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FinnZ [79.3K]
3 years ago
13

Is this statement true or false? The littoral zone is exposed to air.

Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0
It is indeed true. the litoral zone is exposed to air.
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A sound that's produced by a single wave at a constant frequency and with no overtones is called
notka56 [123]
It is called a pure sound
7 0
3 years ago
What is the difference between radial acceleration and tangential acceleration and how do you calculate both of these accelerati
sergey [27]

Answer:

Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion

a = (V2 - V1) / T

Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)

a = m V^2 / R = m ω^2 R   where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not  in direction

8 0
2 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
You pull a rope to the left with 300 N and a friend pulls the rope to the right with 425 N
eimsori [14]
I’m assuming you’re supposed to calculate the resultant force?

425N (right) -300N (left)
=125 N to the right
6 0
3 years ago
Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and
Lubov Fominskaja [6]

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0    so, h = 1/2 (a)(t)²

Then, height fo      h = 1/2 (9.81)(10.7)² = 561.6 m

          height so     h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights =  1074.4 m - 561.6 m = 521.8 m

5 0
3 years ago
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