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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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3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the
const2013 [10]

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car  = 7.65 x 10³N

Unknown:

Acceleration of the car  = ?

Solution:

From Newton's second law of motion:

      Force  = mass x acceleration

   Acceleration  = \frac{Force }{mass}   = \frac{7.65 x 10^{3} }{2.5 x 10^{3} }    = 3.06m/s² to the east

6 0
3 years ago
One of your delivery trucks traveled 1,200 miles on 55 gallons of gas. How many miles per gallon did the truck get? (Round off y
Aleksandr-060686 [28]
The word "Per" means divide

"miles per gallon" is the same as "miles / gallon"

The truck went 1,200 miles
on 55 gallons

1,200 ÷ 55 = 21.81
7 0
3 years ago
Read 2 more answers
Deadpool is doing a superhero landing from a 21 meter tall building what would his velocity be right before he hits the ground?
Morgarella [4.7K]

Answer:

I think he would be dead poggers

Explanation:

4 0
3 years ago
Do quasars reside within or without side of galaxies?
sveticcg [70]

They almost entirely reside within galaxies because quasars are a subset of blackholes with a large and fast enough accretion disk to generate a beam of interstellar material perpendicular to itself. This typically only occurs in the largest black holes at the center of galaxies (supermassive blackholes) or at least stellar black holes---which still occur within galaxies because the material is necessary to form them.

6 0
3 years ago
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
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