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timurjin [86]
3 years ago
6

A boy weighs 400N. what is his mass?

Physics
2 answers:
shepuryov [24]3 years ago
8 0
The mass is 40.8kilograms
Basile [38]3 years ago
4 0

≈40.8kg


this is to take up space


m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

m

=

W

g


=

400

N

9.8

m/s

2


≈

40.8

kg

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3 years ago
An 81.5-kg man stands on a horizontal surface.
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Answer:

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Explanation:

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As we know that

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V= 0.0827 m³

The force exerted by weight = m g

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Area ,A= 4.5 x 10⁻² m²

The Pressure P

P=\dfrac{F}{A}

P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2

P=181.11 x 10²  N/m²

7 0
3 years ago
I need a short answer ?
mars1129 [50]

Answer:

Explanation:

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7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

  or

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7 0
3 years ago
An experiment is designed to investigate the relation between the pressure and temperature in a tank that has a constant volume
viva [34]

Answer:

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Explanation:

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The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias

4 0
3 years ago
A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is
iris [78.8K]
(a) The net flux through the coil is zero.
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(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
\Phi = BA \cos \theta
where \theta is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case \theta=90^{\circ} and so the cosine is zero, therefore the net flux is zero.
5 0
3 years ago
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