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natta225 [31]
3 years ago
11

Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied

halfway between points A and B), and a moment M=49 ft-lbs. Provide your answer in lbs and pay attention to your signs.
Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

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Which techniques can scientists use to determine the characteristics of Earth's layers? Select the two correct answers.
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The temperature of a black body is 500 and its radiation is of wavelength 600 . If the number of oscillators with energy is 100
stiks02 [169]

Answer: An equation is missing in your question below is the missing equation

a) ≈ 8396

b) 150 nm/k

Explanation:

<u>A) Determine the number of Oscillators in the black body</u>

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attached below is the detailed solution

<u>b) determine the peak wavelength of the black body </u>

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applying Wien's law / formula

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T = 20,000 K

b = 3 * 10^6 nm

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4 0
2 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
2 years ago
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