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ale4655
1 month ago
7

Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 ​7 ​

​ (light-years)2, where a light-year is a unit of astronomical distance. Assuming that this region is a square, then a single side of the square would be approximately how many light-years long?
Physics
1 answer:
Sav [3.1K]1 month ago
5 0

Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

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Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by \frac { m{ v }^{ 2 } }{ r }.

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =\frac { m{ v }^{ 2 } }{ r }

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assuming the starting speed is v, after doubling it becomes 2v

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Consequently, the magnitude of the tension increases to four times its original value, 4F.

3 0
22 days ago
(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
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Response:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force applied by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
1 month ago
Read 2 more answers
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