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SIZIF [17.4K]
3 years ago
5

A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric dens

ity of 1.225 kg/m2 (and assuming that altitude has a negligible effect on atmospheric density). The skydiver can change the body orientation from straight-erect with a cross-sectional area of 6 sqft to a belly-flat cross-sectional area of 24 sqft. Calculate the terminal velocity of the person when the body has straight and when the body has belly-flat orientations. Calculate the terminal velocity on these two different orientations.
Physics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

The right solution is:

(a) 89.455 m/s

(b) 44.73 m/s

Explanation:

The given values are:

Mass,

m = 200 lbs

or,

   = \frac{200}{2.205} \ kg

   = 90.7 \ kg

Air's density,

\delta = 1.225 \ kg/m^3

Drag coefficient,

C_d=0.325

When body is straight, area,

A_1=6 \ ft^2

As we know,

Terminal velocity,

⇒  V_t=\sqrt{\frac{2W}{C_d \delta A} }

or,

⇒      =\sqrt{\frac{2mg}{C_d \delta A} }

At straight orientation,

⇒ V_t'=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558} }

⇒      =\sqrt{\frac{1777.72}{0.223}}

⇒      =89.455 \ m/s

When belly flat,

⇒  V_t''=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558\times 4} }

⇒        =\sqrt{\frac{1777.72}{0.889} }

⇒        =44.73 \ m/s

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Answer:

Option c) are perpendicular to the electric field

Explanation:

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Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

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The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?
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Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, B=2.7\ mT=2.7\times 10^{-3}\ T

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

B=\dfrac{\mu_o I}{2r}

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}

r = 0.00238 m

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The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
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Answer: the correct option is B ( on the half of the road that the vehicle is traveling).

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Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

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Momentum (P)= Mass × Velocity

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Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

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Explanation:

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So 200 yards is (200 / 1.0936)

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Hope it helps..

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