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AURORKA [14]
1 year ago
10

Carts, bricks, and bands

Physics
1 answer:
tatiyna1 year ago
6 0

The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.

<h3>What is acceleration?</h3>

Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).

From the table given,

Trial 1 ----> 1 band = 0.24m/s²

Trial 2 ----> 2 bands = 0.51 m/s²

Trial 3 ----> 3 bands = 0.73 m/s²

Trial 4 -----> 4 bands = 1.00 m/s²

This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.

This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.

Learn more about acceleration here:

brainly.com/question/25749514

#SPJ1

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Explanation:

static friction opposes relative motion

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What is the formula for wave velocity?
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Speed = Wavelength x Frequency. 

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What will happen if you increse the amplitude of a drum hit
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The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases.  As the amplitude of the sound wave increases, the intensity of the sound increases. Sounds with higher intensities are perceived to be louder.

5 0
2 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
Lyrx [107]

Answer:

\theta=4.64^{\circ}

Explanation:

It is given that,

The frequency of monochromatic light, f=5\times 10^{14}\ Hz

Slit separation, d=2.2\times 10^{-5}\ m

Let \theta is the angle away from the central bright spot the third bright fringe past the central bright spot occur. The condition for bright fringe is :

d\ sin\theta=n\lambda

n = 3

\lambda=\dfrac{c}{f}

d\ sin\theta=\dfrac{nc}{f}        

sin\theta=\dfrac{nc}{fd}        

sin\theta=\dfrac{3\times 3\times 10^8}{5\times 10^{14}\times 2.2\times 10^{-5}}  

\theta=4.64^{\circ}

So, at 4.64 degrees the third bright fringe past the central bright spot occur. Hence, this is the required solution.

3 0
3 years ago
You have a stopped pipe of adjustable length close to a taut 85.0cm, 7.25g wire under a tension of 4170N. You want to adjust the
Makovka662 [10]

Answer:

The length is L_d= 0.069 \ m

Explanation:

From the question we are told that

               The length of the wire  L = 85cm = \frac{85}{100}  = 0.85m

                The mass is  m = 7.25g = \frac{7.25}{1000}  = 7.25^10^{-3}kg

                The tension is  T = 4170N

Generally the frequency of  oscillation of a stretched wire is mathematically represented as

             f = \frac{n}{2L} \sqrt{\frac{T}{\mu}

Where n is the the number of nodes = 3 (i.e the third harmonic)

             \mu is the linear mass density of the wire

 This linear mass density is mathematically represented as

               \mu = \frac{m}{L}

Substituting values

            \mu = \frac{7.2*10^{-3}}{0.85}

                = 8.53 *10^{-3} kg/m

 Substituting values in to the equation for frequency

            f = \frac{3}{2 80.85} * \sqrt{\frac{4170}{8.53*10^{-3}} }

               = 1234Hz

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire

The fundamental frequency is mathematically represented as

        f = \frac{v}{4L_d}

Where L_d is the  length of the pipe

          v is the speed of sound with a value of v = 343m/s

    Making  L_d the subject of the formula

                      L_d = \frac{v}{4f}

   Substituting values

                   L_d = \frac{343}{(4)(1234)}

                        L_d= 0.069 \ m

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire

6 0
3 years ago
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