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2 years ago

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at maximum extension a bungee cord stores 2.0 X 10^6 J of energy. A 10 kg mass extends the bungee cord 1.3m. what is the maximum

extension of the bungee cord.

1 answer:

Gnoma [55]2 years ago

4 0

Explanation:

A k E increases While P E Decreases

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what branch of Science studies matter and energy and the way they work together with force to cause motion

algol13

The branch of science that studies this is physics

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8 0

3 years ago

What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a

kondor19780726 [428]

<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

s = 0 x t + 0.5 x 9.8 x t²

s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

5 0

2 years ago

An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.

nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

$\epsilon = -L\frac{dI}{dt}$

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

$\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V$

Therefore, the self-induced emf in this inductor is 4.68 mV.

I hope it helps you!

6 0

3 years ago

I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app

Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

initial velocity of the sprinter, u = 18 km/h

final velocity of the sprinter, v = 27 km/h

time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

$initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\$

The time of motion is seconds;

$t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s$

The sprinter’s average acceleration is calculated as follows;

$a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2$

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0

2 years ago

Energy conversion is the change of energy from ____ to another.

Alexeev081 [22]

from one energy form to one

8 0

2 years ago