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xxMikexx [17]
3 years ago
15

I need the answer for both of the questions please

Physics
1 answer:
Lady_Fox [76]3 years ago
5 0
But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain
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What is quantum programing
Komok [63]

Answer: Quantum programming is the process of assembling sequences of instructions, called quantum programs, that are capable of running on a quantum computer. Quantum programming languages help express quantum algorithms using high-level constructs.

Explanation:

7 0
4 years ago
Compare the gravitational acceleration on the following objects compared to the Sun using:
arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0
3 years ago
A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
Rama09 [41]
Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
Using mirror formula: 
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)


Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
3 0
3 years ago
Read 2 more answers
Suppose that the average U.S. household uses 15600 kWh (kilowatt‑hours) of energy in a year. If the average rate of energy consu
sweet [91]

Answer:

a) 166.4 s

b) (2.155 × 10⁷) s

Explanation:

15600 KWh for a year,

1 year consists of 365 × 24 hours = 8760 hours.

So, the power consumed in a year for an average household = (Energy/time)

= (15600/8760) = 1.781 KW = 1781 W

a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take

The power required for this lifting = (mgh/t)

m = 1800 kg

g = 9.8 m/s²

h = 16.8 m

t = ?

P = 1781 W

1781 = (1800×9.8×16.8)/t

t = (296,352/1781)

t = 166.4 s

b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km

The power required for this lifting = (mgh/t)

m = 405000 kg

g = 9.8 m/s²

h = 9.67 km = 9670 m

t = ?

P = 1781 W

1781 = (405000×9.8×9670)/t

t = (38,380,230,000/1781)

t = 21,549,820 s = (2.155 × 10⁷) s

Hope this Helps!!!

8 0
3 years ago
Do magnets have to touch each other in order to experience a magnetic<br> force? Explain
tiny-mole [99]
Magnets don’t need to touch each other in order to be magnetic because all objects have an attraction however, the closer the magnets get, the more magnetic it is.
4 0
3 years ago
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