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harina [27]
3 years ago
11

Fog rolls in and bright headlights make it hard to see in the fog

Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:F., D.

Explanation:If you use low beams it should help and if you use the defogger it should also help.

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What is the minimum mass needed for a star to be on the main sequence? What happens to stars that do not meet the minimum mass?
Mariana [72]
Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available. 

To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.
6 0
3 years ago
Raina is running for student body president and doesn't want to forget her campaign speech. She practices her speech over and ov
mezya [45]

The answer is intentional.

6 0
3 years ago
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An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and
Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

\Delta P.E=mg(h_{2}-h_{1})

\Delta P.E=10000\times9.8\times(10000-0)

\Delta P.E=10000\times9.8\times10000

\Delta P.E=980000000\ J

\Delta P.E=980\ MJ

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)

\Delta K.E=148298642\ J

\Delta K.E=148.3\ MJ

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0
3 years ago
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
3 years ago
Find the temperature
mote1985 [20]

Where is the Temperature bud?

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