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3 years ago

13

Is there change in the force of gravity between two objects when their masses are doubled and the distance between them is also

doubled?

1 answer:

belka [17]3 years ago

3 0

No. The forces of gravity between them would remain unchanged after those three changes.

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1. An electrically charged atom ___________

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Answer:

Is called an Ion

Explanation:

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The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

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Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

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The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases

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Answer: $1.79\ rad/s^2$

Explanation:

Given

Initial angular speed is $\omega_1=26.2\ rad/s$

Final angular speed is $\omega_2=36.5\ rad/s$

Time period $t=5.75\ s$

Magnitude of the fan's acceleration is given by

$\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}$

Insert the values

$\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2$

Thus, fan angular acceleration is $1.79\ rad/s^2$

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A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?

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Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

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A light-year is a measurement of _____.<br> a. time<br> b. gravity<br> c. energy<br> d. distance

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D. distance

A light-year is the distance light would travel in 1 year.

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