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Sonja [21]
3 years ago
9

A person who weighs 509,45 N empties her lungs as much as

Physics
1 answer:
Masja [62]3 years ago
7 0

Answer:

The weight of the girl = 1045.86 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.

⇒ D₁/D₂ = W/U............................... Equation 1.

Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.

Making D₁ the subject of the equation,

D₁ = D₂(W/U)................................... Equation 2

<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>

<em>Substituting these values into equation 2</em>

D₁ = 1000(509.45/487.11)

D₁ = 1045.86 kg/m³

Thus the weight of the girl = 1045.86 kg/m³

<em></em>

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cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your
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We have that the time in seconds, minutes, and hours is

t=1.083*10^{-4}s

T_{min}=1.805*10^{-6}min

T_{hours}=3.0083*10^{-8}hours

From the Question we are told that

Velocity v=13.0m/s

Distance d=120 km

Generally the equation for the Time  is mathematically given as

t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s

Therefore

T_{min}=1.083*10^{-4}s/60

T_{min}=1.805*10^{-6}min

And

T_{hours}=T_{min}/60

T_{hours}=3.0083*10^{-8}hours

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4 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

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