Which change of state is shown in the photo

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

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

or

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

3 0

3 years ago

If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

arlik [135]

<u>In modern physics</u>, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T

as:

$P\alpha T^4$

where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.

then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}$

As

$T_{2}=4T_{1}$

Then

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}$

then

$P_{2}=256*P_{1}$

The factor will the total energy radiated per second per square meter increase = 256

5 0

3 years ago

A heavy flywheel is accelerated (rotationally) by a motor thatprovides constant torque and therefore a constant angularaccelerat

larisa86 [58]

Answer: (a) t1 = omega1/alpha

(b) theta1 = 1/2 * alpha*theta1^2

(c) t2 = omega2/5*alpha

Explanation: see attachment

4 0

3 years ago

A cylinder 0.170 m in diameter rotates in a lathe at 530 rpm . what is the tangential speed of the surface of the cylinder?

Ad libitum [116K]

Diameter = 0.170 meter
Circumference = 0.170 π meters

530 rpm = 530 circumferences / minute

   = (530 x 0.170 π meters) / minute

   =    283.06 meter.minute

   =    4.72 meters/second

8 0

3 years ago