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Oksana_A [137]
3 years ago
14

The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth

er with a tuning fork that is oscillating accurately at a concert A (400Hz). What is the period of the violin string oscillations
Physics
1 answer:
Basile [38]3 years ago
3 0

Answer:

T=2.5*10^{-3}s

Explanation:

From the question we are told that:

Beat frequency F_b=4

Frequency F=400Hz

Generally the equation for Frequency of the violin is mathematically given by

 f_v=F_b+F

 f_v=4+400Hz

 f_v=404Hz

Therefore the period of the violin string oscillations is

 T=\frac{1}{f_v}

 T=\frac{1}{404}

 T=2.5*10^{-3}s

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Bill and amy want to ride their bikes to school which is 14.4 kilometers away. It takes Amy 49 minutes to get to school and bill
Drupady [299]

3.4m/s

Explanation:

Given parameters:

Distance to school  = 14.4km

Time taken by Amy = 49min

Time taken by bill = 20min after Amy = 20+49 = 69min

Unknown parameters:

How much faster is Amy's average speed = ?

Solution:

Average speed is the rate of change of total distance with total time taken.

 Average speed = \frac{total distance }{total time taken}

convert units to meters and seconds

      1000m = 1km

       60s = 1min

Distance to school  = 14.4 x 1000 = 14400m

Time taken by Amy = 49 x 60 = 2940s

Time taken by Bill = 69 x 60 = 4140s

Average speed of Amy = \frac{14400}{2940}  = 4.9m/s

Average speed of Bill = \frac{4140}{2940}  = 1.4m/s

Differences in speed = 4.9 - 1.5 = 3.4m/s

Amy was 3.4m/s faster than Bill

learn more:

Average speed brainly.com/question/8893949

#learnwithBrainly

5 0
3 years ago
How might the distance between the needle and the water droplets have affected the movement of the droplets?
shutvik [7]

Answer:

The behavior of droplets trapped in geometric structures is essential to droplet manipulation applications such as for droplet transport. Here we show that directional droplet movement can be realized by a V-shaped groove with the movement direction controlled by adjusting the surface wettability of the groove inner wall and the cross sectional angle of the groove. Experiments and analyses show that a droplet in a superhydrophobic groove translates from the immersed state to the suspended state as the cross sectional angle of the groove decreases and the suspended droplet departs from the groove bottom as the droplet volume increases. We also demonstrate that this simple grooved structure can be used to separate a water-oil mixture and generate droplets with the desired sizes. The structural effect actuated droplet movements provide a controllable droplet transport method which can be used in a wide range of droplet manipulation applications.

Explanation:

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5 0
3 years ago
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How do the four fundamental forces differ?
Basile [38]
Look at the first person’s answer. Cause I know I’m wrong
4 0
3 years ago
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
3 years ago
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gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure
Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be '\text{P}_2'.

As we know that Boyle's law formula states that;

                    P_1 \times V_1 = P_2 \times V_2

where, P_1 = original pressure of gas in the container = 310 mm hg

           P_2 = required new pressure

            V_1 = volume of gas in the container = 185 ml

            V_2 = desired new volume of the gas = 74 ml

So,  P_2 = \frac{P_1 \times V_1}{V_2}  

       P_2 = \frac{310 \times 185}{74}

            =  775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0
3 years ago
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