An automobile accelerates from rest at 1+ 3* sqrt (t) mph/sec for 9 seconds.

1 answer:

3 0

Writing the acceleration as a function of time:
a(t) = 1 + 3√t

Integrating acceleration, we obtain velocity:
v(t) = t + 2(t)^(3/2) + c;
object at rest so velocity at t = 0 is 0 so c = 0.
v(t) = t + 2(t)^(3/2)

Integrating velocity to obtain an equation for displacement:
d(t) = t²/2 + 4/5 t^(5/2) + c
Applying limits from t = 0 to t = 9
d = 9²/2 + 4/5 9^(5/2)
d = 234.9 m

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To what temperature must you raise a silver wire (c = 0.0038), originally at 20.0°C, to double its resistance, neglecting any ch

balu736 [363]

Answer:

$T=283^{\circ}C$

Explanation:

Given a material with temperature coefficient of resistance c, the equation that relates the resistance $R_0$ at temperature $T_0$ and the resistance $R$ at temperature $T$ is

$\frac{R-R_0}{R_0}=c(T-T_0)$

We want to double our resistance, so $R=2R_0$ , thus having:

$\frac{2R_0-R_0}{R_0}=\frac{R_0}{R_0}=1=c(T-T_0)$

For this T must be:

$1=cT-cT_0$

$T=\frac{1+cT_0}{c}$

which for our values means (with $T=20^{\circ}C=293^{\circ}K$ , remember to write temperature in S.I., and that for silver $c=0.0038^{\circ}K^{-1}$ ):