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3 years ago

How long a star lives and what it becomes at the end of its life depends primarly on what

1 answer:

4 0

I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.

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Seth and Diana were asked to be the leaders for a game of tag during recess at school. Seth and Diana can ensure that all intere

grandymaker [24]

Answer:

Explanation:

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2 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$

Now we can find the linear speed of the ladybug, which is given by

$v=\omega r$

where:

$\omega=6.28 rad/s$ is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

$v=(6.28)(0.65)=4.1 m/s$

Learn more about angular speed:

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4 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

Which system provides structure and support to the organism?

attashe74 [19]

I found this in a physics tab, I do t know why.
But your answer sir/ma’am is skeleton

3 0

3 years ago

Two identical charges are separated by a distance d. If the distance between them is increased to 3d, what will happen to the fo

belka [17]

A-It will be one-ninth the original force

3 0

3 years ago

Read 2 more answers