How is motors and generators alike

Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 10 m/s.

abruzzese [7]

Hi there!

Kinetic energy can be calculated using the following:

$\large\boxed{KE = \frac{1}{2}mv^2}}$

Where:

KE = Kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

$KE = \frac{1}{2}(625)(10^2) = \boxed{31250J}$

5 0

2 years ago

Read 2 more answers

physical changes occur when existing bonds break and new bonds form

Luba_88 [7]

<span>A physical change is any change that alters the form or appearance of a substance but does not change it into another substance
-Some of the physical properties may be altered, but the chemical composition stays the same
-Examples: Bending, Crushing, Cutting, Melting, Freezing, Boiling</span>

6 0

3 years ago

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$

8 0

4 years ago

Does the sun always shine directly overhead at the equator at noon

Nikitich [7]

The sun always shines directly overhead at noon. This is because the equator always gets the equivalent amount of sunlight. The area always get 12 hours of sunlight, because it's 0 degrees north and south and it's at the center of the Earth.

4 0

3 years ago

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$