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zloy xaker [14]
3 years ago
8

Why do the constellations seem to move around in the sky?.

Physics
1 answer:
Gnoma [55]3 years ago
8 0

Answer: As Earth spins on its axis, we, as Earth-bound observers, spin past this background of distant stars. As Earth spins, the stars appear to move across our night sky from east to west, for the same reason that our Sun appears to “rise” in the east and “set” in the west.

Explanation:

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A wooden block of mass 2kg is 20cm thick ,10cm wide and 30cm tall
Kitty [74]

Answer:

A.33333.3N/M2

B.100000N/M2

Explanation:

HAVE A NICE DAY ;]

The formula is  \frac{Newton}{Meter^{2} }

10 Newton=1 kilogram

6 0
3 years ago
Read 2 more answers
Radio waves from an fm station have a frequency of 103.1 mhz. if the waves travel with a speed of 3.00 ´ 10 m/s, what is the wav
ryzh [129]
The frequency of the radio waves from the fm station is:
f=103.1 MHz = 103.1 \cdot 10^6 Hz
And the speed of the waves corresponds to the speed of light:
v=3 \cdot 10^8 m/s
Therefore, the wavelength of the radio waves can be found by using the following equation:
\lambda= \frac{v}{f}= \frac{3 \cdot 10^8 m/s}{103.1 \cdot 10^6 Hz}=2.91 m
3 0
3 years ago
A stone is dropped into water from a bridge 52 m above the water's
dusya [7]

Answer:

Its final velocity and how much time it takes to reach the water

Explanation:

The motion of the stone is a uniformly accelerated motion, so we can use the following suvat equation to determine its final velocity:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

s = 52 m is the distance covered during the fall

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0^2+2(9.8)(52)}=31.9 m/s

We can also find how much time it takes to reach the water, using the equation

v=u+at

where

v = 31.9 m/s is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2

t is the time

And solving for t,

t=\frac{v-u}{a}=\frac{31.9-0}{9.8}=3.26 s

3 0
4 years ago
Sand falls from a conveyor belt at a rate of 30 m3m3/min onto the top of a conical pile. The height of the pile is always 3535 o
Reil [10]

Explanation:

As the given data is as follows.

         h = \frac{3}{5}d

            = \frac{3}{5} \times (2r)

Also, we know that r = \frac{4}{3}h

and       Volume (V) = \frac{1}{3} \pir^{2}h

                         = \frac{1}{3} \pi (\frac{4}{3}h)^{2} h

                         = \frac{16}{27} \pi h^{3}

And,  \frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}

         \frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}

Putting the given values into the above formula as follows.

       \frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}

       30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}    

          [tex]\frac{dh}{dt} = 1.343 m/min

or,                        = 134.3 cm/min     (as 1 m = 100 cm)

thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.

4 0
3 years ago
How do you find the average speed of an entire trip? I could really use some explaining, or......you could just answer the pictu
padilas [110]

This one is easy. You just find the rise over run for the graph. Here, its 3/2 m/s.

5 0
3 years ago
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