How fast, in rpm, would a 5.8 kg, 22-cm-diameter bowling ball have to spin to have an angular momentum of 0.21 kgm2/s

What is the difference between light energy and heat energy

MrRissso [65]

ANSWER:

What is the difference between heat and light? - Physics Stack ... Heat and light are different but they are both forms of energy. Heat is a form of kinetic energy contained in the random motion of the particles of a material. Light is a form of electromagnetic energy. As with other forms of energy, heat energy can be transformed into light energy and vice versa.

4 0

3 years ago

At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni

creativ13 [48]

Answer:

3. 5.0N/kg

Explanation:

Gravitational field strength = gravitational force/mass of astronaut = 350N/70kg = 5.0N/kg

6 0

3 years ago

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of

Anestetic [448]

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

$KE = \frac{1}{2}m* v^{2}$

It is given:

v = 8.5 m/s

m = 91 kg

So:

$KE= \frac{1}{2}*91*8.5^{2} =3,287.4J$

b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height

Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

3 0

3 years ago

What is the cow's average velocity to reach the apples?

docker41 [41]

Answer:1.). -0.65

2.). 0.65

Explanation:

3 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago