Answer:
<h3>2,321.62Joules</h3>
Explanation:
The formula for calculating workdone is expressed as;
Workdone = Force * Distance
Get the force
F = nR
n is the coefficient of friction = 0.5
R is the reaction = mg
R = 46 ( 9.8)
R = 450.8N
F = 0.5 * 450.8
F = 225.4N
Distance = 10.3m
Get the workdone
Workdone = 225.4 * 10.3
Workdone = 2,321.62Joules
<em>Hence the amount of work done is 2,321.62Joules</em>
My calculator is about 1cm thick, 7cm wide, and 13cm long.
Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .
The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.
If its density is actually 1 g/cm³, then its mass is about 91 grams.
The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.
My result of 91 grams is about 100 grams ... about 10² grams.
Your results could be different.
Answer:
Iron
Explanation:
The buoyant force equals to the weight of water being displaced by the object.
Since, the volume of both iron and wood is equal and the wood is not completely submerged, but the iron block is completely submerged i.e more volume of the water is being displaced by the iron block.
Hence, the buoyant force is more on the iron.
Explanation:
<h3>The beam balance is a device used</h3><h3> for the determination of the mass of </h3><h3>a body under gravitation. It consists </h3><h3>of a beam supported at the centre </h3><h3>by an agate knife edge resting on a</h3><h3> support moving inside a vertical </h3><h3>pillar. The beam carries a light</h3><h3>pointer which moves over a scale.</h3>
Answer:
Explanation:
Given data:
v = 220 rms
power factor = 0.65
P = 1250 W
New power factor is 0.9 lag
we knwo that
s = 1923.09 < 49.65^o
s = [1250 + 1461 j] vA
![P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]](https://tex.z-dn.net/?f=P.F%20new%20%3D%20cos%20%5Btan%5E%7B-1%7D%20%5Cfrac%7BQ_%7Bnew%7D%7D%7BP%7D%5D)
solving for 
![Q_{new} = P tan [cos^{-1} P.F new]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%20P%20tan%20%5Bcos%5E%7B-1%7D%20P.F%20new%5D)
![Q_{new} = 1250 [tan[cos^{-1}0.9]]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%201250%20%5Btan%5Bcos%5E%7B-1%7D0.9%5D%5D)
Faraday
