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3 years ago

A car slows down uniformly from a speed of 22.0 m/s to rest in 5.50 s. how far did it travel in that time?

2 answers:

8 0

Since the acceleration is 'uniform', the car's average speed during that time is 11 m/s.

Traveling at an average speed of 11 m/s for 5.5 sec, it covers

(11 m/s) x (5.5 sec) = 60.5 meters. (No calculus. Hardly any algebra. Mostly arithmetic.)

svetlana [45]3 years ago

3 0

First get the acceleration. Since it's uniform we use

$a= \frac{-22m/s}{5.5s}=-4 \frac{m}{s^2}$

Note the acceleration is negative since we start at a positive speed and end at zero.
Now the distance is the acceleration integrated twice. The first integral gives the velocity at any time, t

$v=- \int\limits^{ } _ {}{4 \frac{m}{s^2} } \, dt =-4t \frac{m}{s^2}+22.0\frac{m}{s}$

Notice if you put 5.5s for t in the expression we get 0 m/s as we should. Now to get the distance it traveled over this time we integrate this velocity expression:

$s= \int\limits^{5.5} _0{-4t \frac{m}{s^2} +22.0\frac{m}{s} } \, dt =-2t^2\frac{m}{s^2} +22.0t\frac{m}{s} \\ \\ Evaluating: \\ \\ -2(5.5)^2+22(5.5)=60.5m}$

So it travels 60.5 meters

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work

aksik [14]

It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
$F=k_e \frac{q_e q_e}{d^2}$
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.

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Norma-Jean [14]

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