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vovikov84 [41]
3 years ago
14

A jogger takes five minutes to run a distance of three kilometres. His speed, in metres per second, is approximately A) 5.5 M/S

B) 10 M/S C) 36 M/S D) 4.5 M/S
Physics
1 answer:
just olya [345]3 years ago
5 0

Answer:

His speed, in metres per second, is approximately 10m/s. The correct option is B.

Explanation:

Speed is defined as the rate in which an object cover a distance in a given time. It is measured in meters per second ( m/s). From the question, the jogger covered a distance of 3Km which when converted to meter is 3000meters in a given time 5 minutes which is 300seconds. The calculation of his speed in meter per second is shown below:

Distance= 3Km to meters ( as 1000meters = 1Km). Therefore 3× 1000 = 3000m

Time= 5 minutes to seconds ( as 1 minute = 60 Seconds). Therefore 5× 60= 300.

Speed= distance/ time

= 3000/300

= 10m/s

Therefore his speed, in metres per second, is approximately 10.

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Explanation:

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
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Answer:

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b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

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q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

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3 years ago
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