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scoundrel [369]
3 years ago
5

URGENT FOR FINAL

Physics
1 answer:
uranmaximum [27]3 years ago
3 0

Answer:

I need this 2

Explanation:

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Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa
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Answer:

Capacitance of the second capacitor = 2C

Explanation:

\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              \texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C

Similarly for capacitor 2

               \texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C

Capacitance of the second capacitor = 2C

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3 years ago
Can a body have constant speed and still be acclerating ? Give an Example
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Hi Pupil Here's Your answer :::





➡➡➡➡➡➡➡➡➡➡➡➡➡





An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.





⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅






Hope this Helps . . . . . . . . .
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The map here shows land surface temperatures on earth during one month in 2015
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Latitude is the answer for apex.

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3 years ago
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A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m
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Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

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What term best describes the regular path of a spacecraft or other object around a planetary body?
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