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3 years ago

How high was a brick dropped from if if falls in 2.5 seconds?

1 answer:

7 0

Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m

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The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00

Vera_Pavlovna [14]

Answer:

(a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

Explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

$C_{1}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}$

$C_{1}=2.5075\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=2.5075\times10^{-12}\times6.00$

$Q=1.5045\times10^{-11}\ C$

(b). We need to calculate the initial stored energy

Using formula of initial energy

$E_{i}=\dfrac{1}{2}\times CV^2$

$E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36$

$E_{i}=4.5135\times10^{-11}\ J$

(c). We need to calculate the capacitance

Using formula of capacitance

$C_{2}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}$

$C_{2}=9.403\times10^{-13}\ F$

We need to calculate the final stored energy

Using formula of initial energy

$E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}$

$E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}$

$E_{f}=12.036\times10^{-11}\ J$

(d). We need to calculate the work done

Using formula of work done

$W=E_{f}-E_{i}$

Put the value in the formula

$W=12.036\times10^{-11}-4.5135\times10^{-11}$

$W=7.5225\times10^{-11}\ J$

Hence, (a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

5 0

4 years ago

Complete the hypothesis statement by filling in the blank. The best answer gets brainlist

stepladder [879]

Answer:

<em>If the amount of water in the glass is close to brimming, then the pitch will ring as high-toned. </em>

Hope it helped :)

8 0

3 years ago

Read 2 more answers

Meaning of power in physics

zmey [24]

Answer:

The rate of doing work is called power.

4 0

3 years ago

Read 2 more answers

You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbita

meriva

Answer:

11.9 years

Explanation:

We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:

$\frac{T^2}{r^3}=const.$