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blondinia [14]
4 years ago
6

I have a science test with a bonus question that I’m trying to figure out. Why would it be a bad idea to skydive on the moon. Hi

nt: It has nothing to do with gravity. Get back to me about it please.
Physics
1 answer:
Talja [164]4 years ago
7 0
The no atmosphere on the moon
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Consider that each tick mark represents 1 mile on the vector grid, use the Gizmo to simulate the following scenario: Imagine tha
Alexxandr [17]

Answer:

The best description of the general direction that the blue line points is East direction

4 0
4 years ago
Which of the following would cause an atom to have a positive electric charge? A. having more protons than electrons B. having m
Ostrovityanka [42]
In neutral state the atom got an equal number of protons (+) and electrons (-). If you lose and electron then you will have 1 more proton than electron, so you have a +, but if you gained an electron, then you'd have one more electron than protons, so you'd get a -. 
<span>Fx: carbon has 6 protons and 6 electrons. If carbon loses an electron, then the carbon atom would have 6 protons and only 5 electrons, then you'd have 6+ and 5-, which is 1 more plus and minuses, so you have C+. </span>
<span>If Carbon gained an electron, then you'd have 6+ and 7-, which is 1 more minus than pluses, so you'd have C-. </span>
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5 0
3 years ago
Read 2 more answers
How much work, in Joules. is done when a pressure of 95.000 Pa changes a volume from 15 ma to 43 m2?
gayaneshka [121]

Answer:

-2.66 kJ

Explanation:

At constant pressure, work done is:

W = -P ΔV

W = -(95.000 Pa) (43 m² − 15 m²)

W = -2660 J

W = -2.66 kJ

5 0
4 years ago
The unit of work is called derived unit.Why​
Artemon [7]

Answer:

the unit of work is derived unit because joule is defined the work done by the force aftab 1 newton causing the displacement of one metre something newton metre(n-m) is also used to measuring work.

6 0
3 years ago
You illuminate a slit with a width of 71.7 μm with a light of wavelength 737 nm and observe the resulting diffraction pattern on
miskamm [114]

To solve this problem we will apply the concepts related to the double slit-experiment. Under this concept we understand the relationship between the minimum angle, depending on the order of the fringes, the wavelength and the distance between slits. Therefore we have the following relation,

sin(\theta_{min}) = \frac{m\lambda}{D}

Here,

m = Order of the fringes

D = Distance between slits

\lambda = Wavelength

Replacing with our values we have,

sin(\theta_{min}) = \frac{(1)(737*10^{-9})}{71.7*10^{-6}m}

sin(\theta_{min}) = 0.01028

Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then

Y=Lsin\theta

Y  =2.27*0.01028

Y =0.0233m

Thus the width of the central maximum is

W=2Y=2*0.0233

W = 0.0466m

Therefore the widht is 0.466m

6 0
4 years ago
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