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Vaselesa [24]
3 years ago
15

When is the earth said to be at perihelion

Physics
2 answers:
cluponka [151]3 years ago
6 0
Perihelion is a place ... It's the point in Earth's orbit
where the Earth is closest to the sun.

We reach that point every year some time during
the first few days of January.
Llana [10]3 years ago
5 0
When the earth is closest to the sun it is said to be in perihelion position.
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The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant
AlexFokin [52]

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m =  6,370,000 x 0.001

                       = 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

      = 6.37 x 10³ Km

7 0
3 years ago
Explain the difference between general intelligence and specific intelligence.
Olegator [25]

Answer:

General intelligence refers to the existence of a broad mental capacity that influences performance on cognitive ability measures.

Specific intelligence refers to a person's aptitude in individual 'modalities' or abilities rather than the more general understanding of intelligence.

Explanation:

4 0
3 years ago
Read 2 more answers
When determining the number of significant digits in a measurement,
Black_prince [1.1K]
B) All nonzero digits are significant.
6 0
3 years ago
Read 2 more answers
What simple machine can be described as a shaft that is attached to the center of a wheel?
charle [14.2K]
"Wheel & Axle" <span>can be described as a shaft that is attached to the center of a wheel

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4 0
4 years ago
A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at
Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)

  • Rearranging terms, we have:

        m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)

  • We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

        \Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1}  *v_{1f} ^{2}  + \frac{1}{2}* m_{2}  *v_{2f} ^{2} (3)

  • Rearranging , and simplifying common terms, we have:

        m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2}  *v_{2f} ^{2} (4)

  • Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

        v_{10} + v_{1f} = v_{2f}

  • Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

       m_{1} * v_{10}  = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3}  = \frac{-17.1cm/s}{3} = -5.7 cm/s

7 0
3 years ago
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