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3 years ago

11

according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-

objects energy. C object volume. D- objects mass

1 answer:

puteri [66]3 years ago

3 0

A an objects weight hope this helps! :D

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Why is the position vs. time graph for an object in free fall a curve?

Ymorist [56]

Under free fall, the object is falling with a velocity that is increasing uniformly. Since the gradient of position-time graph reflects the velocity, the gradient is increasing, i.e. curve.

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3 years ago

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

miskamm [114]

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}$

Applying values we get

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)$

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3 years ago

Read 2 more answers

Can someone explain this to me? 20 POINTS!

Anarel [89]

Answer:What is the first law of thermodynamics and how does it relate to energy use? The first law of thermodynamics states that energy is conserved in chemical processes. ... This amount of energy that must be lost to the surroundings for the process to occur is nature's heat tax, an unavoidable cut of every energy transaction.

Explanation:

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3 years ago

A wave that can travel only through what is a mechanical wave

WINSTONCH [101]

The answer is matter

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3 years ago

Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc

tresset_1 [31]

From the definition of apparent magnitude, we know that:
$m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )$
where:
m = apparent magnitude
F = corresponding flux

We also know that the flux is given by the formula:
$F = \frac{L}{4 \pi d^{2} }$
where:
L = luminosity
d = distance

Therefore:
$\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ = \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}$

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
$m - M = 2.5 Log ( \frac{d}{10pc})^{2}$

Now, let's solve for m:
$m = M + 2.5 Log ( \frac{d}{10})^{2}$
= <span> $8 + 2.5 Log ( \frac{100}{10})^{2}$ </span>
= 13

Hence, the apparent magnitude of the star would be m = +13

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3 years ago