PHYSICS COLLEGE

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound produced by the pipe, in SI units, is closest to:

Answers

Answer 1

Answer:

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=(nv_s)/(4L)$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_(3)=((3)(340m/s))/(4(2.5m))=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

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Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?

Answers

Hi, thank you for posting your question here at Brainly.

To solve this problem, we use Coulomb's Law:

F = kQ1Q2/d^2, where k = 9x10^9

Q1 = 3.5 uC
Q2 = -3.5 uC
Q3 = 4.0 uC

But first, we find the distance between Q1 and Q3 and between Q2 and Q3.

d between Q1 and Q2:
d = sqrt[(0-0.4)^2+(0.3-0)^2]
d = 0.5 m

d between Q1 and Q3:
d = sqrt[(0-0.4)^2+(-0.3-0)^2]
d = 0.5 m

Through force balance, F between Q2 and Q3 - F between Q1 and Q3:

$F_(net) = ((9x 10^(9))(-3.5)(4) )/( 0.5^(2) ) -((9x 10^(9))(3.5)(4) )/( 0.5^(2) )=-1.008* 10^(12)$

Thus, the net force is -1 x 10^-12 C

Final answer:

The total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC is zero. This is because the forces due to each of these charges on the third charge are equal in magnitude but opposite in direction, hence they cancel each other completely.

Explanation:

The question asks for the magnitude and direction of the total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC. This is related to Coulomb's Law, which describes the force between charged objects. Specifically, Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1*q2) and inversely proportional to the square of the distance (r) between them. It also depends on the permittivity of free space (ε₀).

First, you would determine the force between each of the point charges and the third charge separately, and then superpose these forces to find the total force. The force in each case can be calculated using the equation F = k*|q1*q2|/r², where k is Coulomb's constant (8.99 * 10^9 N.m²/C²). You would need to make sure you take into account the signs of the charges when deciding the directions of the forces and when superposing the separate forces.

Assume upwards to be the positive direction. The 3.5 uC charge forces and -3.5 uC charge forces on the 4 uC charge would be opposite in direction (one downwards and one upwards) and identical in magnitude. Therefore, they will cancel each other out, and hence, the total electric force on the third charge (4 uC) will be zero.

Learn more about Total Electric Force here:

brainly.com/question/30448827

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A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec

Answers

Maybe try A for your answer

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Answers

Answer:

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?

Answers

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^(-6) T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2d
B) d/√2
C) d/4
D) 2d
E) d/2

Answers

Answer:b

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

$F=(kq_1q_2)/(d^2)---1$

where k=constant

$q_1$ and $q_2$ are charges

d=distance between them

In order to double the force i.e. 2F

$2F=(kq_1q_2)/(d'^2)----2$

divide 1 and 2 we get

$(F)/(2F)=(d'^2)/(d^2)$

$d'=(d)/(√(2))$

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?