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Angelina_Jolie [31]
3 years ago
11

Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

lowing intervals would you expect t to fall? You do not have to calculate t exactly to answer this question; just make an estimate.
Physics
1 answer:
iogann1982 [59]3 years ago
6 0

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

\large{\boxed{\bold{\sum F=0}}

(forces acting as translational motion only, not including rotational forces)

\displaystyle \sum F_x = 0\\\\ \sum F_y = 0

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g

we assume g = 10 m/s², then :

\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}

So the value of T is between 2000 and 4000 N

<h3>Learn more</h3>

Newton's Law

brainly.com/question/13725525

Keywords : the loudspeaker, ropes, Tension, mass, intervals

#LearnWithBrainly

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kogti [31]

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

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Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

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3 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
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Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

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g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
3 years ago
A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c
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First establish the summation of the forces acting int the ladder

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So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

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H = 1.4 m can the 60 kg person climb berfore the ladder will slip

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MariettaO [177]

Answer:

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julia-pushkina [17]

Answer:

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Explanation:

From the question we are told that

    The wavelength of each source is  \lambda  = 720 \ nm =  720 *10^{-9} \  m

     The distance from the pinhole D = 10 \ m

     The diameter is d =  1.2 mm = \frac{1.2}{1000} = 0.0012 \ m

Generally from Rayleigh's criterion  we have that the distance between the sources of light for their diffraction patterns is mathematically represented as '

        y  =  \frac{ 1.22 *  \lambda *  D}{d}

=>      y  =  \frac{ 1.22 * 720 *10^{-9}*   10 }{ 0.0012}

=>      y  =  0.00732 \ m      

6 0
3 years ago
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