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antiseptic1488 [7]
3 years ago
8

If the velocity of a bycycle is 10 m/s, how long does it take to cover a distance of 18 km? plzzz help me​

Physics
2 answers:
insens350 [35]3 years ago
8 0

Answer:

1800 seconds

Explanation:

Given,

Distance ( d ) = 18 km

Velocity ( v ) = 10 m/s

To find : Time ( t ) = ?

Distance in km need to be converted to m.

1 km = 1000 m

18 km

= 18 x 1000

= 18000 m

So,

d = 18000 m

Formula : -

v = d / t

t = d / v

= 18000 / 10

t = 1800 seconds.

Therefore,

1800 seconds long, it will take to cover a distance of 18 km.

pashok25 [27]3 years ago
4 0

Your answer would 1800 seconds.

You're welcome ;)

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And in order to calculate the voltage, we can use a voltmeter that must be in parallel with the resistance, this way it will not affect the circuit.

The correct option that shows an amperemeter in series and a voltmeter in parallel is the fourth option.

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1 year ago
One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

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P_f=2 atm

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(b)if the gas expands by the same amount again isotherm-ally and irreversibly

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4 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

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We use Newton's second law

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        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

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we substitute

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b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

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we substitute

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For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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