All categories

2 years ago

Definition of complex machine

Physics

2 answers:

timama [110]2 years ago

7 0

Answer:A complex system is a system composed of many components which may interact with each other.

Explanation: hope this helped!

ASHA 777 [7]2 years ago

4 0

A complex machine is a machine made up of two or more simple machines that make your work easier to do. There are six simple machines from which all complex machines are made. They include: The lever. The inclined plane

You might be interested in

A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

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

3 years ago

What happens to the force of gravity between two masses if one mass is decreased?

Andrews [41]

<span>B. It stays the same</span>

7 0

3 years ago

Read 2 more answers

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

2380 = 0.0459 x $\frac{v- 0}{0.001}$

0.0459v = 2.38

v = 51.85m/s

8 0

3 years ago

Read 2 more answers

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the

vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

$v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}$

$v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}$

$\mathbf{v_{avg} =12.31 \ m/s}$

8 0

2 years ago