Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Conduction involves physical contact to charge, well induction does not.
Learn more at: <span>www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction</span>
Answer:
K = 588.3 N/m
Explanation:
From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):
Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:
By conservation of energy:
Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:
K = 588.3 N/m
Answer:
D by spinning magnets to induce electric current.
The answer is C.
The question says the potential difference is what is changing, which means we're solving for V.
It tells us that potential difference increases by a factor of two, which just means V doubles.
With this info, we can pick some numbers, plug it into Ohms law and see what happens.
Here's an example where I just picked random numbers that are easy to work with:
V=I*R
10=I*5
I=2
Lets increase the potential difference (V) by a factor of two and see what happens to current:
V=I*R
20=I*5 (all I've done is double the potential difference from 10 to 20)
I=4
When we increase V by a factor of 2, I increases by a factor of 2. We went from I=2 to I=4.
We can increase V by factor of 2 again and see:
V=I*R
40=I*5
I=8
Okay, current just increased by a factor of 2 again when we increased the potential difference by a factor of 2.
It's always good to check work with alternate numbers, so here's one more set:
V=I*R
16=I*4
(remember, we know we're solving for V, so I'm just plugging in random numbers for I and R)
I=4
Increase V by factor of 2:
32=I*4
I=8
So, when we increase V (the potential difference) by a factor of 2, I (current) always increases by a factor of 2 as well.
Hope this helps!
