All categories

3 years ago

What kind of energy does a rubber band have when it is stretched?

Physics

2 answers:

dalvyx [7]3 years ago

6 0

Answer:

elastic potential energy

Explanation:

RideAnS [48]3 years ago

5 0

Elastic potential energy. When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.

You might be interested in

which way would 2 negatively charged balloons naturally move? what would that do to the amount of potential energy stored in the

zheka24 [161]

Answer:

gsg

Explanation:

5 0

3 years ago

I'm bored. ........................

NeX [460]

Answer:

Explanation:

8 0

2 years ago

Read 2 more answers

Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then t

pashok25 [27]

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

4 0

2 years ago

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

tangare [24]

Answer:

Explanation:

Given

speed of Electron $u=2\times 10^7\ m/s$

final speed of Electron $v=4\times 10^7\ m/s$

distance traveled $d=1.2\ cm$

using equation of motion

$v^2-u^2=2as$

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

$(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}$

$a=5\times 10^{16}\ m/s^2$

acceleration is given by $a=\frac{qE}{m}$

where q=charge of electron

m=mass of electron

E=electric Field strength

$5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}$

$E=248.3\ kN/C$

5 0

3 years ago

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

Time = $\frac{\textup{0.016}}{\textup{23}}$

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

q = 9.98 × 10⁻⁹ C

4 0

3 years ago