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3 years ago

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on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t

he peak of mountain

1 answer:

andreyandreev [35.5K]3 years ago

7 0

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).

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Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by 850

x=455/850

x=0.54

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Por una resistencia de 10 Ω fluyen 5A. ¿Cuál será la diferencia de potencial que se le debe aplicar a la resistencia?

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Answer:

V = 50 volts

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Given that,

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What is the safest way to enter a body of water? Enter with you feet first. Dive in.

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2 years ago

Read 2 more answers

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Answer:

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Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

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Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

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$H=19.9*10^{-9}$

Where

$H=\frac{\triangle P}{\rho g}$

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Solution :

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$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

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≈ 0.18

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3 years ago