Kinetic Energy - What does it depend on?

Determine the angular velocity ω of the telescope as it orbits around the Sun.

lara31 [8.8K]

The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun

5 0

2 years ago

The term "ideal gas" refers to a gas for which certain assumptions have been made. Which of the following is not such an assumpt

Jobisdone [24]

Answer:

(A) Consists of a small number of tiny particles that are far apart- relative in their size.

Explanation:

An ideal gas is defined as a simplification of a real gas, with punctual particles, in which all collisions are elastic, with random displacements and with no attractive force between them.

The assumption of the particles being punctual make clear that they do not have size at all. So if they were far apart-relative in their size, they can not collide each other, that is why assumption (B) can not be possible (for that particular case).

It is clear that (A) is not an assumption for an ideal gas, because do not fit in any of its properties.

Elastic collision: It is a case in which the energy is conserved (Kinetic Energy).

Kinetic Energy: It is the energy that will have an object as a consequence of its movement.

6 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0

4 years ago

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

Let's calculate

v = √(1.23² + 2 9.8 1.69)

v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

W = ΔK

-fr x = 0- ½ m v²

Newton's second law

N- W = 0

The equation for the friction is

fr = μ N

fr = μ m g

We replace

μ m g x = ½ m v²

x = v² / 2μ g

Let's calculate

x = 5.89² / (2 0.255 9.8)

x = 6.94 m

6 0

3 years ago

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