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masya89 [10]
3 years ago
12

Periodic trends vary as we move across the periodic table. In general, as you move across a row (from left to right) in the peri

odic table A) ionic size increases. B) atomic radius decreases. C) nuclear charge decreases. D) ionization energy decreases.
Physics
2 answers:
Trava [24]3 years ago
8 0
Pretty sure the answer is B
igor_vitrenko [27]3 years ago
8 0

Answer: Option (B) is the correct answer.

Explanation:

In the periodic table when we move from left to right across a period then there occurs a decrease in atomic radius of the elements.

This is because electrons are added into the same shell and therefore, force of attraction between the nucleus and valence electrons is increases which leads to shrinkage in size.

Hence, atomic radius of the elements decreases.

Thus, we can conclude that in general, as you move across a row (from left to right) in the periodic table atomic radius decreases.

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Which illustration represents low accuracy but high precision?​
Misha Larkins [42]

Answer:

D i think

Explanation:

7 0
3 years ago
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A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
3 years ago
As the wave interacts with a wall, which kind of wave interaction is shown?
kow [346]
 The answer would be a reflection. This is because, t<span>he color of an object is actually the wavelengths of the light reflected while all other wavelengths are absorbed. Color, in this case, refers to the different wavelengths of light in the </span>visible light spectrum<span>perceived by our eyes. The physical and chemical composition of matter determines which wavelength (or color) is reflected.</span>
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3 years ago
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PLSS HELP AND I WILL GIVE BRAINLY TO THE ONE WHO ANSWERS CORRECTLY
Ierofanga [76]

The potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The energy in terms of the charge and potential difference is;

E= qV

60=5 C × V

V= 12 V

The electric current is found as;

\rm I= \frac{q}{t} \\\\ I= \frac{5}{1.25}\\\\ I=4 \ A

From the ohm's law;

V=IR

12=4 ×R

R=3Ω

If the voltage is constant and the resistance is doubled, then the new electric current is half of the previous condition;

\rm I'=\frac{I}{2} \\\\ I'=\frac{4}{2} \\\\\ I'=2 \ A

Hence, the potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

To learn more about the resistance, refer to the link;

brainly.com/question/20708652

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7 0
2 years ago
A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
GalinKa [24]

Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

                =\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}

rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}

n_l = 1.97

7 0
3 years ago
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