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2 years ago

Explain whether the unit of work is a fundamental or derived unit

Physics

2 answers:

lesya [120]2 years ago

7 0

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

klasskru [66]2 years ago

6 0

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

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A stone is dropped from a cliff. What will be its speed when it was fallen 100 m?

Mars2501 [29]

Answer:

final velocity will be44.72m/s

Explanation:

HEIGHT=h=100m

vi=0m/s

vf=?

g=10m/s²

by using third equation of motion for bodies under gravity

2gh=(vf)²-(vi)²

evaluating the formula

2(10m/s²)(100m)=vf²-(0m/s)²

2000m²/s²=vf²

√2000m²/s²=√vf²

44.72m/s=vf

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3 years ago

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Electromagnetic radiation consists of particles called:

Westkost [7]

Photons are particles of electromagnetic radiation.

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2 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

2 years ago

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

4 0

3 years ago

Describe how to make an electromagnet

Black_prince [1.1K]

First you get a tool and remove some of the warping from the tip.hehe. Then you neatly wrap the wires around the nail. Wrap more if you want to make it stronger. Make sure your able to fit the battery. connect the end of the wire to the positive side. Be careful.

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2 years ago