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4 years ago

Drag the tiles to the correct boxes to complete the pairs.

Physics

1 answer:

marishachu [46]4 years ago

7 0

Answer:

Air pollution---> smog

Water pollution---> eutrophication

Land pollution---> contaminated soil

Light pollution---> sky glow

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a school bus takes 0.50 hours to reach the school from your home. if the average speed of the bus is 20km/h, what is displacemen

Kazeer [188]

Answer:

10 km

Explanation:

.5 hr * 20 km/hr = 10 km

8 0

1 year ago

Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and t

Alecsey [184]

Known values; mass per / volume and mass of each atom

average volume= volume per atom. 
density*mass per atom = volume per atom 

1cm^3 / 7.87 gram * 9.27 *10^-26 kg /atom* 1000 gram / kg=
=1.178 * 10^-22 cm^3 per atom 

Convert that to mass 
1 cm^3 * (1 m / 100 cm)^3 
average volume= 1.178 *10^-29 m^3 per atom 

distance: 

Volume of a cube is L^3 

so L = (1.178*10^-29)^(1/3) 

Half way through the cube is distance 1/2 L. 
Second cube is 1/2 L 
So distance between the center of 2 cubes = L 

=2.28 *10^-10 m

4 0

3 years ago

Please help, I don’t get the question.

aalyn [17]

All you have to do is get the culmative effect of the forces by predicting the direction in which each object will move. Just write down what you know and put down questions for your teacher.

6 0

3 years ago

Is the strange solid/liquid changing behavior of Oobleck the SAME as an ice cube melting?

Dmitrij [34]

I don’t think so because an ice cube melting needs heat and relies on temp while ooblecks transition from solid to quickly depends on force and speed

6 0

3 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s