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2 years ago

A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.

Physics

1 answer:

Vika [28.1K]2 years ago

4 0

The change in the object’s gravitational potential energy is 2.11 J.

<h3>Change in the object's gravitational potential</h3>

ΔP.E = mg(hf - hi)

where;

m is mass of the object
hf is final height
hi is initial height

ΔP.E = 0.54 x 9.8(1.1 - 0.7)

ΔP.E = 2.11 J

Thus, the change in the object’s gravitational potential energy is 2.11 J.

Learn more about gravitational energy here: brainly.com/question/15896499

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During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the e

cupoosta [38]

The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

<h3>What is an atomic bomb?</h3>

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

energy released, W,

the elapsed time, t, and

the air density.

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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7 0

3 years ago

What is kepler's law??

Elan Coil [88]

<h2>QUESTION:- </h2>

➜what is kepler's law??

$\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}$

Kepler gave the three laws or theorems of motion of the orbitals bodies

${\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}$

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

${\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}$

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

$\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL$

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

${\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}$

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

${T}^{2} = {a}^{3}T2=a3$

where:-

T = time of revolution

a is the distance between the planet and star.

$\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}$

8 0

3 years ago

1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom.

oksian1 [2.3K]

1. $1.0\cdot 10^{-6}m$

First of all, let's convert the energy of the absorbed photon into Joules:

$E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J$

The energy of the photon can be rewritten as:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m$

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

$\lambda=1.66\cdot 10^{-6}m$

So the energy of the emitted photon is given by the formula used previously:

$E=\frac{hc}{\lambda}$

and using

$\lambda=1.66\cdot 10^{-6}m$

we find

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J$

converting into electronvolts,

$E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV$

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.

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3 years ago

Angel hits a ball with a ball. The action is the ball . What what is reaction force?

Mumz [18]

When Angel hits a ball with the ball. As the action is a ball as it is hitted. The reaction force is the force exerted on another ball.

Newton’s third law of motion describes the two forces, action and reaction forces. This states that for every action force, there is an equal and opposite reaction force.

As Angel hits a ball with another ball, the action is the ball as it is hit. The ball exerts a force on the ball. This is the action force. The ball exerts an equal and opposite force on the bat, which is known as the reaction force.

Learn more about Newton's Law of motion:

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2 years ago

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

bonufazy [111]

Answer:

(a) $P=33000W$