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trapecia [35]
3 years ago
13

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

a 2m/s headwind. How large is the drag force that acts on him? Assume that the density of the air is 1.2kg/m^3, his cross sectional area is 0.5m^2, and the coefficient of drag is 1.1.
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

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If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
3 years ago
What is the main reason water from the ocean turns to water vapor , and then evaporates into the air
melamori03 [73]

Hello! :)

C) heat from the sun

Why?

Because the sun’s heat makes the water turn to steam—or what we call “water vapor.”

Hope this helped and I hope I answered in time!

Good luck!

~ Destiny ^_^

3 0
3 years ago
Read 2 more answers
What do an earthquake wave and a light wave have in common?
rusak2 [61]
It’s c.) I think so at least
5 0
3 years ago
Read 2 more answers
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0
3 years ago
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