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2 years ago

Which constellation is marked by a W or M? a. Cassiopeia b. Draco c. Ursa Minor d. Ursa Major

Physics

2 answers:

yuradex [85]2 years ago

8 0

Answer:

Ursa Minor Constellation

Explanation:

Ursa Minor constellation lies in the northern sky. The constellation's name means “the smaller bear,” or “the lesser bear,” in Latin. The Great Bear constellation is represented by its larger neighbor Ursa Major.

Anestetic [448]2 years ago

6 0

I believe it is the Cassiopeia. It is visible during winter in the early part of the night. It look like a distorted letter W or M.

I am not very sure of it though, I apologize if this is not what you meant.

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3 years ago

Read 2 more answers

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

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2 years ago

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Answer:

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Answer:

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