 0.48 H
Explanation:
let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax be the maximum voltage and XL be the be the reactance of the inductor.
Vrms = Vmax/(√2)
= (3.00)/(√2)
= 2.121 V
then:
XL = Vrms/I
= (2.121)/(2.50×10^-3)
= 848.528 V/A
that is L = XL/(2×π×f)
= (848.528)/(2×π×(280))
= 0.482 H
Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.