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True [87]
3 years ago
7

Bungee jumping is an example of

Physics
2 answers:
Shtirlitz [24]3 years ago
5 0
I think it’s c gravitational and elastic energy
muminat3 years ago
4 0
I would say C. Because of elastic energy that’s what bungee jumping correlates to.
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How far will a rubber ball fall in 10 seconds?
Inessa [10]

If it's not moving at all at the beginning of the 10 seconds, then it falls 490 meters straight down in 10 seconds.

(Note: This is true of all objects on Earth . . . rubber balls, feathers, grains of sand, school buses, battle ships . . . everything.  As long as air doesn't hold them back.  Anything falling from rest falls 490 meters in the first 10 seconds.)

4 0
3 years ago
an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0
3 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
yuradex [85]

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

     V=4x8=32V

k.e=eV= 2X32=64eV

3 0
4 years ago
The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started
bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
4 years ago
A friend tells you that a rowboat is propelled forward by the force of its oars against the water. First explain whether the sta
stellarik [79]
It is correct, the action is paddling, where you move the water backwards, and the reaction is the boat moving forwards.
7 0
3 years ago
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