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salantis [7]
3 years ago
10

What are 90 percent of the stars in space called?

Physics
1 answer:
Naddik [55]3 years ago
6 0
Yes there called stars because when a star in space goes nova it shines in the sky like a little twinkle because it far away the sun is also a star too.<span />
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Which of the following is an example of homeostasis?
Bogdan [553]
The answer is : D

Reasoning:
Homeostasis is the body’s balance
3 0
3 years ago
Read 2 more answers
Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k
tino4ka555 [31]

Answer:

\mathbf{ current(I) =1766.67 \ A}

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

13 \times  current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}

current(I) \times 10.92= 19292

current(I) = \dfrac{19292}{10.92}

\mathbf{ current(I) =1766.67 \ A}

8 0
3 years ago
What term best describes the regular path of a spacecraft or other object around a planetary body?
Law Incorporation [45]
B. Orbit. The planets orbit the sun, the moon orbits earth, etc.
7 0
3 years ago
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
You kick a soccer ball off the ground with a velocity of 15 m/s at an angle of 40° above the horizontal
jeyben [28]

At its maximum height, the ball will have zero vertical velocity, so the ball's velocity at this point is exactly equal to its horizontal velocity.

At any time <em>t</em>, the horizontal component of its velocity is

<em>v</em> = (15 m/s) cos(40°) ≈ 11.49 m/s

so at the highest point of its trajectory, the ball has a velocity of about 11.49 m/s pointed in the positive horizontal direction.

8 0
3 years ago
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