Answer:
ΔP = (640 N/cm^2)
Explanation:
Given:-
- The volume increase, ΔV/V0 = 4 ✕ 10^-3
- The Bulk Modulus, B = 1.6*10^9 N/m^2
Find:-
Calculate the force exerted by the moonshine per square centimeter
Solution:-
- The bulk modulus B of a material is dependent on change in pressure or Force per unit area and change in volume by the following relationship.
B = ΔP / [(ΔV/V)]
- Now rearrange the above relation and solve for ΔP or force per unit area.
ΔP = B* [(ΔV/V)]
- Plug in the values:
ΔP = (1.6*10^9)*(4 ✕ 10^-3)
ΔP = 6400000 N/m^2
- For unit conversion from N/m^2 to N/cm^2 we have:
ΔP = (6400000 N/m^2) cm^2 / (100)^2 m^2
ΔP = (640 N/cm^2)
Explanation:
When m=<em>mass</em>
G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>
<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>
<em>M</em><em>g</em><em>h</em>
<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>
6×10×h
=60joules
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
Answer:
b. The current stays the same.
Explanation:
In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .
Now, when an identical bulb is connected in parallel to the original bulb .
Therefore, both the resistance( bulb) are in parallel.
We know, when two resistance are in parallel , current through them is same and voltage is divided between them.
Therefore, in this case current stays same in the original bulb.
Hence, this is the required solution.