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Fofino [41]
10 months ago
9

When a pulse travels on a taut string, does it always invert upon reflection? Explain.

Physics
1 answer:
sineoko [7]10 months ago
6 0

If pulse travels on a taut string, it may or may not invert upon reflection because it is one that tends to depends on the place the wave reflects.

Hence, If reflecting is obtained from a less dense string, the reflected part of the wave will tend to be from be right side up. A wave inverts if only  it reflects off via a means in which the wave speed is said to be smaller.

<h3>When a wave pulse on a string reflects?</h3>

If a pulse on string is said to show or reflects from free end, the outcome of the resultant pulse is said to be made in such a way that the said slope of string is seen at free end is zero.

Therefore, If pulse travels on a taut string, it may or may not invert upon reflection because it is one that tends to depends on the place the wave reflects.  If reflecting is obtained from a less dense string, the reflected part of the wave will tend to be from be right side up. A wave inverts if only  it reflects off via a means in which the wave speed is said to be smaller.

Learn more about taut string from

brainly.com/question/13740518

#SPJ4

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2 years ago
Rank the five terrestrial worlds in order of size from smallest to largest: Group of answer choices Mercury, Venus, Earth, Moon,
dimaraw [331]

Answer: Moon, Mercury, Mars, Venus, Earth

Explanation:

7 0
2 years ago
What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

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4 0
3 years ago
A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
PLEASE HURRY
Vikentia [17]

Answer: B superconductors

Explanation:

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2 years ago
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