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3 years ago

The ability to attract electrons is known as:

Physics

1 answer:

german3 years ago

8 0

The ability to attract electrons is called electronegativity.

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What is the displacement of the runner, whose velocity versus time graph is shown in the Figure, in the first 15.5 s?

muminat

Answer:

10 displacement of the runner

6 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

3 years ago

A scooter is accelerated from rest at the rate of 8m/s

algol13

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

8 0

3 years ago

A high ____will have a short wavelength.

Sphinxa [80]

Question:

A high ____will have a short wavelength

Answer:

That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths

Explanation:

Hope it help

4 0

3 years ago

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