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Xelga [282]
2 years ago
8

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Physics
1 answer:
mamaluj [8]2 years ago
3 0

Answer:

sup qwertyasdfghjk

Explanation:

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2 years ago
The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I
erma4kov [3.2K]

Answer:

Part a)

I_{end} = \frac{mL^2}{3}

Part b)

I_{edge} = \frac{2ma^2}{3}

Explanation:

As we know that by parallel axis theorem we will have

I_p = I_{cm} + Md^2

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

I = \frac{mL^2}{12}

now if we need to find the inertia from its end then we will have

I_{end} = I_{cm} + Md^2

I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2

I_{end} = \frac{mL^2}{3}

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

I = \frac{ma^2}{6}

now if we need to find the inertia about an axis passing through its edge

I_{edge} = I_{cm} + Md^2

I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2

I_{edge} = \frac{2ma^2}{3}

7 0
3 years ago
How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?
Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

Q = ms\Delta T

We know that

melting temperature of copper = 1085 degree C

Specific heat capacity of copper = 385 J/kg C

now we have

Q = 3(385)(1085 - 83)

Q = 1157310 J

Q = 1157.3 kJ

now in order to melt the copper we know the heat required is

Q = mL

here we know that

L = 205 kJ/kg

now from above formula

Q = 3(205) kJ

Q = 615 kJ

now total heat required will be

Q = 1157.3 kJ + 615 kJ

Q = 1772.3 kJ

As we know that

1 Cal = 4.18 kJ

now we have

Q = \frac{1772.3}{4.18} = 430 KCal

6 0
2 years ago
Which coordinate system would be most useful for two observers (one in Michigan and one in Florida) who wants to observe the sam
Mkey [24]

Answer : Celestial or azimuth - altitude

Explanation :  

Celestial : The celestial coordinates that are analogous to longitude and latitude are called RA and Dec.

RA = Right Ascension

Dec = Declination

RA is the measured in unit of time and Dec is measured in degree.

The equatorial coordinate system is the projection of the latitude and longitude coordinate system on the celestial sphere.

Azimuth - altitude :  Azimuth - altitude define the location of an object in the sky.

The altitude is the distance of an object appears to be above the horizon.

The Azimuth of the object is the angular distance  along the horizon to the location of the object.

3 0
3 years ago
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\. A mixture of gases con-tains oxygen, nitrogen, and water vapor. What physical process could you use to remove the water vapor
Eddi Din [679]

Answer:

Condensation.

Explanation:

The boiling point of water is much higher than that of either nitrogen or oxygen gas . So when the mixture is condensed  to a temperature lower than

100°C , water vapor will come out first in the form of water leaving other

elements of mixture in gaseous phase. In this way, water vapor will get separated from others.

3 0
3 years ago
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