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dem82 [27]
2 years ago
9

How far did an object travel if they rode a bike at 15m/s for 5 seconds?

Physics
2 answers:
Nady [450]2 years ago
6 0

Answer:

A) 75m

Explanation:

distance = speed x time taken

Here given:

speed: 15 m/s

time: 5 second

So, distance covered:

15 x 5

75 m

s2008m [1.1K]2 years ago
6 0
  • Speed=15m/s
  • Time=5s

Distance

  • Speed×Time
  • 15(5)
  • 75m
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olga2289 [7]

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

Vf=14+(5*20)

Vf=114 m/s.

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An object in a certain direction with an acceleration in the perpendicular direction
geniusboy [140]

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An object moving in certain direction with an acceleration in the perpendicular direction. The above condition is possible . Example of such situation in life would be when stone tied to a string whirling in a circular path

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Aneli [31]

Answer:

c

Explanation:

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3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

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

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

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What is pressure?
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Answer:

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