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Greeley [361]
3 years ago
7

A long string is pulled so that the tension in it increases by a factor of four. If the change in length is negligible, by what

factor does the wave speed change?
Physics
1 answer:
rusak2 [61]3 years ago
3 0

To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is

v = \sqrt{\frac{T}{\mu}}

Here

v = Wave speed

T = Tension

\mu = Linear mass density

From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension

v \propto \sqrt{T}

Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion

v \propto \sqrt{4} = 2  

The factor that the wave speed change is 2.

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Calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
Neko [114]

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball  = 4kg

Height above the floor  = 1.5m

g  = 9.8n/kg

Unknown:

Potential energy  = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

  Potential energy  = 4 x 9.8 x 1.5  = 58.8J

5 0
3 years ago
Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain
Luden [163]

Answer:

P=1362\ W

t'=251.659\ s is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

Q_v=1.1282\times 10^{6}\ J

Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

P=1362\ W

Now the time required to heat to boiling point form initial temperature:

t'=\frac{Q_s}{P}

t'=\frac{342760}{1362}

t'=251.659\ s

6 0
3 years ago
Measurements of two electric currents are shown in the chart. A 3-column table with 2 rows titled Electric Currents. The first c
Jobisdone [24]

The answer is B - Current Y has a greater potential difference, and the charges flow at a slower rate.

I just took the quiz

4 0
3 years ago
Read 2 more answers
A motorcyclists starts from rest and reaches a speed of 6m/s after traveling with uniform acceleration for 3s .What is his accel
siniylev [52]
2m/s is the answer ! hopefully this helps
7 0
3 years ago
What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman = \frac{95.08}{100 } = 0.9508

The density of the woman is calculate as;

Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman  = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0
3 years ago
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