If a positively charged rod is brought near the knob of a positively charged electroscope, the leaves of the electroscope will...
A) converge, only
B) diverge, only
C) first diverge, then converge
D) first converge, then diverge

Answers

Answer 1
I think it’s gonna be A.
Answer 2

Answer:

B

Explanation:

if a positively charged rod is brought near the knob of a neutral electroscope, it will attract some electrons up from the leaves onto the knob. That process causes both of the leaves to be positively charged (excess protons), and the leaves will diverge.


Related Questions

Physicist ____ was found to have completely made up some of the data on which his astounding discoveries were made.

Answers

Answer:

Jan Hendrik Schön I believe

What would happen to the distance between the bright fringes produced by a diffraction grating if the entire interference apparatus (light source, grating, and screen) were immersed in water

Answers

Answer:

Since the wavelength would be reduced by the factor of N where N is the index of refraction, fringes would be be closer to one another.

(Similar to the difference of fringes of red light and blue light.)

What are the si units

Answers

Answer:

The uniy which is accepted all over the world is called SI unit.

Explanation:

The system of measurement that is agreed by the international convention if scientists that is held in paris of France to adopt an international unit is called SI unit unit.

You have just learned that the planet core of Mars is no longer radiating heat like the core of Earth. With this information what would you communicate to building designers as the lead geologist in regards to suitable building locations on the surface of Mars? Remember your informing builders about locations that would best avoid natural disasters like marsquakes and volcanoes? What evidence or theories do you have to support your statements?

Answers

Answer:

Explanation:

That is a fun question!

Without a hot radiating core like Earth, Mars will have very different geological and seismic events. The Mars core will be relatively cold and there will not be any molten magma. So Mars will not have earthquakes or volcano activities. Both only occur when there is magma flowing or tectonic plate motion and they will not occur with a cold core.

A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What is the magnitude of the electric field strength at a distance of 20 cm

Answers

Answer:

[tex]E_2=80N/C[/tex]

Explanation:

From the question we are told that:

Initial Distance [tex]d_1=40cm=>0.4m[/tex]

Initial Electric field strength [tex]E_1=100N/C[/tex]

Final Distance [tex]d_2=20cm=>0.20m[/tex]

Generally the equation for Electric field is mathematically given by.

 [tex]E=\frac{kq}{d^2}[/tex]

 [tex]q=\frac{100*(0.4)^2}{K}[/tex]

Substituting q for [tex]d=20cm[/tex]

 [tex]E_2=\frac{k}{0.2}*\frac{100*(0.4)^2}{K}[/tex]

 [tex]E_2=80N/C[/tex]

Explain how frequency and amplitude affect sound vibrations?

Answers

Answer:

The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.

Explanation:

La resistividad de cierto alambre es 1.72x10 -8 Ωm a 20 oC .Una batería de 6v se conecta a una bobina de 20 m de este alambre, con un diámetro de 0.8 mm. ¿Cuál es la corriente en el alambre?

Answers

Answer:

   i = 1.09 10⁴ A

Explanation:

For this exercise we will look for the resistance of the wire

          R = ρ L / a

the area of ​​the wire is

          a =ππ r² = π πd² / 4

we substitute

         R = ρ L 4 / π d²

          R = 1.72 10⁻⁸   20  4/π 8 10⁻⁴

         

         R = 5.47 10⁻⁴ Ω

to calculate the current we use ohm's law

           V = R i

            i = V / R

            i = 6 / 5.47 10⁻⁴

            i = 1.09 10⁴ A

An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What is the length of the rod at the freezing point of water (0 0C)? (b) What is the temperature if the length of the rod is 10.009 cm? /4mks

Answers

Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

8× +5+9×+3 how can I solve this​

Answers

Answer:

collect like terms then add

=>8x+9x+5+3

=>8x+9x+5+3=>17x+8

A (200 g) of water at (80 °C) is mixed with (100 g)of water at (20 °C). What is the final temperature of the water?

Answers

Answer:

60 °C

Explanation:

From the question given above, the following data were obtained:

Mass of 1st sample (M₁) = 200 g

Temperature of 1st sample (T₁) = 80 °C

Mass of 2nd sample (M₂) = 100 g

Temperature of 2nd sample (T₂) = 20 °C

Equilibrium temperature (Tₑ) =?

NOTE: Since the sample are the same, the specific heat capacity is constant.

We can obtain the equilibrium temperature as follow:

Heat lost by 1st = heat gained by the 2nd

M₁C(T₁ – Tₑ) = M₂C(Tₑ – T₂)

Cancel out C

M₁(T₁ – Tₑ) = M₂(Tₑ – T₂)

200 (80 – Tₑ) = 100 (Tₑ – 20)

Clear bracket

16000 – 200Tₑ = 100Tₑ – 2000

Collect like terms

16000 + 2000 = 100Tₑ + 200Tₑ

18000 = 300Tₑ

Divide both side by 300

Tₑ = 18000 / 300

Tₑ = 60 °C

Therefore, the equilibrium temperature (i.e the final temperature) of the mixture is 60 °C.

Which object will take the most force
to accelerate? *
4 kg
6 kg
8 kg
02 kg

Answers

Answer:

I think it might be 8kg grams because it is bigger

A bird in flight pushes itself upward with
a 7.28 N force. If the bird is climbing at a
constant rate of 1 m/s (no acceleration),
what is the weight of the bird?
[?] N

Answers

Answer:

The weight of the bird is equal to 7.28 N.

Explanation:

The upward force acting on the bird = 7.28 N

The bird is climbing at a constant rate of 1 m/s.

We need to find the weight of the bird.

We know that the weight of an object is the force of gravity acting on it. It can be calculated as follows :

W = mg

In this case, 7.28 N of force is acting on the object. Hence, the weight of the bird is equal to 7.28 N.

A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.

Answers

Answer:

3.504 MeV

Explanation:

Given that;[tex]\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q[/tex]

Also;

[tex]Q= KE_{\alpha } (M_{Y} + M_{\alpha } /M_{Y[/tex])

Mass number of X = 229

Mass number of Y = 225

Mass number of alpha particles = 4

Kinetic energy of alpha particles = 3.443 MeV

Q = 3.443 MeV (225 + 4/225)

Q= 3.504 MeV

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