Answer:
a. The Sun is 110.24 cm.
b. The distance of the nearest star at 4.24 light years is 3.1586 × cm.
Explanation:
Given a scale :
12700 Km : 1 cm
a. How large is the Sun of diameter 1.4 × Km?
Thus using the given scale, we have to compare the diameter of the Sun to that of the Earth.
=
= 110.2362
The Sun is 110.24 cm.
b. how far away is the nearest star at a distance of 4.24 light-years on this scale?
A light year is a term that is used to express the total distance that light would travel in an empty space in a year.
One light year = 9.461 × Km
So that,
4.24 light years = 4.24 × 9.461 × Km
= 40.11464 × Km
Using the given scale,
= 3.1586 ×
The distance of the nearest star at 4.24 light years is 3.1586 × cm.
Answer:
The answer is "".
Explanation:
Using formula for calculating the Voltage:
Hence the range of accelerating in voltage is
Answer:
The formula i use is called, Product over Sum. Which means it is figured by their multiplied resistances divided by their sum. It is applied by pairs of known resistances. Starting with 20 and 30 Ohms, 600 is divided by 50. Using a quick mental calculation, the first pair has a resistance of 12 Ohms. Then, do that with 12 Ohms and 10 Ohms. 120 Ohms divided by 22. The answer is, about 5.5 Ohms. By this interesting development, we are reminded that resistances in parallel are effectively never more than the least one.
The students decide to assemble a convenient experiment and will run one amp through them all in parallel and measure their voltage. Watching the Amperage gauge on their teacher’s power supply. As one begins to turn it up to an Amp, another is watching its voltage till an Amp is perfectly applied. But as they carefully do that, watching the Amp gauge, another screams, their 10 Ohm resistor turns black and smokes as they were only pumping out 2 or 3 tenths of an Amp. What happened? What did they need, to make this simple experiment not so embarass-king?
Buy room air freshener?
Answer:
a) u = 6 m/s
b) a = 4 m/s²
c) d(3) = 16 m
Explanation:
equation for the first second
distance will be the average velocity times the time of travel
8 = ½(u + (u + at))t t is one second, so reduces to
8 = u + ½a
velocity at the end of the first second is
v = u + at = u + a
position equation for the second period is
12 = ½((u + a) + (u + a + at))t t is one second so reduces to
12 = u + 3a/2
subtracting the first position equation from the second
12 - 8 = u + 3a/2 - (u + ½a)
a = 4 m/s²
8 = u + ½4
u = 6 m/s
in the third second
d = 6(3) + ½(4)(3²) - 8 - 12
d = 16 m
Answer:
The two space stations are 644.1653 miles far apart.
Explanation:
The distance of Huston and the International Space Station is 323 miles.
The distance of Huston and the Chinese space station Tiangong is 462 miles.
Angle between the two stations = 109⁰
So, applying the law of cosines as;
c² = a² + b² - 2abcosC
From the picture shown below, a = 323 miles and b = 462 miles
So,
c² = 323² + 462² + 2(323)(462)cos109 = 104329 + 213444 - 298452cos109
cos109 = -0.3256
So,
c² = 414948.9712
<u>c = 644.1653 miles</u>