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3 years ago

Which element of the third period has the highest first ionization energy?

Physics

1 answer:

nataly862011 [7]3 years ago

5 0

D. argon (Ar), a noble gas

Explanation:

Argon is the element on the third period with the highest first ionization energy.
Ionization energy is a measure of the readiness of an atom to lose an electron.
Weakly held electrons in the valence shell are easier to lose compared to the inner shell electrons.
Across a period, ionization energy increases and since Argon is a noble gas on the periodic table, it has the highest first ionization energy.
The energy required to remove electron from a filled orbital or inert-gas configuration is very great.

Learn more:

Noble gases brainly.com/question/1781595

#learnwithBrainly

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During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one c

Alexandra [31]

Answer:

138.18 minutes

Explanation:

$L_v$ = Latent heat of water at 0°C = 80 cal/g

m = Mass of water = 570 g

Heat removed for freezing

$Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal$

Let N be the number of cycles and each cycle removes 56 cal from the freezer.

So,

$55\times N=45600\\\Rightarrow N=\frac{45600}{55}$

Each cycle takes 10 seconds so the total time would be

$\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes$

The total time taken to freeze 138.18 minutes

6 0

2 years ago

A train moving at a constant speed of 50.0 km/h moves east for 45.0 min, then in a direction 45.0° east of due north for 10.0 mi

Yuki888 [10]

Answer:

3.47km/h with a direction of 67.8 degrees north of west.

Explanation:

First we need to calculate the displacement on the X axis, so:

$d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km$

then on the Y axis:

$D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km$

The magnitud of the displacement is given by:

$D=\sqrt{D_x^2+D_y^2} \\D=6.37km$

and the angle:

$\alpha =arctg(\frac{5.90}{2.41})=67.8^o$

that is 67.8 degrees north of west.

$v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h$

8 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$