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monitta
3 years ago
11

Suppose an astrophotographer hands you a picture with star trails taken looking toward the north celestial pole. If the star tra

ils are 1/6 of a complete circle, about how many hours was the picture exposed? Hint: How many hours would it take the stars to make a full circle? Report your answer to the nearest whole number
Physics
1 answer:
Reil [10]3 years ago
4 0

Answer:

Our answer will be 4 hours.

Explanation:

It takes 24 hours to complete a circle(360° rotation)

Hence to complete 1/6th of complete circle it will take 24/6= 4 hours

Exact in integers= 4 hrs. Hence our answer will be 4 hours.

You might be interested in
A car is traveling at a speed of 45 km/h into town. It takes the car 2 hours to get there. How far has the car traveled?
hjlf

Answer:

90 km

Explanation:

45 km per hr

2 hrs

45 x 2 = 90

90 km/2 hrs

Hope this helps!

7 0
3 years ago
Read 2 more answers
a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration
slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0
2 years ago
Which of these processes describes the effect Earth's atmosphere has on Earth's hydrosphere?
Ede4ka [16]
Warm, moist air increasing ocean temp
6 0
3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
Why does the moon orbit the earth as the earth orbits the sun?<br><br>will give brainliest​
AysviL [449]
As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky.
6 0
3 years ago
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