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zepelin [54]
3 years ago
9

A bell is ringing inside of a sealed glass jar that is connected to a vacuum pump. Initially, the jar is filled with air. What d

oes one hear as the air is slowly removed from the jar by the pump
Physics
1 answer:
Mila [183]3 years ago
4 0

Answer:

See Explanation

Explanation:

Sound is a mechanical wave. A mechanical wave requires a material medium for propagation. This means that sound waves must be carried in air. If there are no air molecules, sound waves can not travel.

When air is gradually removed from the jar by the pump, the sound intensity from the bell gradually decreases owing to the fact that air which is the medium through which sound waves are propagated is gradually being removed from the jar.

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g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

6 0
3 years ago
If we push a piece of furniture to move it but we don't manage to move it, are forces acting on
zubka84 [21]

Answer:

Yes

Explanation:

Forces are pushing such as gravity, and forces from your own body and the couch are pushing against each other.

6 0
3 years ago
What is the length of the orbit of the earth around the sun?
Rzqust [24]

Answer:

S = 2 π R

R (mean) = 92.9E6 miles

S = 2 * 3.14 * 92.9E6 miles = 5.84E8 miles

5 0
2 years ago
If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?
Bess [88]

This is an incomplete question, here is a complete question.

The air in the cylinder with a piston has a volume of 220 mL and a pressure of 650 mmHg.

If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?

Answer : The final volume of the cylinder is, 117.6 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas = 650 mmHg = 0.855 atm   (1 atm = 760 mmHg)

P_2 = final pressure of gas = 1.6 atm

V_1 = initial volume of gas = 220 mL

V_2 = final volume of gas = ?

Now put all the given values in the above equation, we get:

0.855atm\times 220mL=1.6atm\times V_2

V_2=117.6mL

Therefore, the final volume of the cylinder is, 117.6 mL

8 0
3 years ago
A wire of radius 5 x 10⁻⁴ m is needed to prepare a coil of resistance 40 Ω. The resistivity of the material of the wire is 3.14x
SIZIF [17.4K]

Answer:

100 m

Explanation:

From the question,

R = Lρ/A.................... Equation 1

Where R = resistance of the wire, L = length of the wire, ρ = resistivity of the wire, A = cross sectional area of the wire.

But,

A = πr².................... Equation 2

Where r = radius of the wire.

Substitute equation 2 into equation 1

R = Lρ/πr²

Make L the subject of the equation

L = Rπr²/ρ...................... Equation 3

Given: R = 40 Ω, r = 5×10⁻⁴ m, ρ = 3.14×10⁻⁷ Ωm

Constant: π = 3.14

Substitute these values into equation 3

L = [40×3.14×( 5×10⁻⁴)²]/ (3.14×10⁻⁷)

L = 40×3.14×25×10⁻⁸/(3.14×10⁻⁷)

L = 100 m

Hence the length of the wire is 100 m

4 0
2 years ago
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