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2 years ago

Order the speed of sound through these materials from the slowest to the fastest.

1 answer:

8 0

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

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A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum

NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars $g_{m}=3.7\ m/s^2$

Time Period of pendulum on earth $T=1\ s$

Time period of Pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

for earth

$1=2\pi\cdot \sqrt{\frac{L}{9.8}}$

$L=\frac{9.8}{(2\pi )^2}$

$L=0.498\ m$

(b)For same time period on mars length is given by

$L'=\frac{g_m}{(2\pi )^2}$

$L'=\frac{3.7}{39.48}$

$L'=0.0936\ m$

$L'=9.36\ cm$

3 0

3 years ago

A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten

Stella [2.4K]

She can climb 0.92 m without losing weight.

<u>Explanation</u>:

Gravitational potential energy is the energy consisting of the product of mass, gravity and height.

1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J, m = 65.0 kg = 65000 g, Efficiency = 20 %

GPE = mgh

where m represents the mass

g represents the gravity,

h represents the height.

585760 = 65000 $\times$ 9.8 $\times$ h

h = 0.92 m.

7 0

3 years ago

An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?

Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

To know more about speed

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5 0

1 year ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.