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MA_775_DIABLO [31]
2 years ago
6

The solid shaft with a 20 mm radius is used to transmit the torques applied to the gears. Determine the maximum torsional shear

stress developed in in the shaft.
Physics
1 answer:
AysviL [449]2 years ago
3 0

Answer:

τ = (7.96 x 10⁴ m⁻³)T

This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.

Explanation:

The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:

τ = Tc/J

where,

τ = Maximum Allowable Shear Stress = ?

T = Maximum Torque Applied to the Shaft

c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m

J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴

Therefore,

τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)

<u>τ = (7.96 x 10⁴ m⁻³)T</u>

<u>This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.</u>

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3 years ago
An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached
Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0
3 years ago
A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr
Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

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1 year ago
You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of
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2 years ago
Noooooooooooooooooooooooo
Afina-wow [57]

Answer:

yes

Explanation:

5 0
3 years ago
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