A vector has an x-component

A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

Artyom0805 [142]

Answer:

$f=81.96 \ Hz$

Explanation:

Givens

$L=95cm$

$m_{sculpture} =13kg$

$m_{wire}=5g$

The frequency is defined by

$f=\frac{v}{\lambda}$

Where $v$ is the speed of the wave in the string and $\lambda$ is its wave length.

The wave length is defined as $\lambda = 2L = 2(0.95m)=1.9m$

Now, to find the speed, we need the tension of the wire and its linear mass density

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

Where $\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}$ and the tension is defined as $T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N$

Replacing this value, the speed is

$v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s$

Then, we replace the speed and the wave length in the first equation

$f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz$

Therefore, the frequency is $f=81.96 \ Hz$

3 0

3 years ago

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When the Moon orbits Earth, what is the centripetal force?

Troyanec [42]

Answer:

D

Explanation:

pls mark brainliest

7 0

3 years ago

What would make oppositely charged objects attract each other more? increasing the positive charge of the positively charged obj

kykrilka [37]

Answer:

A.

Explanation:

I did the test :D

3 0

2 years ago

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A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

7 0

3 years ago