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The speed of sound is measured to be 340 m/s on a certain day.

ziro4ka [17]

Answer: 1,224 km/h

Explanation:

To do this, we pick the first unit and convert
Picking m first and converting to km:
Since we're converting from a non-prefix to a prefix, we divide the value by the prefix were taking it to. In this case, kilo = 10³ which means we're going to divide our value by 1000 to convert it from m to km
340 m/s ÷ 1000 = 0.34 km/s
Now, let's convert our seconds to hour:
We'll need to calculate how many hours is equivalent to one second first;
1 hr = 60×60 seconds
X hr = 1 second
*Cross multiply*
1 × 1 = X × 60 × 60
1 = 3,600 X
X = 1 / 3,600
X = 2.778×10⁻⁴ hour
So, in the place of "1 Second", we're going to be inserting 2.778×10⁻⁴ hour instead
0.34 km / s = 0.34 km / 2.778×10⁻⁴ hour
(0.34 / 2.778×10⁻⁴) km/hour
1,224 km/h.
340 m/s = 1,224 km/h

6 0

2 years ago

A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz

Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, $\bar f$ is given by

$\bar f=f\frac {V}{V-V_s}$ where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and $V_s$ is the speed of car which is unknown, \bar f is given as 440 Hz

$440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s$

8 0

3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?

Hunter-Best [27]

She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.

7 0

3 years ago

In this experiment you will investigate which of the following properties of Faraday's law of electromagnetic induction? (Select

11Alexandr11 [23.1K]

Answer:

Answered

Explanation:

Part A

According to Faraday's law the induced emf in coil is equal to negative of its rate of change of magnetic flux time the number of turns in the coil.

$\epsilon = -N\frac{d\phi}{dt}$ = $-N\Delta\frac{BA}{\Delta t}$

When an emf generated by a change of magnetic flux, produced current of whose magnetic field opposes the change which produces it.

By the above equation the correct options are 1,2 and 4

Part B

Large signals of frequency of 60Hz are measured by osciloscope.

Hence the correct option is part 1.

3 0

3 years ago