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3 years ago

If the bus driver is the last one off the bus then who closes the door?

Physics

2 answers:

attashe74 [19]3 years ago

7 0

they leave it open actually I see the doors left open once the bua driver leaves they just leave it open

lakkis [162]3 years ago

4 0

They just close the door but they don’t lock the door so they can get in the bus the next time it’s going to be used.

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If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means

lilavasa [31]

Answer:

M V = m v conservation of momentum (Caps-cannon Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J

6 0

3 years ago

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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°.

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3 years ago

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IrinaVladis [17]

<span>The tendency of an atom to pull electrons towards itself is referred to as its electronegativity. When it comes to chemical properties, electronegativity is use to describe the frequent behavior of atoms to attract shared pairs of electrons towards themselves.</span>

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KengaRu [80]

It comes from the sun, and then it is converted to energy/electricity (by solar panels)

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ikadub [295]

The correct answer is False

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