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Maksim231197 [3]
3 years ago
8

An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside th

e pipe are
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

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Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 2x2 − 3x
liq [111]

Answer:

Rate of change of potential V at the point P (2,6,5) in the direction (u = i + j - k) is

(û.∇V) = 6.928

Explanation:

V(x,y,z) = (2x² − 3xy + xyz)

Rate of change of a potential, V, in a direction, u, at a point P, is given as (û.∇V)

where û = unit vector in the direction of u = (vector u)/(magnitude of u)

u = v = i + j − k

Magnitude of u = √[(1²) + (1²) + (-1)²] = √3

û = (i + j − k)/(√3)

∇V = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)k](V) at point (2,6,5)

V(x,y,z) = (2x² − 3xy + xyz)

(∂V/∂x) = 4x - 3y + yz = 4(2) - 3(6) + (6)(5) = 20

(∂V/∂y) = - 3x + xz = -3(2) + (2)(5) = 4

(∂V/∂z) = xy = (2)(6) = 12

∇V = (20î + 4j + 12k)

Rate of change of potential V = (û.∇V)

û = (1/√3) (i + j − k)

(û.∇V) = (1/√3) [(i + j − k).(20î + 4j + 12k)]

(û.∇V) = (1/√3) [20 + 4 - 12) = 12/√3 = 6.928

3 0
2 years ago
a projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. what
QveST [7]

A projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. projectile's horizontal speed after it has fallen 25 meters is 35 m/s.

<h3>what is speed ?</h3>

Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

The formula of speed is represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

There are four types of speed such as Uniform speed, Variable speed, Average speed, Instantaneous speed

Uniform speed is an uniform speed when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the uniform speed the ratio of total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as an object is moving with variable speed, then the speed at any instant of time is known as instantaneous speed.

For more details regarding speed, visit

brainly.com/question/13263542

#SPJ2

3 0
9 months ago
Read 2 more answers
A cylindrical wire of radius 2 mm carries a current of 3.0 A. The potential difference between points on the wire that are 44 m
BARSIC [14]

Answer:

a. E = 86.36 x 10⁻³ V/m = 86.36 mV/m

b. ρ = 3.6 x 10⁻⁷ Ωm

Explanation:

a.

The electric field in terms of the voltage is given by the following formula:

E = V/d

where,

E = Electric Field in the Wire = ?

V = Potential Difference = 3.8 V

d = distance between the points = 44 m

Therefore,

E = 3.8 V/44 m

<u>E = 86.36 x 10⁻³ V/m = 86.36 mV/m</u>

<u></u>

b.

Now, from Ohm's Law:

V = IR

R = V/I

where,

R = Resistance of wire = ?

I = Current = 3 A

Therefore,

R = 3.8 V/3 A

R = 1.27 Ω

Now, the resistance of a wire can be given as:

R = ρL/A

where,

ρ = resistivity of material = ?

L = Length = 44 m

A = Cross-sectional area = πr² = π(0.002 m)² =  1.25 x 10⁻⁵ m²

Therefore,

1.27 Ω = ρ*44 m/1.25 x 10⁻⁵ m²

(1.27 Ω)(1.25 x 10⁻⁵ m²)/44 m = ρ

<u>ρ = 3.6 x 10⁻⁷ Ωm</u>

5 0
3 years ago
The high spark voltage supplied to the spark plugs of an automobile is increased from the battery voltage by the principle of .
Furkat [3]
<span>A coil is used to step up the electric potential difference using the principle of induction.</span>
8 0
3 years ago
An electric field of magnitude e is measured at a distance r from a point charge q. if the charge is doubled to 2q, and the elec
pav-90 [236]
Electric field is proportional to the charge and inversely proportional to the square of distance:
E\propto\frac{Q}{r^{2}}
With charge 2q and distance 2r, the electric field is proportional to:
E_{q}\propto\frac{2q}{(2r)^{2}}=\frac{2q}{4r^{2}}=\frac{1}{2}\frac{q}{r^{2}}
The new electric field is half of that of the original measured field.
5 0
3 years ago
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