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3 years ago

A 1000-kg aircraft going 25 m/s collides with a 1500-kg aircraft that is parked. They stick

Physics

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

48.54 m

Explanation:

Mass of moving aircraft = 1000 kg

Mass of stationary aircraft = 1500 kg

Velocity of moving aircraft = 25 m/s

Velocity by which they both move together = 10 m/s

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow 0=10+a\times 9.7\\\Rightarrow \frac{-10}{9.7}=a\\\Rightarrow a=-1.03\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-10^2}{2\times -1.03}\\\Rightarrow s=48.54\ m$

Both planes skid a distance of 48.54 m

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

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Answer:

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Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

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3 years ago