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3 years ago

A string of length 0.67 m and mass 2.0 g is held at both ends under tension. The string

Physics

1 answer:

ikadub [295]3 years ago

7 0

Answer:

The value is $\lambda = 0.335 \ m$

Explanation:

From the question we are told that

The length of the string is $l = 0.67 \ m$

The mass is $m = 2.0 \ g = 0.002 \ kg$

The frequency produced at fourth harmonic vibration(i.e n =4 ) is $f = 700 \ Hz$

Generally the wavelength of the string is mathematically represented as

$\lambda = \frac{ 2L }{n }$

=> $\lambda = \frac{ 2 * 0.67 }{ 4 }$

=> $\lambda = 0.335 \ m$

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Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

Leviafan [203]

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

5 0

4 years ago

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

A wave with a wavelength of 15m travel at 330m/s.Calculate its frequency.

katovenus [111]

ANSWER:

22 Hz.

Explanation:

Frequency = (speed) divided by (wavelength) = (330) / (15) =

8 0

3 years ago

Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to

abruzzese [7]

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

6 0

3 years ago

As a recreational boat operator, what actions must you take when in a narrow channel?

alexandr402 [8]

One should never anchor in a narrow channel, until unless required very importantly. One should stay to the starboard side, and use a prolonged blast. The announcement must be done to alarm the nearby vessels, about your approach. The vessel should be kept at the outer limit of the starboard side.

6 0

3 years ago

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