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Irina18 [472]
3 years ago
11

The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the co

st? In three to five sentences, provide two specific examples of things we have learned from space exploration, and explain how these examples influence your opinion.(4 points)
Physics
1 answer:
-BARSIC- [3]3 years ago
5 0

Sunday, July 20, marked 45 years since the United States put the first two astronauts safely on the moon. The cost for the Mercury, Gemini and Apollo programs was more than $25 billion at the time more like $110 billion in today’s world. The ensuing U.S. space efforts have cost an additional $196 billion for the shuttle and $50 billion for the space station. NASA’s total inflation-adjusted costs have been more than $900 billion since its creation in 1958 through 2014 (more than $16 billion per year). Looking back, have we gotten our money’s worth from the investment?

IamSugarBee

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Which of these would not improve accuracy? 
fgiga [73]
I think it’s D, mostly just because all the other ones seem like they couldn’t be the answer.
6 0
3 years ago
Read 2 more answers
Suppose you have a polarized light source in your lab that has an intensity of 1288 W/m2, where the electric field oscillates in
IRINA_888 [86]

Answer:

644 W/m²

0 W/m²

Explanation:

I_0 = Initial intensity of light = 1288 W/m²

\theta = Angle of filter

Intensity of light is given by

I=I_0cos^2\theta\\\Rightarrow I=1288\times cos^2(45)\\\Rightarrow I=644\ W/m^2

The intensity of the transmitted light is 644 W/m²

I=I_0cos^2\theta\\\Rightarrow I=1288\times cos^2(90)\\\Rightarrow I=0\ W/m^2

The intensity of the transmitted light is 0 W/m²

4 0
2 years ago
Please help!!!!!!!!!!
pshichka [43]

Answer:

D

Explanation:

5 0
3 years ago
What must the charge (sign and magnitude) of a particle of mass 1.42 g be for it to remain stationary when placed in a downward-
Elena L [17]
We have two forces acting on the particle: the weight of the particle, downward, with intensity W=mg, and the Coulomb's force due to the electric field, with intensity F=qE. 

In order to keep the particle in equilibrium, F must point upward. The direction of F depends on the sign of the charge. The electric field's direction is downward, so if we want F to point upward, the charge q must have negative sign.

Then, to find the magnitude of the charge, we should require that the intensity of the two forces acting on the particle is equal:
mg=qE
from which we find q:
q= \frac{mg}{E} = \frac{(1.42 \cdot 10^{-3}kg)(9.81 m/s^2)}{610 N/C}=2.28 \cdot 10^{-5}C
8 0
3 years ago
A 500-kg elevator is pulled upward with a constant force of 6500 n for a distance of 50.0 m. what is the work done by the 6500 n
umka21 [38]
We know
work=force*distance=6500*50=?
plz tell me the result !
and feel free to ask if u hv any more confusion
3 0
3 years ago
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