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A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago

For pea plants, T represents a dominant allele for tall pea plants and t is the recessive allele for short pea plants. If two pl

JulsSmile [24]

Answer:

75%(TT,Tt,tt) Hope that helps

4 0

2 years ago

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

How to do this question?

marissa [1.9K]

Answer:

i3 =11.014A

i5 = 3.15A

Explanation:

Here according to k'chofs first law

i1 =i2 + i3

i3 = i4 + i5

For determine the i1 you have to consider the resultant resistor of the system

4 , 1 and 3 resistors are in pararel

Then, Resultant is

1/4 + 1/1 + 1/3 = 1/ R

R = 12/19

For get total we have to add another remaining 3 resistor because of serious

Then Resultant is = 12/19 + 3

= 69/19

Then using V = IR

40 =i3* 69/19

i3 = 11.014 A

Other 3 resistors are parrarel because of this voltage of those resistors are same.

Then i inversely propotional to its resistor

Then ,

i5 * 2 = (i3-i5)*4/5

i 5 = 3.15 A

7 0

3 years ago

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