50 BP

Calculate the size of the image of a tree that is 8m high and 80 m a pinhole camera that is 20 cm long . what is its magnificati

Vladimir79 [104]

1) Size of the image: 2 cm

In order to calculate the size of the image, we can use the following proportion:

$p:q = h_o : h_i$

where

p = 80 m is the distance of the tree from the pinhole

q = 20 cm = 0.2 m is the distance of the image from the pinhole

$h_o$ = 8 m is the heigth of the object

$h_i$ is the height of the image

By re-arranging the proportion, we find

$h_i = \frac{h_o \cdot q}{p}=\frac{(8 m)(0.2 m)}{80 m}=0.02 m=2 cm$

2) Magnification: 0.0025

The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

$M=\frac{h_i}{h_o}$

so, in this problem we have

$M=\frac{0.02 m}{8 m}=0.0025$

4 0

2 years ago

Sort the items based on whether they are simple machines or compound machines.

Tom [10]

Answer:

Simple machine: pair of tongs, seesaw and wheelbarrow

Compound machine: sewing machine, fishing rod and reel and crane.

Explanation:

Simple machine is the simplest device to use mechanical advantages. It has simplest mechanism to multiply the magnitude of force.

So, pair of tongs, seesaw and wheelbarrow are the example of simple machine because their mechanism is simplest.

Compound machine is a device that is made up of more than one simple machine.

so, sewing machine, fishing rod and reel and crane are example of compound machine because these devices shows more than one machine in it.

Thus, classification is as follows:

Simple machine: pair of tongs, seesaw and wheelbarrow

Compound machine: sewing machine, fishing rod and reel and crane.

3 0

3 years ago

Read 2 more answers

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )