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2 years ago

11

Assume the following vehicle are all moving at the same speed.it would be harder to change the velocity of which vehicle.a.bicyc

le.B.car.c.motorcycle.D.semi-tractor trailer.

1 answer:

Anuta_ua [19.1K]2 years ago

6 0

Answer:im just guessing d but i think its d though

Explanation:

it pretty obvious

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Which car has the most kinetic energy?

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50kg driving at 30kph because it’s the heaviest yet fastest!

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3 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

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3 years ago

9. Ripening is a _______________ and spoiling is a ______________.

IrinaK [193]

Answer:

C

Explanation:

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1 year ago

Read 2 more answers

Se aplica una fuerza neta de 25 N a una masa de 10 kg. ¿Cuál es la aceleración dada a la masa?

lys-0071 [83]

Answer:

Se queda

Explanation:

Porque 10 kg es más que 25N

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3 years ago

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther

diamong [38]

Answer:

$T_{2} = 606.69 K$

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

How the boiling water is under regular atmospheric pressure, then

$T_{1} = 373 .15 K$

Thus

$\frac{286000}{373.15} = \frac{465000}{T_{2} }$

$T_{2} = 606.69 K$

5 0

2 years ago