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valkas [14]
2 years ago
10

The force stretching the D string on a certain guitar is 130 N. The string's linear mass density is 0.006 kg/m. What is the spee

d of the waves on the string?
Physics
1 answer:
inysia [295]2 years ago
3 0

Answer:

The value is  v  = 147.2 \  m/s

Explanation:

From the question we are told that

    The force is   F =  130 \  N

     The linear mass density is  \mu =  0.006 \  kg/m

     

Generally the speed of the wave on the string is mathematically represented as  

            v  = \sqrt{\frac{F}{\mu } }

=>        v  = \sqrt{\frac{130}{0.006} }

=>        v  = 147.2 \  m/s

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Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free
sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

           ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

           ΔE = W

internal energy can be related to the change in temperature

            ΔE = 3/2 n K ΔT

we substitute

           3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

            W = 3/2 n (T₁- T₂)

3 0
3 years ago
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0
2 years ago
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

or, if taking significant digits into account,

m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
3 years ago
A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fas
Likurg_2 [28]

Answer:

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

8 0
2 years ago
Read 2 more answers
What is the product of the cellular respiration process
zlopas [31]
When oxygen is present, cellular respiration produces carbon dioxide, water, and energy in the form of ATP.

When oxygen is not present, cellular respiration produces lactic acid or alcohol.

Hope this helps! :)
4 0
3 years ago
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