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3 years ago

Which element in Figure 2 is a transition metal? Which is a noble gas?

Physics

1 answer:

quester [9]3 years ago

5 0

<span>V is the transition metal. He is the noble gas. The periodic chart is arranged with metals on the left and non-metals on the right and the noble (non-reactive) gases form the rightmost column. The transition metals are the group of elements from columns 3 through 12. Although a strict definition of transition metals is those elements that have an incomplete d sub-shell, in which case column 12 is excluded since zinc, cadmium, and mercury do have complete d sub-shells. With that in mind, since the element V is the only element listed in the range of a transition metal, that's the answer. And since He is the only element listed in the noble gas column, that too is the answer.</span>

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A 3-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the a

Kipish [7]

Answer:

Energy, 9 kWh or 32400 kJ

Explanation:

Given that,

The power of heater, P = 3 kW

It runs for 3 hours to raise the water temperature to the desired level. We need to find the amount of electric energy used. We know that the electrical power of an object is given by total energy delivered per unit time. It is given by :

$P=\dfrac{E}{t}$

$E=P\times t$

$E=3\ kW\times 3\ h$

E = 9 kWh

Since, 1 kWh = 3600 kJ

E = 32400 kJ

So, the amount of electric energy used is 9 kWh or 32400 kJ. Hence, this is the required solution.

4 0

3 years ago

Hello please help i’ll give brainliest

Alja [10]

Answer:

First answer.

Explanation:

There may be a 5N force, but if the frictional force also equals 5N, than they cancel eachother out, resulting in the brick still staying still, as it is resting on a (perfectly) level surface, but any amount of force would make the brick move.

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3 years ago

What is the production function? The production function is the relationship between

Ludmilka [50]

Answer:

The Production Function is the relationship between the quantity or variables of output and the different inputs quantities used in the production process.

Explanation:

Generally in quality, production or any business, we need to know what is the output with respect to different input variables. Suppose if an output is about bottle produce per hour, then we look at the inputs such as temperature, raw material/inventory, power supply, manpower, etc which are needed to produce bottle. any changes, for example in temperature could change the production rate.

Furthermore, a graph is made to see the production function.

7 0

3 years ago

The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100

Tema [17]

Answer:

drag force $F_D = 1.5 \ N$

Velocity (V) = 40.169 m/s

Explanation:

The drag force $F_D$ is given by the formula:

$F_D = C_D * \frac{1}{2}* \rho * V^2*A$

where:

$C_D$ = drag coefficient depending on the Reynolds number

Reynolds number Re = $\frac{\rho *V*D}{ \mu}$

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air $\rho$ = 1.1845 kg/m³

viscosity of fluid or air $\mu$ = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

$Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5$

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

$C_D$ = 0.3

Projected Area A = $\frac{\pi D^2}{4}$

A = $\frac{\pi 0.074^2}{4}$

A = 0.0043 m²

Finally, drag force is then calculated as ;

$F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N$

$- F_D = ma$

since acceleration a = $\frac{dV}{dt}$

Then;

$-F_D = m \frac{dV}{dt}$

Also;

velocity (V) = $\frac{dx}{dt}$

Then;

$- F_D = \frac{md_2x}{dt^2}$

$\frac{d_2x}{dt^2} = \frac {- F_D}{m}$

$F_D = 1.5 \ N\\m = 0.15 \ kg$

Then;

$\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}$

$\frac{d_2x}{dt^2} =- 10$

Integrating the above equation ; we have :

$\frac{dx}{dt}= - 10 t + C\\$

when time (t) = 0 ; then $\frac{dx}{dt}= V = 44.3$

44.3 = - 10 × 0 + C

C = 44.3

$\frac{dx}{dt}= V = -10 t + 44.3$

Time (t) =

$\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s$

∴ Velocity ; $\frac{dx}{dt}= V = - 10t +44.3$

$\frac{dx}{dt}= V = - 10(0.413 s) +44.3$

Velocity (V) = 40.169 m/s

6 0

3 years ago

Fast-moving glaciers that can move up to 6 kilometers per year

AlladinOne [14]

The answer for this question would be the term SURGING. Surging is the fast-moving glacier that can move up to 6 kilometers per year. I<span>t flows more quickly, sometimes moving 10 to 100 times faster than it normally does. This is one of the classification of a glacier aside from the normal type. Hope this answers your question.</span>

7 0

3 years ago

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