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Blababa [14]
2 years ago
12

Describe a situation where an object would have potential energy transformed into kinetic energy?

Physics
2 answers:
tigry1 [53]2 years ago
6 0

Answer:

When an object falls, its gravitational potential energy is changed to kinetic energy.

Explanation:

Alinara [238K]2 years ago
5 0
A prime example is a rollercoaster. As the rollercoaster goes up the hill it has a ton of potential energy and as it falls it is turned into kinetic energy which keeps the coaster moving.
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A way to charge insulators and conductors (the answer in 7 alphabets)
Fudgin [204]

Answer

RUBBING

Explanation:

3 0
3 years ago
1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len
igomit [66]

Answer:

a) \alpha = -0.233 rad/s^{2}

b) the rotational acceleration will remain the same,  \alpha = -0.233 rad/s^{2}

c) When r = 36 m, a_{c} = 157.25 m/s^{2}

   When r = 18 m, a_{c} = 78.63 m/s^{2}

d) When r = 36 m, a_{c} = 39.31 m/s^{2}

   When r = 36 m, a_{c} = 19.66 m/s^{2}

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60  = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, a_{c} = w^{2} r

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}

At the halfway point, r = 18 m

a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}

When r = 18 m

a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}

8 0
3 years ago
A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f
scoray [572]

Answer:

BA

0

\dfrac{NBA}{\Delta t}

Explanation:

B = Magnetic field

A = Area

\theta = Angle

t = Time taken

Before rotation the magnetic flux is given by

\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin90\\\Rightarrow \phi_i=BA

Magnetic flux is BA

After rotation the magnetic flux is given by

\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin0\\\Rightarrow \phi_i=0

The magnetic flux is 0

Magnitude of emf is given by

\epsilon=NA\dfrac{d\phi}{dt}\\\Rightarrow \epsilon=\dfrac{NBA}{\Delta t}

The magnitude of the average emf induced in the entire coil is \dfrac{NBA}{\Delta t}

4 0
3 years ago
Sug<br>. To repeat the experiment what is it?<br>​
statuscvo [17]

Answer:

....more foam

Explanation:

7 0
3 years ago
To test the performance of its tires, a car
Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

\mu mg=m\frac{v^2}{r}

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

\mu is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

g=9.8 m/s^2

And re-arranging the equation for \mu, we can find the coefficient of static friction:

\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0
3 years ago
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