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2 years ago

If the sun is a medium sized star, why does it look bigger than others?

Physics

2 answers:

hjlf2 years ago

8 0

Answer:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller. The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Explanation:

nata0808 [166]2 years ago

5 0

Answer:

The sun looks bigger than other stars because it is closer to the Earth, distance makes it look larger

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A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

2 years ago

How could you use the microscopes from this activity to determine

OlgaM077 [116]

Answer:

The magnification is a function of the lenses in the objective and the eyepiece, so the magnification of the two must be multiplied to obtain the total magnification possible. So, for example, if the objective lens was 4X and the eye piece lens was 10X, the total magnification would be 40. (4 x 10 = 40)

Explanation:

5 0

2 years ago

How does the orbital speed of an asteroid in a circular solar orbit with a radius of 4.0 AU compare to a circular solar orbit wi

Murrr4er [49]

Answer:

The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU

Explanation:

am = mvr= angular momentum

am4= 4mvt

am1= mvp1

Vt=1/4vp

Vp=4vt

am1= 4mvt

am1=am4

The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU

6 0

3 years ago

Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha

GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

Thus, the solutions are;

t1 = (1/(18 × 10³)) cos^(-1) (2.5/5)

And

t2 = (1/(18 × 10³)) cos^(-1) (-2.5/5)

Angle of the cos function is in radians, thus;

t1 = 58.18 × 10^(-6) s

t2 = 116.36 × 10^(-6) s

So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

4 0

2 years ago

What change should be expected in the velocity of a body to maintain the same

Readme [11.4K]

Answer:

firstly,

ke=1÷2mv^2

on putting same ke by increasing mass by 16 times new velocity becomes v'

then

ke=1÷2×16m×v'^2

from above we can write

1÷2mv^2=1÷2×16m×v'^2

v^2=16v'^2

1÷4v=v'

hence original velocity should be decreased by 4 times to keep same ke

4 0

3 years ago