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3 years ago

How many constellations does the international astronomical union recognize?

Physics

1 answer:

Aliun [14]3 years ago

3 0

The answer is 88 the international astronomical union recognized 88 constellations in the sky

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QUICK!!!!!!!!!!!! QUICKKKK!!! HELP!!!! What is the slope of the line????

Contact [7]

Answer:

4 m/s

Explanation:

If you pick the points (2,8) and (3,12) on the graph, y2-y1 over x2-x1, you get 4 over 1 which equals 4.

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4 years ago

What is 1.0147 rounded to the nearest thousandths place?

Agata [3.3K]

Answer:

1.0147 rounded to the nearest thousandths place is 1.015

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3 years ago

Read 2 more answers

Increasing excitatory signals above threshold levels for neural activation will not affect the intensity of an action potential.

ahrayia [7]

Answer:

Option C

An all-or-none response

Explanation:

Increasing excitatory signals above threshold levels for neural activation will not affect the intensity of an action potential indicates that the reaction of a neuron is an all-or-none response

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3 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

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3 years ago

What is the acceleration of a car at rest that speeds up to 75m/s over 25 seconds?

Yuki888 [10]

It’s a because I just did it

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3 years ago