The correct answer is <span>3)
.
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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
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<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
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But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
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Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x 
x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as 
)
where lift coefficient (Cl)= 
=
= 0.245
therefore
induced drag coefficient (Cdi) = = 
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
Answer:
10061.56 m/s
Explanation:
Gravitation potential of a body in orbit from the center of the earth is given as
Pg = -GM/R
Where G is the gravitational constant 6.67x10^-11 N-m^2kg^-2
M is the mass of the earth = 5.98x10^24 kg
R is the distance from that point to the center of the earth = r + Re
Where r is the distance above earth surface, Re is the earth's radius.
R = 1510 km + 6370 km = 7880 km
Pg = -(6.67x10^-11 x 5.98x10^24)/7880x10^3
Pg = -50617512.69 J/kg
The negative sign means that the gravitational potential is higher away from earth than it is at the earth's surface (it shows convention).
This indicates the kinetic energy per kilogram that the chest of jewel will fall with to earth.
For the jewel chest, the velocity V will be
0.5v^2 = 50617512.69
V^2 = 101235025.4
V = 10061.56 m/s
