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Dennis_Churaev [7]

2 years ago

14

A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star

ting velocity of 2V, what would the period be? Half as long, remains constant, 1/4 as long , 4 times as long, or 2 times as long?

1 answer:

LiRa [457]2 years ago

8 0

Answer:

Equation for SHM can be written

V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0

V1 = w1 A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1 since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

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Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi

Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

$\delta = m\lambda$ (1)

where

$\delta =|d_1 -d_2|$ is the path difference, with

$d_1$ is the distance of the point from the first source

$d_2$ is the distance of the point from the second source

m is an integer number

$\lambda$ is the wavelength

In this problem, we have

$d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m$

So let's use eq.(1) to see if the resulting m is an integer

$\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5$

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

$\delta = (m+\frac{1}{2})\lambda$ (2)

If we apply the same procedure to eq.(2), we find

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2$

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

$d_1 = 91.0 m\\d_2 =221.0 m$

So the path difference is

$\delta =|91.0 m-221.0 m|=130.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5$

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

$d_1 = 44.0 m\\d_2 =135.0 m$

So the path difference is

$\delta =|44.0 m-135.0 m|=91.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5$

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3$

which is an integer: so, this point is a point of destructive interference.

3 0

3 years ago

Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

Bezzdna [24]

Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2

7 0

3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago

The train is accelerated from rest at a speed of 0.11 m/s2 to reach a speed of 15.6 m/s. After reaching that speed the train mov

9966 [12]

Answer:

cliff

Explanation:

7 0

2 years ago

How many atoms are in 3MgCl2?

Bogdan [553]

Answer:

3MgCl2 has 9 atoms.

Explanation:

The Element Magnesium (Mg) has 3 atoms.

The Element Chloride (Cl) has 6 attoms.

Their fore 6 + 3 is 9 of course. 3MgCl2 has 9 atoms.

BTW: 3MgCl2 is a molecular compound as well as H2O and CO2.

8 0

2 years ago