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3 years ago

Describe the motion of the moon in space to produce the different phases of the moon.

Physics

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

Moon slowly moves eastward, rising later each day and passing through its phases:

Explanation:

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If A = 5 and C = 13, what is sin x?

lisabon 2012 [21]

By pythagorean theorem then, the vertical side of the right triangle must be 12.
Then if x is the angle between the horizontal side and the hypotenuse, sin(x) = 12/13 but also the anser should be in this sentences.

6 0

3 years ago

Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto

mars1129 [50]

I’ve answered this before so I know the question is missing an important given and that given is: 1 has an empty trailer and the other has a fully loaded one.

So, it would be the fully loaded trailer that would take a longer distance to stop because a lot of weight is being pulled, and when the brakes are started, the fully loaded trailer is more like pushing against the truck.

6 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

Which classification scheme is based on the brightness of a star as it appears to you in the night sky?

anastassius [24]

The apparent magnitude scale is a classification scheme which is based on the brightness of stars. The range of brightness values is from 1 to 6.
The stars which are the most brightest are ranked as number 1 and also called first magnitude stars, stars which are little dimmer than number 1 are ranked as number 2 and also called second magnitude stars. Similarly the most faintest stars are ranked number 6 and also called as the sixth magnitude stars.

3 0

3 years ago