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Lelu [443]
3 years ago
15

What is the correct description for an element?

Physics
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

B

Explanation:

Elements are made up of one type of atom only

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PART ONE
Helga [31]

Answer:

1.129×10⁻⁵ N

1.295 m

Explanation:

Take right to be positive.  Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.

5 0
3 years ago
What is the magnitude of the net force on the block (the moment it is released?
GarryVolchara [31]
What are ur answers questions
3 0
2 years ago
What will be the ME of a machine that produces a 240 N work with a 300
Elden [556K]

Answer:

Efficiency = 80%

Explanation:

Given the following data;

Work output = 240 N

Work Input = 300 N

To find the mechanical efficiency of a machine;

Efficiency = \frac {Out-put \; work}{In-put \; work} * 100

Substituting into the equation, we have;

Efficiency = \frac {240}{300} * 100

Efficiency = 0.8 * 100

Efficiency = 80%

Therefore, the mechanical efficiency of the machine is 80 percent.

3 0
3 years ago
Identify the law, write the equation and calculate the answer to the problem below.
lyudmila [28]

Find refractive index first

\\ \rm\Rrightarrow \mu=\dfrac{c}{v}

\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}

\\ \rm\Rrightarrow \mu =0.75

Now

\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu

\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75

\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr

\\ \rm\Rrightarrow sinr=0.94

\\ \rm\Rrightarrow r=sin^{-1}(0.94)

\\ \rm\Rrightarrow r=70^{\circ}

5 0
2 years ago
A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.
NARA [144]

1 watt = 1 joule per sec

11,000 Watts = 11,000 joules per sec

The frequency doesn't matter.

3 0
3 years ago
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