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andre [41]
2 years ago
14

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

beam to be balanced, a 2.9-kg mass must be placed on the left-end, and a 8.00-kg mass must be placed at another location on the beam. How far from the left-end is the location of the 8.00-kg mass
Physics
1 answer:
Shalnov [3]2 years ago
8 0

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

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Answer key says b but i think it is false please help me
klasskru [66]
Energy in a spring:

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5 0
3 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
Strike441 [17]

Answer:

E=3.5(8.98*10^{6}x-2.69*10^{15}t)

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)

Explanation:

The electric field equation of a electromagnetic wave is given by:

E=E_{max}(kx-\omega t) (1)

  • E(max) is the maximun value of E, it means the amplitude of the wave.
  • k is the wave number
  • ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

k=\frac{2\pi}{\lambda}            

k=8.98*10^{6} [rad/m]      

And the relation between λ and f is:                

c=\lambda f

f=\frac{c}{\lambda}

f=\frac{3*10^{8}}{700*10^{-9}}

f=4.28*10^{14}

The angular frequency equation is:

\omega=2\pi f

\omega=2\pi*4.28*10^{14}

\omega=2.69*10^{15} [rad/s]

Therefore, the E equation, suing (1), will be:

E=3.5(8.98*10^{6}x-2.69*10^{15}t) (2)

For the magnetic field we have the next equation:

B=B_{max}(kx-\omega t) (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

E_{max}=cB_{max}

B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}

B_{max}=1.17*10^{-8}T

Putting this in (3), finally we will have:

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t) (4)

I hope it helps you!

8 0
3 years ago
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