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3 years ago

A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?

Physics

1 answer:

vovangra [49]3 years ago

5 0

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

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A vector is parallel to the y axis, what is it x component?<br>

slavikrds [6]

Answer:

Explanation:

A vector is parallel to the y axis .

Let its magnitude be A . So the vector can be represented as A j .

where i and j are unit vectors in x and y axis direction .

The x component of A j will be dot product of A j with i

The x component of A j = A j . i

= A x 0 [ Since j . i = 0 ]

= 0

4 0

2 years ago

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

$(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}$

$k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}$

$\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}$

$\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}$

$\Delta T_{air} = 44.9 \Delta T_{glass}$

There are two layers of Glass and one layer of Air so the total temperature would be given as,

$\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}$

$\Delta T = 2\Delta T_{glass} +\Delta T_{air}$

$20\°C = 46.9\Delta T_{glass}$

$\Delta T_{glass} = 0.426\°C$

Finally the rate of heat flow through this windows is given as,

$\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}$

$\Delta {Q}{t} = 0.84*24*10 -3*0.426$

$\Delta {Q}{t} = 179W$

Therefore the correct answer is D. 180W.

3 0

2 years ago

A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²

Then we know that k is constant for both trials, we have:
k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0

3 years ago

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$