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2 years ago

if you stretch back a rubber band and release it, it shoots across the room. What type of energy conversion has occurr?

Physics

2 answers:

matrenka [14]2 years ago

8 0

I believe the answer is potential energy if i remember correctly.

kati45 [8]2 years ago

7 0

Potential energy :)))

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The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. $\omega _0=0 rad/s$

after $t=10 s$

$\omega =10 rad/s$

using $\omega =\omega _0+\alpha t$

$10=0+\alpha \cdot 10$

$\alpha =1 rad/s^2$

Revolutions turned in 2 s

$\theta =\omega _0t+\frac{\alpha t^2}{2}$

$\theta =0+\frac{1\times 2^2}{2}$

$\theta =2 rad$

To get revolution $\frac{\theta }{2\pi }$

= $\frac{2}{2\pi}=0.318\ revolutions$

3 0

3 years ago

Read 2 more answers

a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s

Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

= 6/15

So,

15x = 6x + 6d

9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

= (2/3) (d/dt)

Also,

Given:

d(d)=dt

= 7 ft/s

Thus,

d(d + x)=dt

= (7/3)d (d/dt)

Substitute, d= 7

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0

3 years ago

In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol

igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s$

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m$

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0

3 years ago

Please help me with this question

ladessa [460]

Answer:

conduction

Explanation:

5 0

3 years ago

Read 2 more answers

As the distance between two charged objects increases, the strength of the electrical force between the objects

GuDViN [60]

Answer:

I believe the answer is It increases

4 0

3 years ago