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3 years ago

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What is your average velocity if you drive a distance of 113 km at a speed of 47 km/h, then the same distance at a speed of 66 k

m/h? Answer in units of km/h.

2 answers:

Sauron [17]3 years ago

8 0

Answer:

$V_{avg}=54.90\ km/h$

Explanation:

Given that

Drive 113 km at 47 km/h and another 113 km at 66 km/h.

When distance is same then average velocity can be found by using below formula:

$V_{avg}=\dfrac{2V_1V_2}{V_1+V_2}$

Here

$V_1= 47\ km/h$

$V_2= 66\ km/h$

Now by putting the values

$V_{avg}=\dfrac{2V_1V_2}{V_1+V_2}$

$V_{avg}=\dfrac{2\times 47\times 66}{47+66}$

So

$V_{avg}=54.90\ km/h$

Svetlanka [38]3 years ago

3 0

Answer:

Average velocity v = 54.90 km/h

Explanation:

We have given the distance d = 113 km

First speed = 47 km/h

Time taken to cover 113 km with speed of 47 km/h

$t=\frac{distance}{speed}=\frac{113}{47}=2.4042h$

Time taken to cover 113 km with speed of 66 km/h

$t=\frac{distance}{speed}=\frac{113}{66}=1.7121h$

So total time = 2.4042+1.7121 =4.1163 hour

Total distance = 113+113=226 km

So average velocity $v=\frac{distance}{time}=\frac{226}{4.1163}=54.90km/h$

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A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

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The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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