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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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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

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So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

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