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3 years ago

1) Convert the following:I) 65 kg into gii) 87570 seconds into houriii) 7.5 km into m

Physics

2 answers:

DiKsa [7]3 years ago

4 0

Answer:

65000g

24.325 hrs

7500m

Natalka [10]3 years ago

3 0

Answer:

I) 65000 g II)24.325 hours III)7500 m

Explanation:

I) 1 kg = 1000 g

65 kg = 65 * 1000 g = 65000 g

II) 1 hour = 3600 seconds

1 second = 1/3600 hours

87570 seconds = 87570 * 1/3600 hours =24.325 hours

III) 1 km = 1000 m

7.5 km = 7.5 * 1000 m = 7500 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 <em>c</em>, 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 $c=0.0038^{\circ}K^{-1}$ ):

$T=\frac{1+(0.0038^{\circ}K^{-1})(293^{\circ}K)}{(0.0038^{\circ}K^{-1})}=556^{\circ}K=283^{\circ}C$

3 0

3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s