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Oksanka [162]
2 years ago
13

If a swimmer is swimming 200m at 4 m/s how long would it take them to complete the race

Physics
2 answers:
DENIUS [597]2 years ago
7 0

Answer:

An intermediate swimmer would probably take about four to five minutes, while a trained swimmer could take less than three. The world record is 1:42:00, held by Paul Biedermann. If you are a good swimmer you would swim it easily in 3 and a half minutes.

Explanation:

bonufazy [111]2 years ago
4 0
  • Distance=200m
  • Speed=4m/s

\\ \sf\longmapsto Speed=\dfrac{Distance}{Time}

\\ \sf\longmapsto Time=\dfrac{Distance}{Speed}

\\ \sf\longmapsto Time=\dfrac{200}{4}

\\ \sf\longmapsto Time=50s

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A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen
Sveta_85 [38]

Answer:

v_x=34 m/s

v_y=53.9\ m/s

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

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8 0
2 years ago
What is the name given to this group of elements?
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6 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
3 years ago
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