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Ipatiy [6.2K]
3 years ago
9

A total solar eclipse is visible from

Physics
2 answers:
zzz [600]3 years ago
7 0
Only within the umbra of the moon's shadow.
guajiro [1.7K]3 years ago
3 0

Answer:

b. only within the moon’s umbra.

Explanation:

Solar eclipse occurs when Moon comes between the Sun and the Earth in straight line and blocks the sunlight. A shadow of moon castes on the surface of the Earth. The region passing through the moon's umbra would view a total solar eclipse. The moon completely hides the surface of the sun. It would be visible to the people residing in the bright side of the Earth within the Umbra of the moon.

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A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55
Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0
3 years ago
A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.​
Greeley [361]

Answer:

will

Explanation:

3 0
3 years ago
Which layer of the atmosphere is the top layer of the thermosphere?
joja [24]
The thermosphere is a layer of Earth's atmosphere. The thermosphere is directly above the mesosphere and below the exosphere. It extends from about 90 km (56 miles) to between 500 and 1,000 km (311 to 621 miles) above our planet.
8 0
4 years ago
Read 2 more answers
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
4 years ago
What is the acceleration due to gravity?<br> А<br> B<br> 9.2<br> 12.3<br> С<br> 8.83<br> D<br> 9.81
SCORPION-xisa [38]
Correct answer:

D) 9.81

8 0
3 years ago
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