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Norma-Jean [14]
3 years ago
8

Which of the following elements is in Group 2?

Physics
1 answer:
Rzqust [24]3 years ago
6 0

Answer:

calcium

Explanation:

if you have a periodic table that would be very helpful to you

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Is freezing outside on a cold day conduction convection or radiation?
harina [27]
<span>Conduction is the process of freezing outside on a cold day. Conduction is usually the process of transfer of heat that is temperature from one body to another body which the difference in their heat energy. This transfer of energy which makes it freeze is also considered as conduction.</span>
7 0
4 years ago
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. You are sitting on a beach watching the waves roll in. In 12 seconds, 6 waves roll by. What is the period of the waves (7)? Wh
Andrew [12]

Answer:

Explanation:

6 waves in 12 seconds

The period of a wave is defined as the time taken by the wave to complete one oscillations.

Here, 6 waves are completed in 12 seconds

So, one wave is completed in 2 seconds

So, the period of the wave is 2 second.

frequency is defined as the number of waves completed in one second. It is the reciprocal of period of the wave.

frequency = 1 / 2 = 0.5 Hertz

Let the wavelength is λ.

f = 660  kHz = 660000 Hz

the relation for the wave speed and the wavelength and the frequency is given by

v = f λ

where v is the speed of wave = 3 x 10^8 m/s

So, 3 x 10^8 = 660000 x λ

λ = 454.55 m

7 0
3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

6 0
3 years ago
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Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe
kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

A_1v_1=A_2v_2

We want v_2, and take into account that the areas are circular:

v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2

Since the tube is horizontal h_1=h_2 and those terms cancel out, so the pressure of the water in the pipe will be:

P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa

3 0
3 years ago
A current of 1.40 a flows in a wire. how many electrons are flowing past any point in the wire per second?
ikadub [295]
The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of \Delta t:
I= \frac{Q}{\Delta t}

So, by using this relationship and I=1.40 A, we can find the charge passing any point in the wire in 1 second:
Q=I \Delta t=(1.40 A)(1.0 s)=1.40 C

To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron (q=1.6 \cdot 10^{-19}C):
N= \frac{Q}{q}= \frac{1.40 C}{1.6 \cdot 10^{-19} C}=8.75 \cdot 10^{18} electrons
3 0
4 years ago
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