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FinnZ [79.3K]
1 year ago
9

A sinusoidal wave in a string is described by the wave functiony=0.150 sin (0.800x - 50.0t)where x and y are in meters and t is

in seconds. The mass per length of the string is 12.0 g/m(a) Find the maximum transverse acceleration of an element of this string
Physics
1 answer:
AleksAgata [21]1 year ago
6 0

The maximum  transverse acceleration of an element of this string is 375 m/s

<h3>What is the  maximum transverse acceleration?</h3>

For a transverse wave, the velocity is known to be perpendicular to the direction of the propagation of the particular wave. To get the acceleration, what is necessary is to take the partial  derivative as it compares to time and the velocity.

With this in mind, we would have

transverse acceleration

a(t) = d²y/dt² = -0.150*50.02sin(0.8x - 50t)

= 0.150*502 = 375 m/s2

Hence we can say that the maximum acceleration of the element on the string is given as 375 m/s²

Read more on acceleration here

brainly.com/question/605631

#SPJ4

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Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
soldier1979 [14.2K]

Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

h = 55.5 m

since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

<u>h = 57.6 m</u>

7 0
3 years ago
A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.
NARA [144]

1 watt = 1 joule per sec

11,000 Watts = 11,000 joules per sec

The frequency doesn't matter.

3 0
3 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0
3 years ago
Simple physics (Final) (Pic provided)
Igoryamba

Answer:

15m/s

Explanation:

Divide distance by time

3 0
3 years ago
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