Answer:
the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.
That is,
Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100
Explanation:
Let the intensity of light be represented by I
Let the distance of the star be d
I ∝ (1/d²)
I = k/d²
For the earth,
Iₑ = k/dₑ²
k = Iₑdₑ²
For the other planet, let intensity be Iₒ and distance be dₒ
Iₒ = k/dₒ²
But dₒ = 10dₑ
Iₒ = k/(10dₑ)²
Iₒ = k/100dₑ²
But k = Iₑdₑ²
Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100
Iₒ = Iₑ/100
Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.
Answer:
The answer is "Option C".
Explanation:
It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.
hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.
A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
I’m pretty sure it’s true x
They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>
