a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?

1 answer:

3 0

We are given the following conditions:
$v_{0} = 28m/s

t = 4s

v_{f} = 0m/s$

Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.

So we look at our kinematics equation that contains the given variables.

$v_f = v_0 + at$

This contains our target variable and our initial conditions. Plug and chug, we get:

$(0m/s) = (28m/s) + a(4s) -28m/s = (4s)*a a = -7m/s^2$

Thus the answer is:

a = -7m/s^2

Hope this helps!

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5 0

3 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

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Answer:

For left = 0 N/C

For right = 0 N/C

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Considering the two thin metal plates to be non conducting sheets of charges.

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$E = \frac{\sigma }{2\varepsilon }$

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$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

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3 years ago

Let two objects of equal mass m collide. Object 1 has initial velocityv, directed to the right, and object 2 isinitially station

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the equation 2 would look like this
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