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photoshop1234 [79]

3 years ago

15

The intensity of a certain siren is 0.060 W/m^2 from a distance of 25 meters. What is the intensity of this siren at a distance

of 50 meters?

1 answer:

Nady [450]3 years ago

7 0

<h2>The intensity of siren is 0.015 W m⁻²</h2>

Explanation:

The intensity of sound is inversely proportional to the square of distance between source and the observer .

Thus Intensity at a distance 25 m is I₁ ∝ $\frac{1}{r_1^2}$ I

Similarly the intensity at a distance 50 m is I₂ ∝ $\frac{1}{r_2^2}$ II

Dividing II by I , we have

$\frac{I_2}{I_1}$ = $\frac{r_1^2}{r_2^2}$ = $\frac{25^2}{50^2}$ = $\frac{1}{4}$

Thus I₂ = $\frac{1}{4}$ x 0.06 = 0.015 W m⁻²

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When lights are wired such that if one bulb fails, the entire string of lights

Viktor [21]

Answer:

series

Explanation:

In a series circuit all the components are attached to one branch, so that if one component fails, all the others stop working. In a parallel circuit, however, the components are wired in separate branches, so that even if one branch fails, the rest are not disrupted.

8 0

2 years ago

Read 2 more answers

If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

3 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

A device that converts high voltages to lower voltages is a(n) ________.

AysviL [449]

Answer: A) Transformer

Explanation: A Transformer is an apparatus, device or a component in a system that is used to convert high voltage into low voltage. It is used to either increase or decrease the voltage of an alternating current.

A transformer uses the basic principle of electro magnetic induction, having two or more coils, the voltage is changed from one coil to another but with thesame frequency as alternating current energy passes through them.

7 0

3 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago