As you read this in your chair, how fast are you moving relative to the chair? relative to the sun?

1 answer:

5 0

To answer this question, you need to know the definition of Relative Motion:
The motion is relative when it depends on a reference point or referencial system. If you know the reference point, you can determine the velocity of an object.
If you are sitting on your chair, you are not moving relative to it (Your speed is 0 km/s); but as you know, our planet moves around the Sun (Traslation Movement) with a speed of 30.0 km/s. Therefore, you are moving 30.0 km/s relative to the sun.

You might be interested in

Last one I swear, 50 points again! plz help this is schoolwork

Anit [1.1K]

It is a stretch of the atmosphere ranging from the upper mesosphere to the lower parts of the thermosphere. It’s useful to us in radio communication.

8 0

2 years ago

Read 2 more answers

A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil

emmainna [20.7K]

(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = 5/6 mile

(5/6 mile) x (1,760 yd/mile) = 1,466 and 2/3 yards

3 0

3 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$