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vovangra [49]
2 years ago
13

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

between the boxes and the horizontal surface.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
5 0
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
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Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is
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the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t
LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

4 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

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Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

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Now,

This is relativistic in nature

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E = m_{p}c^{2}

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This energy of proton is \simeq 250 GeV

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T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

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\Delta t_{o} = \Delta t

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3 0
2 years ago
Question 1
Vesna [10]

Answer:

1)  g = 4π² / m, 3) xaxis the  length of the pendulums and the y axis the period squared

Explanation:

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We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

        y = (4π² / g)  l

so the slope is

         m = 4π² / g

clearing

         g = 4π² / m

where the slope can be found with the values ​​of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

    T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

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5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

5 0
2 years ago
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