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AnnZ [28]
3 years ago
5

I have another one of these similar to it, if you need another 20 points then look in my profile.

Physics
2 answers:
zimovet [89]3 years ago
8 0

Answer:

NEW

Explanation:

As the moon moves to position Z it becomes a New moon due to the light of the sun shining on it

mina [271]3 years ago
3 0

Answer:

waxing

Explanation:

waxing is when a 1st quarter moon is transitioning to a full moon

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Match each example to its type of energy.
PtichkaEL [24]
Solar energy - A

nuclear energy - B

fossil fuel energy - C

wind energy - D

geothermal energy - E

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All of the outer planets are much larger than the inner planets true or false
Rama09 [41]
That statement is false because yes Jupiter and Saturn are large, however, planets like Uranus and Neptune are quite small if not smaller than some of the inner planets. 
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A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a
Kobotan [32]

Answer:

Explanation:

A

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3 years ago
Please help, all about Mirrors and Lenses
romanna [79]
Answer: I think it’s 20cm.
3 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
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