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leva [86]
3 years ago
15

In the 2016 Olympics in Rio, after the 50 mm freestyle competition, a problem with the pool was found. In lane 1 there was a gen

tle 1.2 cm/scm/s current flowing in the direction that the swimmers were going, while in lane 8 there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 mm in 25.0 ss in the absence of any current.
Required:
a. How would the time it took the swimmer to swim 50.0 m change in lane 1?
b, How would the time it took the swimmer to swim 50.0 m change in lane 8?
Physics
1 answer:
gogolik [260]3 years ago
6 0

Answer:

Explanation:

still water speed is 50 m / 25.0 s = 2.00 m/s or 200 cm/s

In lane 1 the effective speed would be 201.2 cm/s

5000 cm / 201.2 cm/s = 24.85 s

The change is 25.00 - 24.85 = 0.15 s decrease in time

In lane 8, the effective speed would be 198.8 cm/s

5000 cm / 198.8 cm/s = 25.15 s

The change is 25.00 - 25.15 = 0.15 s increase in time

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Newtons second law in words??​
Dennis_Churaev [7]

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

Explanation:

it's newton's second law

5 0
2 years ago
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

d_0=5m=500cm

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm

By the formula of magnification

\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm

The height of the image formed is 18.89cm.

5 0
3 years ago
PLEASE HELP 50 POINTS SPACE QUESTION
rodikova [14]

Earth's distance from the sun doesn't change enough to cause seasonal differences. Instead, our seasons change because Earth tilts on its axis.

hope it helps.

5 0
3 years ago
Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l
liberstina [14]

Explanation:

Given: \lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}

\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}

\:\:\:\:\:= 6×10^{14}\:\text{Hz}

The work function \phi is then

\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})

\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}

3 0
2 years ago
An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away
katrin [286]

Answer:

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

          W = F = 9.11 10⁻³¹  9.8

          W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

         Fe = k q₁ q₂ / r²

         Fe = - k q² / r²

They ask us that W = Fe

         W = k q² / r²

          r = √ k q² / W

Let's calculate

        r = √ 8.99 10⁹ (1.6 10⁻¹⁹)² /8.9 10⁻³⁰

        r = √ 25.86

        r = 5,085 m

Let's look for the relationship of this distance with the harmonic distance

        R / R_atomic = 5,085 / 10⁻¹⁰

        R / R_Atomic = 5 10¹⁰

We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale

8 0
3 years ago
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