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3 years ago

The distance between adjacent nodes in a standing wave pattern is 25.0 cm. What is the

Physics

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

Answer:

Speed of the wave in the string will be 3.2 m/sec

Explanation:

We have given frequency in the string fixed at both ends is 80 Hz

Distance between adjacent antipodes is 20 cm

We know that distance between two adjacent anti nodes is equal to half of the wavelength

So \frac{\lambda }{2}=20cm

=20cm

\lambda =40cmλ=40cm

We have to find the speed of the wave in the string

Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec

So speed of the wave in the string will be 3.2 m/sec

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Luis wanted to know which of his 4 toy cars is the fastest. He conducted an experime and recorded his results in the table shown

Murrr4er [49]

The dependent variable was the time and the independent variable was thecars

6 0

3 years ago

A boat takes 3.0 hours to travel 50 km down a river, then 5.4 hours to return. Determine the speed of the water in the river.

Nutka1998 [239]

Answer:

3.7 km/h

Explanation:

Let's call v the proper speed of the boat and v' the speed of the water in the river.

When the boat travels in the direction of the current, the speed of the boat is:

v + v'

And it covers 50 km in 3 h, so we can write

$v+v' = \frac{50 km}{3 h}=16.7 km/h$ (1)

When the boat travels in the opposite direction, the speed of the boat is

v - v'

And it covers 50 km in 5.4 h, so

$v-v'=\frac{50 km}{5.4 h}=9.3 km/h$ (2)

So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

$v+v'=16.7 \\v-v'=9.3$

Subtracting the second equation from the first one we get:

$(v+v')-(v-v')=16.7-9.3\\2v'=7.4\\v'=3.7$

So, the speed of the water is 3.7 km/h.

5 0

3 years ago

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

$acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\$ .

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

$\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\$

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354\\$

$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$

3 0

3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i

Mekhanik [1.2K]

Answer:

See below

Explanation:

See attached diagram

280 km east then 190 km north

Use Pythagorean theorem to find the resultant displacement

d^2 = 280^2 + 190^2

d = 338.4 km

Angle will be arctan ( 190/280) = 34.16 °

4 0

2 years ago