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3 years ago

Gravitational potential energy = the energy due to an objects

Physics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

Gravitational energy is the energy of a body when it is taken to a certain height with respect to the earth surface.It is one of the part of potential energy that comes in thr part of the mechanical energy. the work done in taking a body upto a certain height get stored in the body in form of gravtitational potential energy. potential energy is directly proportional to height or altitude upto which the body is taken, mass of the body and gravtational acceleration(9.8m/s^2)So P.E. = mgh

Hope it helps

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A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

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3 years ago

High energy waves have what

GarryVolchara [31]

High energy waves have Gamma rays

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3 years ago

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What is the height of the image? Round the answer

Maru [420]

It’s a concave mirror

8 0

3 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

4 0

3 years ago

Shareen finds that when she drives her motorboat upstream she can travelwith a speed of only 8 m/s, while she moves with a speed

ANEK [815]

Let vb be the velocity of the motorboat and let vs be the velocity of the stream.

We know that when she drives upstream the velocity is 8 m/s, in this scenario the velocities point in opposite directions, then we have the equations: