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Ksivusya [100]
2 years ago
13

What feature of a planet's orbit does Kelper's first law of motion describe?

Physics
1 answer:
IRISSAK [1]2 years ago
5 0
The answer is going to be letter B
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Which is an example of potential en
SIZIF [17.4K]

a yoyo in someones hand is an example of potential energy

8 0
3 years ago
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Why do hot stars look bluer than cool stars?
brilliants [131]

Answer:

(B) The wavelength that a star radiates the most energy is inversely proportional to the temperature.

Explanation:

As we know that

According to Wien's law wavelength is inverse proportional to the temperature .

  λ.T = Constant.

λ.∝ 1 /T

As we know that star radiates wavelength and this wavelength is inverse proportional to the temperature of the star.

The temperature of cool star is cooler than the temperature of hot star that is cool star looks red and hot star looks blue.Cool star have low energy and hot star have high energy.

So option B is correct.

(B) The wavelength that a star radiates the most energy is inversely proportional to the temperature.

5 0
3 years ago
A wave with a period of 0.008 second has a frequency of
coldgirl [10]

The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

         1 / 0.008 sec  =  125 per sec .  (125 Hz)

8 0
3 years ago
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A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of
Soloha48 [4]

Answer:

The length of her shadow is changing at the rate  -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 =  (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

7 0
3 years ago
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A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete
ale4655 [162]

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0
2 years ago
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