Answer:
Explanation:
1. We can find the temperature of each star using the Wien's Law. This law is given by:
(1)
So, the temperature of the first and the second star will be:
Now the relation between the absolute luminosity and apparent brightness is given:
(2)
Where:
- L is the absolute luminosity
- l is the apparent brightness
- r is the distance from us in light years
Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂
If we use the equation (2) we have:
So the relative distance between both stars will be:
(3)
The Boltzmann Law says, (4)
- σ is the Boltzmann constant
- A is the area
- T is the temperature
- L is the absolute luminosity
Let's put (4) in (3) for each star.
As we know both stars have the same size we can canceled out the areas.
I hope it helps!
Answer:
(a) 300 W
(b)0.90 J
Explanation:
(a) The electric power for any circuit element is given as
Here I is current passing in the circuit and V is the voltage.
Given I = 12 A and V =25 V.
Substitute the given values, we get
P = 300 W.
Thus, the power defibrillator deliver to the body is 300 W.
(b) To calculate the electric energy use the relation between electric energy, power and time as
Given t = 3 ms.
Substitute the values, we get
E = 0.90 J.
Thus, the energy transferred is 0.90 J.
When it is not moving or on the ground
Answer: An IR remote (also called a transmitter) uses light to carry signals from the remote to the device so it can be controlled. It emits pulses of invisible infrared light that correspond to specific binary codes. These codes represent commands, such as power on, volume up, or channel down.
Explanation:
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s