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GaryK [48]
2 years ago
8

Please help with this one

Physics
1 answer:
olasank [31]2 years ago
6 0

Answer:

OPTION A is the correct answer

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
algol13

Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

Using formula of linear density

\mu_{s}=\rho_{s}A

\mu_{s}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{s}=5.00\times10^{-3}\ kg/m

We need to calculate the linear density of copper wire

Using formula of linear density

\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{c}=5.67\times10^{-3}\ kg/m

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}

v_{s}=170.3\ m/s

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

v_{c}=159.9\ m/s

We need to calculate the time taken by the wave to travel  along the combination of two wires

t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

4 0
3 years ago
A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.
madam [21]
<h2>Answer:5ms^{-1},133.6m,51.18ms^{-1}</h2>

Explanation:

Let v_{x},v_{y} be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is \frac{235}{47}=5ms^{-1}

So,v_{x}=5ms^{-1}

Question b:

Time of flight=\frac{2v_{y}}{g}

So,v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}

Maximum height is given by \frac{v_{y}^{2}}{2g}

So,maximum height is \frac{51.18^{2}}{2\times 9.8}=133.6m

Question c:

The vertical velocity is already calculated in Question b.

v_{y}=51.18ms^{-1}

7 0
3 years ago
b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept
olasank [31]

Answer:

fftgjjjuhgybcrgjjuhg ui

Explanation:

trggfyh

4 0
3 years ago
Electrons are made to flow in a wire when there is
Mila [183]

Answer:

The answer is C.

Explanation:

8 0
3 years ago
Read 2 more answers
Coal burns in a furnace, producing light and heat. This reaction is
olasank [31]

The reaction is Exothermic

8 0
3 years ago
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