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Ksju [112]
3 years ago
13

The 9-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point Clocated on the vertica

l wall. If the tension in the cable is 3 kN, determine the moment about A of the force exertedby the cable at B.
Physics
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

27000 Nm

Explanation:

The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m

M = FL = 3000 * 9 = 27000 Nm

So the moment about A is 27000 Nm

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A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm
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Answer:

v(0)=2m/s

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

v(t)=\omega  *A*cos(\omega t + \phi)

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Express the amplitude in meters:

10cm*\frac{1m}{100cm} =0.1m

The angular frequency can be found using the next equation:

\omega=\sqrt{\frac{k}{m} }

Using the data provided:

\omega=\sqrt{\frac{400}{1} } =20

At the equilibrium position:

\phi=0

v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s

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3 years ago
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Answer:

Compound.

Explanation:

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4 years ago
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A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring
borishaifa [10]

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

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where

m1 = 1.14 kg

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v2  = 0 m/s

putting all value in the above equation

1.14 *2.0+ 0 =(1.14+1.14)v^{2}

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When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin
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Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

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λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

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m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

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