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Yanka [14]
3 years ago
10

If the amplitude of the resultant wave is twice as great as the amplitude of either component wave, and this wave exhibits reinf

orcement, the component waves must
Physics
2 answers:
worty [1.4K]3 years ago
8 0
<span>If the amplitude of the resultant wave is twice as great as the amplitude of either component wave, and this wave exhibits reinforcement, the component waves must BE IN PHASE WITH EACH OTHER</span>
dimaraw [331]3 years ago
7 0
This question seems to really be incomplete. You must include the options but fortunately I have already encountered this problem.
Here are the choices:

<span>A. have a different frequency than the resultant wave.
B. be in phase with each other.
C. be traveling in the opposite direction of the resultant wave.
D. have a different wavelength than the resultant wave

The correct answer is letter B. be in phase with each other</span>
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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0
2 years ago
A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan
Triss [41]

Answer:

<em>The distance covered by comet is </em>200,000 km

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= \bold{\frac{distance}{time}}

Thus distance= speed*time

In this problem it is given that speed of comet= \frac{50,000km}{hr}

time travelled by the comet= 4 hours

Thus distance= speed*time

                            = 500000*4

                            = \bold{200,000km}

5 0
2 years ago
Which type of interference occurs when two waves exactly cancel out? NEED AN ANSWER ASAP
Nostrana [21]
If the two waves combine to produce ANY wave that smaller
than either of the originals, that's destructive interference.
4 0
2 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
2 years ago
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