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zimovet [89]
4 years ago
11

If

Physics
1 answer:
nadya68 [22]4 years ago
8 0

Answer:

hhjjkkkksksksjskkskakakkskskksksksoao

Explanation:

hiiii look forward but I don't know how to do it

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2.) Explain why the starting angle doesnt impact the time it takes the pendulum to swing back and forth?​
melomori [17]

The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.

The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)

−gsinθ=lθ¨−gain⁡θ=lθ¨

For small angles, θ≪1,θ≪1, and hence sinθ≈θsin⁡θ≈θ. Hence,

θ¨=−glθθ¨=−glθ

This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos⁡(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.

For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause 

7 0
3 years ago
The electron structures of atoms are not involved in the emission of:
nekit [7.7K]
The correct one is gamma rays. Lets go over them one by one.

Gamma rays are rays that arise from gamma decay, a type of radioactive decay. Often, after another decay, the nucleus is still unstable and it gives off energy in the form of gamma rays to stabilize itself. Hence, gamma rays have nothing to do with the electron structure, only with the nucleus of the atom.

X-rays are the product of accelerating electrons, hence only specific atoms can emit a specific energy of X-rays; similarly for the photoelectric phenomenon, the energy which is needed for photoelectrons to be created depends on the electron structure of the atom (in both cases, it is important to see how strong the bond between electron and atom is).
Finally, spectral lines differ depending on the electron structure of the atoms since electrons with different energies absorb different frequencies of light.
7 0
3 years ago
| The electric field 5.0 cm from a very long charged wire is (2000 N/C, toward the wire). What is the charge (in nC) on a 1.0-cm
Serggg [28]

Answer:

The charge is 0.056 nC.

Explanation:

Given that,

Electric field = 2000 N/C

Distance = 5.0 cm

We need to calculate the charge density

Using formula of charge density

E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}

\lambda=2\pi\times\epsilon_{0}\times r\times E

Put the value into the formula

\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000

\lambda=5.56\times10^{-9}\ C/m

We need to calculate the charge in 1.0 cm

Using formula of charge

Charge = \lambda\times\text{length of segment}

Charge =5.56\times10^{-9}\times1.0\times10^{-2}

Charge=0.056\times10^{-9}\ C

Charge=0.056\ nC

Hence, The charge is 0.056 nC.

3 0
3 years ago
In this experiment, you will use a natural indicator to indirectly determine the H+ concentration of some common household solut
Vlad1618 [11]
Some of the scientific questions that may be answered through the experiment are:

(1) What are the physical changes that may occur in the solution or the indicator when added with acidic/basic solution?

(2) How much of the indicator is needed in order to bring about a significant physical change in the solution to identify its H+ concentration?
7 0
3 years ago
Read 2 more answers
2. A small piece of dust is located in between two oppositely charged parallel plates where there exists a uniform electric fiel
Alla [95]

Answer:

a

The total charge is q = 1.6*10^{-4}C

b

The change in potential energy is \Delta U = -4*10^{-2}J

c

The height is  h= 8.2cm

Explanation:

From the question we are told that the

       The magnitude of electric field is E = 5000N/C

        The number of proton is  N_p = 1*10^{15}

        The distance between the plates is  d = 5cm = \frac{5}{100} = 0.05m

The total charge can be mathematically represented as

            q = N_p * e

Where e is the charge on one proton which has a value of =1.6*10^{-19}C

      Substituting values

           q = 1 *10^{15} * 1.6*10^{-19} = 1.6 *10^{-4}C

The change in potential energy is mathematically represented ads

         \Delta U = -(qE)d

where the negative sign shows that the work done by the electric force is against the electric field

                Substituting values

         \Delta U = - 1.6*10^{-4} * 5000 * 0.05

                = -4*10^{-2}J

 

   The mass of the bead is given as 0.05kg

  The change in potential due to gravity is mathematically given as

                    \Delta U = -mgh

the negative sign is  due to the fact that the height is decreasing

And g =9.8m/s^2

                  Making h the subject

        h = \frac{\Delta U}{mg}

Substituting values

                 h = \frac{4*10^{-2}}{0.05 * 9.8}

                   =0.081m = 0.082 *100 = 8.2cm

4 0
4 years ago
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