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Cloud [144]
3 years ago
14

I really need help, dose anybody know the answer?

Physics
2 answers:
Keith_Richards [23]3 years ago
8 0

Answer: no  

Explanation: you add the right side and if it equals the same as the left side then it is balanced.

Anni [7]3 years ago
6 0

no. it is not balanced.

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A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh
Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket  train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}.....(I)

Where, \delta t' = proper time

\delta t = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}

For the second interval,

Put the value of \Delta t' in the equation (I)

\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value in the equation

\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}

\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}

\Delta t_{2}=-5.77\times10^{8}\ s

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0
3 years ago
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general
valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength \lambda_i

• The wave does not satisfy the boundary conditions y_i(0;t)=0


Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions,y(0,1)=0


and y(L,t)=0

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength \lambda_i  can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, y_i(0;t)=0 and y_i(L;t)=0, that are satisfied only for particular value of \lambda_i  .

Given below are the incorrect options about the wave in the string.

•  A_i must be chosen so that the wave fits exactly o the string.

• Any one of  A_i or \lambda_i  or f_i  can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies f_i and the wavelength \lambda_i  must be such that y_i(0;t) = y_i(L;t)=0


(C)

Expression for the wavelength of the various normal modes for a string is,

\lambda_n=\frac{2L}{n} (1)

When n=1 , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

\lambda_n=\frac{2L}{1}\\\\2L

When n=2 , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

\lambda_n=\frac{2L}{2}\\\\L

When n=3, this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

\lambda_n=\frac{2L}{3}

Therefore, the three longest wavelengths are 2L,L and \frac{2L}{3}.

(D)

Expression for the frequency of the various normal modes for a string is,

f_n=\frac{v}{\lambda_n}

For the case of frequency of the i^{th} normal mode the above equation becomes.

f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

Therefore, the frequency of i^{th} normal mode is  f_i=\frac{v}{\lambda_i}

.

6 0
3 years ago
How many board feet of lumber are needed assuming 200 lin. Ft. of 2""x 6"" lumber a. 12 b. 16.67 c. 120 d. 200
Svetach [21]

Answer:

B) 16.67

Explanation:

If the dimension of one lumber is 2" × 6", the total area of one lumber will be 12inch²

If the total board feet of lumber there is 200in, therefore the total board of lumber that will be needed is 200/12 which gives 16.67 lumbers

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3 years ago
A goose waddles 18.0 m north and then 5.00 m east. He then walks 2.00 m south and 2.00 m west. What is the magnitude and directi
olya-2409 [2.1K]
Your answer should be 16.3 m 79.4º east of north
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3 years ago
"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second
Leni [432]
<span>Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculations of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:

d = v0t + at^2/2

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