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Ipatiy [6.2K]
2 years ago
11

Find then circumference

Physics
1 answer:
Keith_Richards [23]2 years ago
5 0

Answer:

A. 13 cm

Explanation:

here the total diameter is given 4 cm,

here the radius then will be 2 cm.

use the formula of circumference of circle = 2πr ........where r is the radius

using the formula: 2 * π * 2 = 4π = 12.57 = 13 cm.

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Los resortes tienen masa, ¿El periodo y la frecuencia reales son mayores que los dados en las ecuaciones para una masa oscilante
MrRa [10]

Answer:

me no speek spanish

Explanation:

4 0
3 years ago
Calculate the change in entropy that occurs in the system when 3.10 mole of isopropyl alcohol (C3H8O) melts at its melting point
ozzi

Answer:

3.10 mole of C3H8O change in entropy is 89.54 J/K

Explanation:

Given data

mole = 3.10 moles

temperature = -89.5∘C = -89 + 273 = 183.5 K

ΔH∘fus = 5.37 kJ/mol =  5.3 ×10^3 J/mol

to find out

change in entropy

solution

we know change in entropy is ΔH∘fus / melting point

put these value so we get change in entropy that is

change in entropy 5.3 ×10^3 / 183.5

change in entropy is 28.88 J/mol-K

so we say 1 mole of C3H8O change in entropy is 28.88 J/mol-K

and for the  3.10 mole of C3H8O change in entropy is 3.10 ×28.88  J/K

3.10 mole of C3H8O change in entropy is 89.54 J/K

4 0
3 years ago
Read 2 more answers
A spectroscope prism separates the white light from a star into a very wide spectrum. When widely spread, black lines appear in
topjm [15]

Answer:

C. It helps scientists detect dark matter.

Explanation:

I got a 100% on my test. hope it help and I hope your having an awesome day :)

5 0
3 years ago
Read 2 more answers
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
rusak2 [61]

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

8 0
2 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
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