A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.
<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>
C. middle layer, density increases with depth as pressure increases
Explanation:
The mantle is the middle layer of the earth. It's density increases with depth as pressure increases.
- Based on the compositional division of the earth, our blue planet is divided into crust, mantle and the core.
- The crust is the lightest and the outermost layer of the earth.
- The mantle is the middle layer. Its density lies between that of the core and crust.
- The mantle is the largest layer and density increases with depth here.
- The core is a metallic ball. It is the most dense part of the earth.
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We shall convert all of the densities to lbs/gal, so the product of
BTU/lbs and lbs/gal gives us the basis of comparison, which was "ratio of energy to volume".
grams / ml x 1 lbs/454 grams → 1 lbs/ 454 ml
1 lbs/454 ml x 3785.41 ml/gal → 3785.41 lbs/454gal
Conversion of g/ml = 8.34 lbs/gal
Looking at each fuel:
Kerosene:
18,500 x (8.34 x 0.82) = 126,517 BTU/gal
Gasoline:
20,900 x (8.34 x 0.737) = 128,463 BTU/gal
Ethanol:
11,500 x (8.34 x 0.789) = 75,673 BTU/gal
Hydrogen:
61,000 x (8.34 x 0.071) = 36,120 BTU/gal
The best fuel in terms of energy to volume ratio is Gasoline.
Gallons required:
BTU needed / BTU per gallon
= 85.2 x 10⁹ / 128,463
= 6.6 x 10⁵ gallons
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
