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3 years ago

What happened from the mid 1700s to the mid 1800s that contributed to the change in the global human population?

Physics

1 answer:

marin [14]3 years ago

4 0

Industrial revolution is the reason of human population increase in 1700's and 1800's.

<h3><u>Explanation:</u></h3>

Industrial revolution is the event that took in the 18th and early 19th century in Europe and then to other parts of the world. In this revolution, there was huge implementation of machines and artificial objects in the daily lives which hugely contributed to the development of quality of product, greatly decreased the production time and costing, etc.

This led to increase in luxury in the lives of people and there was widespread change in the mode of living of the people all over the world.

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A 1700kg rhino charges at a speed of 50.0km/h. what average force is needed to bring the rhino to a stop in 0.50s?

uranmaximum [27]

From 50km/h to 0km/h in 0.5s we need next acceleration:
First we convert km/h in m/s:
50km/h = 50*1000/3600=13.8888 m/s
a = v/t = 13.88888/0.5 = 27.77777 m/s^2

Now we use Newton's law:

F=m*a

F=1700*27.7777 = 47222N

6 0

3 years ago

PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

2 years ago

मारवतन गनुहास (What Is MKS Syste<br>Convert 5 solar days into second.)

mr_godi [17]

Answer:

5 Days to Seconds = 432000

Explanation:

7 0

3 years ago

A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

Bumek [7]

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s speed of ball in the moving frame (the truck)

U = -20 m/s speed of primed (rest) frame with respect to moving frame

So V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

3 0

3 years ago

5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.

Volgvan

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

note : electric field strength ∝ current density

since current density of first segment is > current density of second segment and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because ; vd ∝ current density

7 0

3 years ago