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serg [7]
3 years ago
7

An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.60 m

below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the +x-direction and that the +y-direction is up.)
Physics
1 answer:
sergij07 [2.7K]3 years ago
8 0

Answer:

9.495 m/s

Explanation:

initial horizontal speed, ux = 2.60 m/s

initial vertical speed, uy = 0

height, h = 4.60 m

acceleration due to gravity, g = 9.8 m/s^2

Let the speed of fish is v as it hits the water level.

Use third equation of motion in vertical direction

v^{2}= u^{2}+2as

v^{2}= 0^{2}+2\times 9.8\times 4.60

v = 9.495 m/s

Thus, the fish hits the water level with the speed of 9.495 m/s.

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A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

V = \displaystyle \frac{Q}{C}.

The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

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How are strong nuclear forces and weak nuclear forces alike
Nesterboy [21]
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If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of
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Answer:

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The SI units of the “B” is m/s^2

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Time taken to travel = t (seconds)

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Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

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