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mr Goodwill [35]
3 years ago
5

Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , wherea

s Jill, the faster swimmer, covers 10 lengths in the same time interval. Find the average velocity and average speed of each swimmer.
Physics
1 answer:
Leya [2.2K]3 years ago
5 0

Answer:

Jill average velocity is  0

Jack  average velocity is 0.159337

Jill average speed = 1.593372

Jack average speed = 1.434034

Explanation:

given data

long swimming pool = 25.0 m

9 lengths of the pool = 156.9 s ( 2 min and 36.9 s )

10 lengths = same time interval

to find out

average velocity and average speed

solution

we know that average velocity that is express as

average velocity = \frac{displacement}{time}    .....................1

Jill come back where she start

so here velocity will be = 0

and

Jack ends up on the other end of pool

so average velocity =  \frac{25}{156.9}

average velocity = 0.159337

now we get here average speed that is express as

average speed = \frac{distance}{time}      .............2

jack speed = 9 × \frac{25}{156.9}

jack speed = 1.434034

and

Jill speed = 10 × \frac{25}{156.9}

Jill speed = 1.593372

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Explanation:

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Work = force x distance

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3 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
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Answer:

+1.11\mu C

Explanation:

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Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

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Answer:

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