All categories

4 years ago

What is transformational energy?

Physics

1 answer:

jenyasd209 [6]4 years ago

6 0

It is the process of energy being converted into another type of energy.

You might be interested in

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per

Brilliant_brown [7]

Answer:

$a_{cp}=7.77m/s^2$

Explanation:

The equation for centripetal acceleration is $a_{cp}=\frac{v^2}{r}$ .

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference $C=2\pi r$ and the period T, then we have:

$v=\frac{C}{T}=2\pi r f$

Putting all together:

$a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2$

4 0

3 years ago

A 4.5 kg box slides down a 4.3-m-high frictionless hill, starting from rest, across a 2.0-m-wide horizontal surface, then hits a

Rainbow [258]

Answer:

speed before reaching rough surface = 9.18 m/s

speed before hitting spring = 8.70 m/s

spring compression = 82 cm

number of complete trip = 9

Explanation:

Lets say

Position 1: On top of hill

Position 2: down the hill

Position 3: after the rough surface

Position 4: after hitting the spring

We'll strictly use conservation of energy for this equation

Potential energy on top of energy is full converted into kinetic energy down the hill (since surface is frictionless)

Hence, PEg1 = KE2

mgh = (1/2)mv2^2

(4.5)(9.8)(4.3) = (1/2)(4.5)v2^2

189.63 = (1/2)(4.5)v2^2

v2^2 = 2(9.8)(4.3) = 84.28

v2 = sqrt(84.28) = 9.18 m/s

After down the hill, it passes a rough surface. So some of the energy is loss due to friction forces

Friction force, Ff = u (coeff of kinetic friction ) x N (normal force)

Normal force, N = weight of box = mg = 4.5 x 9.8

Ff = 0.22 x 4.5 x 9.8

Work done / Energy loss = Wf = Ff x d (distance)

Wf = 0.22 x 4.5 x 9.8 x 2 = 19.404

Energy after passing the rough surface is totally kinetic energy

KE3 = KE2 - Wf = 189.63 - 19.404 = 170.226

speed after rough surface,

(1/2)mv3^2 = 170.226

v3 = sqrt((2 x 170.226)/4.5) = 8.70 m/s

After hitting the spring, all the kinetic energy is converted into potential energy of spring

170.226 = (1/2)kx^2

x^2 = 2 x 170.226 / 510 {note that constant of spring, k = 510}

x^2 = 0.668

x = sqrt(0.668) = 0.82m (82 cm)

To calculate complete trip before the box coming to rest, note that the only place where it loss energy is at the rough surface.

Energy before the first time pass rough surface = 189.63

Energy loss each time passing rough surface = 19.404

189.63 / 19.404 = 9.773 (9 complete with balance of 0.773)

That mean, the box will pass the rough surface 9 complete trip before coming to rest

8 0

3 years ago

An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

Nimfa-mama [501]

Explanation:

It is given that,

Speed of the electron in horizontal region, $v=1.9\times 10^7\ m/s$

Vertical force, $F_y=4.9\times 10^{-16}\ N$

Vertical acceleration, $a_y=\dfrac{F_y}{m}$

$a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}$

$a_y=5.37\times 10^{14}\ m/s^2$ ..........(1)

Let t is the time taken by the electron, such that,

$t=\dfrac{x}{v_x}$

$t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}$

$t=1.26\times 10^{-9}\ s$ ...........(2)

Let $d_y$ is the vertical distance deflected during this time. It can be calculated using second equation of motion:

$d_y=ut+\dfrac{1}{2}a_yt^2$

u = 0

$d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2$

$d_y=0.000426\ m$

$d_y=0.426\ mm$

So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

How fast must an airplane be moving if it makes a turn with a 7200-m radius with a

nadezda [96]

Answer:

180 m/s

Explanation:

The centripetal acceleration a for an object on a circular orbit with radius r and velocity v is given by:

$a=\frac{v^2}{r}$

Solving the equation for velocity:

$v=\sqrt{ar}$

5 0

4 years ago

At a rock concert, a dB meter registered 130 dB when placed 2.5 m in front of a loudspeaker on stage. (a) What was the power out

kipiarov [429]

Answer:

785.398 W

444.5698 m

Explanation:

I = Intensity of sound

r = Distance

The intensity of sound is given by

$\beta=10log\frac{I}{I_0}\\\Rightarrow 130=10log\frac{I}{10^{-12}}\\\Rightarrow 13=log\frac{I}{10^{-12}}\\\Rightarrow 10^{13}=\frac{I}{10^{-12}}\\\Rightarrow I=10^{-12}\times 10^{13}\\\Rightarrow I=10\ W/m^2$

Power

$P=IA\\\Rightarrow P=I4\pi r^2\\\Rightarrow P=10\times 4\pi\times 2.5^2\\\Rightarrow P=785.398\ W$

The power output of the speaker is 785.398 W

If $\beta=85\ db$

$\beta=10log\frac{I}{I_0}\\\Rightarrow 85=10log\frac{I}{10^{-12}}\\\Rightarrow 8.5=log\frac{I}{10^{-12}}\\\Rightarrow 10^{8.5}=\frac{I}{10^{-12}}\\\Rightarrow I=10^{-12}\times 10^{8.5}\\\Rightarrow I=10^{-3.5}\ W/m^2$

$P=IA\\\Rightarrow P=I4\pi r^2\\\Rightarrow r=\sqrt{\frac{P}{I4\pi}}\\\Rightarrow r=\sqrt{\frac{785.398}{10^{-3.5}\times 4\pi}}\\\Rightarrow r=444.5698\ m$

The distance would be 444.5698 m

8 0

4 years ago

What is transformational energy?​

What is transformational energy?