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3 years ago

.physics wave. Please don't dare to answer irrelevant

Physics

1 answer:

MAVERICK [17]3 years ago

3 0

Answer:

1a) 857143 m

1b) 414 m

2a)

2b)

3) the medium of air has a wavelength of 0.334 m, the medium of water has a wavelength of 1.493 m, and the medium of 5.130 m.

Explanation:

Question 1a)

Given the velocity/speed, and frequency of the wave, the length can be calculated using these two quantites.

[ λ = v / f ] wavelength = velocity of the wave / frequency of the wave in Hz.

Since 3 × 10^8 × ms^-1 is the velocity, and 350Hz is the frequency.

Anything to the negative power is reciprocated. i.e ms^-1 = m/s.

The wavelength is 300000000m/350Hz = 857142.8571428..... m ≈ 857143 m

Question 1b) Given that the frequency of the second wave in water is 1% of the first wave, and the speed of the second wave is 1450ms^-1

Therefore the second wave has a frequency of 1% of 3.5 = 350/100 Hz = 3.5 Hz

The wavelength is found using the same

formula: wavelength = 1450m/3.5Hz = 414.2857142857.... m ≈ 414 m

Question 2a)

Question 2b)

Question 3) Remember, the speed of sound of the medium = frequency of the medium × wavelength of the medium.

Therefore the wavelength of the medium = speed of sound of the medium / frequency of the medium. This has a similar correlation to the wavelength formula. We are given that all these mediums have a frequency of 1KHz = 1000Hz, where So the wavelength of each medium =

Question 4)

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Answer: The correct answer is "metalloid".

Explanation:

Metal is the material which can conduct electricity as it contains free electrons. It is good conductor of electricity. For example, copper, silver.

Metal can be drawn into wires. This property of material is called ductile.

Metal can be beaten into sheets. This property of material is called malleable.

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3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

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Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

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Object A has more mass than object B

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Net force = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

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