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cluponka
11 days ago
8

The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress

criterion, determine the magnitude of the torque T for which yield occurs when P = 56 kips. (Round the final answer to two decimal places.)
Physics
1 answer:
Keith_Richards [3.2K]11 days ago
7 0

Answer:T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42 ksi = 42 x 6.89 = 289.58 MPa

To find Torque = T =?

Solution:σ = (P x r) / T

Thus, T = (P x r) / σ

Calculating T = (386.11 x 0.02) / 289.58

results in T = 0.03 Nm.

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Suppose a new asteroid was recently discovered which takes 557 months to orbit the Sun once (that's equal to 16,700 days or 46.4
Keith_Richards [3271]

Answer:

The average distance of the new asteroid from the Sun is estimated to be (2.02 × 10⁶) km.

Explanation:

The orbital speed of planets varies based on their distance from the Sun, which also affects their orbital period.

With its 557 months, equivalent to 46.4 years for an orbit around the Sun, the new asteroid's speed is situated between the orbital speeds of Saturn and Uranus.

Uranus orbits the Sun in 84 years at 24.61 km/hour,

while Saturn completes its orbit in 29.4 years at 34.82 km/hour.

To interpolate the speed for our asteroid at 46.4 years,

we denote its speed as x.

84 years ----> 24.61 km/h

46.4 years ----> x km/h

29.4 years -----> 34.82 km/h

Setting up the proportion:

(84 - 46.4)/(46.4 - 29.4) = (24.61 - x)/(x - 34.82)

Solving for x gives the asteroid's speed as 31.64 km/hr.

To find the average speed, use the formula:

Average speed = (total distance)/(time taken).

The total distance covered equals the circumference of the orbit around the Sun = 2πR,

where R = distance from the asteroid to the Sun.

Time taken = 16700 days = 16700 × 24 hours = 400800 hours.

Thus, we find that 31.64 = (2πR)/400800.

From this, we get 2πR = 31.64 × 400800 = 12681312 km.

And, R = 12681312/(2π) = 2018293.5 km = (2.02 × 10⁶) km.

8 0
1 month ago
When θ= 0 ̊, the assembly is held at rest, and the torsional spring is untwisted. if the assembly is released and falls downward
Sav [3153]
The rod measures 450mm in length, while the disk has a radius of 75mm. An upward-supporting pin holds the assembly in place when Θ=0, and there exists a torsional spring with a constant of k=20N m/rad at the pin. One end of the rod connects to the pin, while the other connects to the disk.


7 0
1 month ago
. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it
Keith_Richards [3271]

Ans    specifically, 4

Explanation:

7 0
1 month ago
Recall that weight is a force and is equal to m*g, where g is the acceleration due to gravity exerted by the Earth near the Eart
Sav [3153]
145.43 N
3 0
1 month ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
ValentinkaMS [3465]

Complete Question:

Imagine an asteroid on a direct collision course with the Earth at the equator, embedding itself just beneath the surface. What mass must this asteroid possess, in terms of Earth's mass M, to extend the length of the day by 25.0% due to the collision? Assume the asteroid's mass is negligible relative to the Earth and that the Earth is homogenous throughout.

Answer:

m = 0.001 M

For detailed calculations, refer to the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/[[TAG_14]]

6 0
20 days ago
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