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2 years ago

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If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ

urring. The wave is breaking. Destructive interference is occurring. Save

2 answers:

Kazeer [188]2 years ago

7 0

The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.

soldi70 [24.7K]2 years ago

7 0

The answer is: A. Constructive interference is occurring.

Hope this helped! :)

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A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.

scZoUnD [109]

Explanation:

(a) It is given that two-third of weight is over the drive wheels. So, mathematically, w = $\frac{2}{3}mg$ .

Hence, maximum force is expressed as follows.

$F_{max} = \mu_{s} \times w$

$m \times a_{max} = \mu_{s} (\frac{2}{3} mg)$

Hence, the maximum acceleration is calculated as follows.

$a_{max} = \frac{2}{3} \mu_{s} \times g$

= $\frac{2}{3} \times 1.00 \times 9.8 m/s^{2}$

= 6.53 $m/s^{2}$

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 $m/s^{2}$ .

(b) Since, 30% of the power is lost in the drive train. So, the new power is 70% of $P_{max}$ .

That is, new power = $0.7 \times P_{max}$

Now, the expression for power in terms of force and velocity is as follows.

P = $F_{max} \nu$

$0.7 P_{max} = ma_{max} \nu$

Therefore, speed of the Porsche at maximum power output is as follows.

$\nu = 0.7 \times \frac{P_{max}}{ma_{max}}$

= $0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}$

= 11.568 m/s

= 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c) The time taken will be calculated as follows.

time = $\frac{\text{velocity}}{\text{acceleration}}$

= $\frac{11.57 m/s}{6.53 m/s^{2}}$

= 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

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2 years ago

At the type of plate boundary shown in the image above, two plates are colliding. Which type of plate boundary is shown in this

serg [7]

Convergent plate boundary

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2 years ago

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The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who

densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

$L = \frac{y \times d}{1.22 \times \lambda} = \frac{1.2 \times 0.004}{1.22 \times 400 \times 10^{-9}} = 9836.066 \ km$

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)

$= 1.22 \times \frac{400 \times 10^{-9}}{0.004}$

sinθ = 1.22×10⁻⁴ rad

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2 years ago

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ <span> -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>

Plug in the values in Equation (A)

<span>so </span> $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ <span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>

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2 years ago

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A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric

Tcecarenko [31]

Answer:

a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

Fe + Fm = m a

a = (Fe + Fm) / m

the electric force is

Fe = q E k ^

Fe = 4.25 10-4 60 k ^

Fe = 2.55 10-2 k ^

the magnetic force is

Fm = q v x B

Fm = 4.25 10⁻⁴ $\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]$

fm = 4.25 10⁻⁴ (-j ^ 30 4)

fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

aₓ = 0

there is no force

Axis y

ay = -5.10 10²/5 10⁻⁵ j ^

ay = -1.02 107 j ^ / s2

z axis

az = 2.55 10⁻² / 5 10⁻⁵ k ^

az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

x = vo t + ½ aₓ t₂2

x = 30 0.02 + 0

x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

y = I + go t + ½ ay t²

y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

y = 1 - 2.04 10³

y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

z = zo + v₀ t + ½ az t²

z = 0 + 0 + ½ 5.1 10² 0.02²

z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

since there is no acceleration the speed remains constant

vₓ = 30.0 m / s

Axis y

let's use the equation v = v₀ + $a_{y}$ t

$v_{y}$ = 0 + -1.02 10⁷ 0.02

v_{y} = 2.04 10⁵ m / s

z axis

v_{z} = vo + az t

v_{z} = 0 + 5.1 10² 0.02

v_{z} = 1.02 10⁻¹m / s

8 0

2 years ago