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3 years ago

How does Earth's rotation affect our view of stars

1 answer:

4 0

Due to rotation of Earth on each day, our East & west gets change in every half day, and comes in it's initial state after a day. So, Stars which are in east relative to a stationary object in the Universe, will be in west relative to Earth after some time, and it changes our view of stars

Hope this helps!

You might be interested in

If the frequency of oscillation of the wave emitted by an fm radio station is 92.4 mhz, determine the wave's period of vibration

lapo4ka [179]

The basic relationship between the frequency of a wave and its period is
$f= \frac{1}{T}$
where f is the frequency and T the period of vibration.

In our problem, the frequency is
$f=92.4 MHz = 92.4 \cdot 10^6 Hz$
so, by re-arranging the previous formula, we can find the period of the wave:
$T= \frac{1}{f}= \frac{1}{92.4 \cdot 10^6 Hz}=1.1 \cdot 10^{-8} s$

3 0

3 years ago

A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0

3 years ago

Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg

n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

$\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N$

Thus, the net force over the body is:

$F=(913.14N)\hat{i}+(274.87N)\hat{j}$

Next, you calculate the magnitude of the force:

$F=\sqrt{(913.14N)+(274.87N)^2}=953.61N$

and the direction is:

$\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°$

7 0

4 years ago

What happens to acceleration when mass is increased

Lubov Fominskaja [6]

Answer:

Newtown's second law of motion

F= ma

a = F/m

if mass is increased then acceleration get decrease

because acceleration is inversely proportional to mass

8 0

3 years ago

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

6 0

3 years ago