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yarga [219]
3 years ago
5

What are 7 elements from the periodic table that are named after objects in outer space

Physics
1 answer:
suter [353]3 years ago
5 0
Helium... from the greek word helios... the sun 
<span>selenium... from the greek word selene... the moon </span>
<span>palladium.. after the asteroid pallas </span>
<span>tellurium...from the greek word tellus... the earth </span>
<span>mercury...after the planet mercury </span>
<span>cerium... after the asteroid ceres </span>
<span>uranium...after the planet uranus </span>
<span>neptunium.. after the planet neptune </span>
<span>plutonium.. after the planet pluto</span>
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Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

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3 years ago
En una fundición hay un horno eléctrico con capacidad para fundir totalmente 540 kg de cobre. Si la temperatura inicial del cobr
11Alexandr11 [23.1K]

Answer:

Q = 2.95*10^5 kJ

Explanation:

In order to calculate the energy required to melt the cooper, you first calculate the energy required to reach the boiling temperature. You use the following formula:

Q_1=mc(T_b-T_1)     (1)

m: mass of cooper = 540 kg

c: specific heat of cooper = 390 J/kg°C

Tb: boiling temperature of cooper = 1080°C

T1: initial temperature of cooper = 20°C

You replace the values of the parameters in the equation (1):

Q_1=(540kg)(390\frac{J}{kg.\°C})(1080\°C-20\°C)=2.23*10^8J

Next, you calculate the energy required to melt the cooper by using the following formula:

Q_2=mL_f         (2)

Lf: melting constant of cooper = 134000J/kg

Q_2=(540kg)(134000\frac{J}{kg})=7.24*10^7J

Finally, the total amount of energy required to melt the cooper from a temperature of 20°C is the sum of Q1 and Q2:

Q=Q_1+Q_2=2.23*10^8J+7.24*10^7J=2.95*10^8J=2.95*10^5kJ

The total energy required is 2.95*10^5 kJ

3 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

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What two factors could the students change to investigate how to increase and decrease the magnetic force between the paperclip
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Answer:

Pencil is to pen

Step by step explanation:

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3 years ago
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