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alex41 [277]
3 years ago
12

A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t

o the zebra. What is the lion’s speed?
Question 1 options:

1 mi


1 mi/h


25 mi


25 mi/h
Physics
1 answer:
Helen [10]3 years ago
5 0

To solve this exercise, the following speed equation must be applied:

 V = d / t (mi/hr)

 V: speed (mi/hr)

 d: distance (mi)

 t: time (s)

We know the distance (d= 5 miles) and the time (t= 0.2 hours), so we proceed to replace these values in the formula to calculate the lion’s speed:

 V = d / t

 V = 5 mi/0.2 hr

 V = 25 mi/hr

 The lion’s speed is 25 mi/hr, so the last option is the right answer

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Answer:

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Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

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