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Effectus [21]
3 years ago
7

A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered

to the string is tripled, by what factor does the amplitude change
Physics
1 answer:
Serjik [45]3 years ago
4 0

Answer:

n = 1,732    the amplitude must be increased by a factor of 1,732

Explanation:

The power delivered by a wave is given by

                  P = E / t

                  P = ½ μ w² v A²

let's apply this expression to our case the power tripled

                3P₀ = ½ μ w² v A’²

                 

let's write the amplitude function of a initial amplitude

               A ’= n A₀

where n is a number

               3 P₀ = (½ μy w² v  A₀²) n²

               3P₀ = P₀ n²

               n = √ 3

               n = 1,732

therefore the amplitude must be increased by a factor of 1,732

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In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to
lubasha [3.4K]

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%

negative indicates less than actual value.

3 0
4 years ago
A certain cable of an elevator is designed to exert a force of 4.5x 104 N. If the maximum acceleration that a loaded car can wit
meriva

Answer: 3,383.5 kg

Explanation:

from the question we were given the following

tension (T) =  4.5 x 10^4 N

maximum acceleration (a) = 3.5 m/s^2

acceleration due to gravity (g) = 9.8 m/s^2  ( it's a constant value )

mass of the car and its contents (m) = ?

we can get the mass of the car and it's contents from the formula for tension which is T = ma + mg

    T = m (a + g)

therefore m = T / (a+g)

m = (4.5 x 10^4 / ( 3.5 + 9.8 )

m = 3,383.5 kg

5 0
3 years ago
Which of the following is a definite indicator of a chemical change?
Andrew [12]

Answer:

Some signs of a chemical change are a change in color and the formation of bubbles. The five conditions of chemical change: color chage, formation of a precipitate, formation of a gas, odor change, temperature change.

Explanation:

3 0
4 years ago
What was the maximum speed of the car in your experiment?
MA_775_DIABLO [31]
The measurements used in the experiment is the amount of speed over time.

The measurement of speed is indicated along the “y” axis.

Upon viewing the graph, the highest point along the “y” axis shown is 25 m/s. This would be the maximum.

The maximum speed of the car would be 25 m/s.
7 0
3 years ago
A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will
gtnhenbr [62]
1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t
y(t)=h- \frac{1}{2}gt^2
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2, and initial height h=1.85 m.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0:
0=h- \frac{1}{2}gt^2
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }=  0.61 s

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one: 
v_x=18.1 m/s
On the y-axis, the velocity is given by
v_y(t)=gt
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s
And the speed of the ball is the magnitude of the resultant of the two components:
v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s
8 0
3 years ago
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