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2 years ago

11

Which of the following characteristics of stars is affected by a stars temperature?

1 answer:

vodomira [7]2 years ago

7 0

Answer:

(2) Color

Explanation:

The color--the wavelength of the visible waves emitted by a star-- is tied to the star's temperature. In general, the hotter the surface the shorter the emitted wavelengths (and the stronger will the blue component be represented in the visible band), and the cooler the surface the longer the wavelength (with a shift toward the red end of the visible band).

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When catching a baseball, the ball applies a force of 39.6 N to a catcher's glove. If the work done on the catchers glove is 4

cricket20 [7]

Answer:

d = 1.19 m

Explanation:

Given that,

The force applied by the ball, F = 39.6 N

The work done on the catchers glove is 47.5 J

We need to find the distance traveled by the ball. We know that,

Work done, W = Fd

Where

d is the distance traveled

$d=\dfrac{W}{F}\\\\d=\dfrac{47.5 }{39.6 }\\\\d=1.19\ m$

So, it will cover 1.19 m.

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Deimos completes one (circular) orbit of Mars in 1.26 days. The distance from Mars to Deimos is 2.35×107m. What is the centripet

Akimi4 [234]

Answer:

The centripetal acceleration of Deimos is $0.077m/s^{2}$ .

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity of Deimos and r is the orbital distance.

Notice that is necessary to determine the velocity first.

The speed of the Deimos can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (2)

Then, replacing Newton's second law in equation 2 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (3)

However, a is the centripetal acceleration since Deimos almost describes a circular motion around Mars:

$a = \frac{v^{2}}{r}$ (4)

Replacing equation 4 in equation 3 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Where v is the orbital speed, G is the gravitational constant, M is the mass of Mars, and r is the orbital radius.

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(6.39x10^{23}kg)}{2.35x10^{7}m}}$

$v = 1346m/s$

Finally, equation 4 can be used:

$a = \frac{(1346m/s)^{2}}{2.35x10^{7}m}$

$a = 0.077m/s^{2}$

Hence, the centripetal acceleration of Deimos is $0.077m/s^{2}$ .

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