Energy conversion is the change of energy from ___

A racecar on a straight track, starting from rest*, steps on the

kirill115 [55]

I think the answer is C

6 0

2 years ago

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are char

Marta_Voda [28]

Answer:

KE = 2.535 x 10⁷ Joules

Explanation:

given,

angular speed of the fly wheel = 940 rad/s

mass of the cylinder = 630 Kg

radius = 1.35 m

KE of flywheel = ?

moment of inertia of the cylinder

$I = \dfrac{1}{2}mr^2$

= $\dfrac{1}{2}\times 630\times 1.35^2$

= 574 kg m²

kinetic energy of the fly wheel

$KE = \dfrac{1}{2}I\omega^2$

$KE =\dfrac{1}{2}\times 574\times 940^2$

KE = 2.535 x 10⁷ Joules

the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules

5 0

2 years ago

A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

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

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

1 year ago

A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

3 0

2 years ago

How is velocity ratio of wheel and axle calculatedaad<br>

iogann1982 [59]

Answer:

VR = $\frac{Radius of wheel}{Radius of axle}$

Explanation:

A machine is a device that can be used to overcome a load by the application of an effort through a pivot. Examples are: pulleys, wedge, screw jack, wheel and axle etc.

The wheel and axle is a simple device that can be used to lift a load through a height. Its velocity ratio (VR) can be determined by:

VR = $\frac{Radius of wheel}{Radius of axle}$

Note that for a practical wheel and axle, the radius of the wheel is greater than the radius of the axle.

3 0

2 years ago

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