<span>Let's convert the speed to m/s:
speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds)
speed = 24.59 m/s
Let's convert the mass to kilograms:
mass = (2135 lb) (0.45359 kg / lb)
mass = 968.4 kg
We can find the kinetic energy KE:
KE = (1/2) m v^2
KE = (1/2) (968.4 kg) (24.59 m/s)^2
KE = 292780 joules
The kinetic energy of the automobile is 292780 joules.</span>
As per the question, the velocity of the airplane [v] = 660 miles per hour.
The total time taken by airplane [t] = 3.5 hours.
We are asked to determine the total distance travelled by the airplane during that period.
The distance covered [ S] by a body is the product of velocity with the time.
Mathematically distance covered = velocity × total time
S = v × t
= 660 miles/hour ×3.5 hours
= 2310 miles.
Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.
Answer:
24 Coulumbs
Explanation:
Given data
time= 1 minute= 6 seconds
P=2 W
R= 12 ohm
We know that
P= I^2R
P/R= I^2
2/12= I^2
I^2= 0.166
I= √0.166
I= 0.4 amps
We know also that
Q= It
substitute
Q= 0.4*60
Q= 24 Columbs
Hence the charge is 24 Coulumbs
Answer:
Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:
Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium
Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:
The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
