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Illusion [34]
3 years ago
13

A standing wave of the third harmonic is induced in a stopped pipe of length 1.2 m. The speed of sound through the air of the pi

pe is 340 m/s. How many antinodes form in the standing wave pattern in the pipe?
Physics
1 answer:
stira [4]3 years ago
7 0

Answer:

Answer is 3.

Explanation:

Start by putting the values in the formula i.e.

Lenght 1.2m

The speed of sound through the pipe 340m/s.

Put it in the formula and the answer is 3.

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An 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55
Sindrei [870]

Answer:

a)Yes will deform plastically

b) Will NOT experience necking

Explanation:

Given:

- Applied Force F = 850 lb

- Diameter of wire D = 0.15 in

- Yield Strength Y=45,000 psi

- Ultimate Tensile strength U = 55,000 psi

Find:

a) Whether there will be plastic deformation

b) Whether there will be necking.

Solution:

Assuming a constant Force F, the stress in the wire will be:

                       stress = F / Area

                       Area = pi*D^2 / 4

                       Area = pi*0.15^2 / 4 = 0.0176715 in^2

                       stress = 850 / 0.0176715

                       stress = 48,100.16 psi

      Yield Strength < Applied stress > Ultimate Tensile strength

                        45,000 < 48,100 < 55,000

Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.

6 0
3 years ago
A system of 4 electrons, 18 protons, and 4 neutrons has a net charge of?
slamgirl [31]

Answer:

+ 14

Explanation:

18 protons make a positive 18 charge (+18)

4 electrons make a negative 4 charge (-4)

both combined give + 18 - 4 = + 14

The four neutrons don't carry net charge, so the addition of the electrons doesn' affect the net charge found above which still gives + 14.

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3 years ago
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Explanation:

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3 years ago
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Simora [160]
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5 0
2 years ago
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The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen
Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer:  A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life =  5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 =  5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing   N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0
3 years ago
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