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Nostrana [21]
2 years ago
9

1. You walk along a long straight school corridor for 55

Physics
1 answer:
Sedaia [141]2 years ago
3 0

Explanation:

14 m is the displacement and towards the north

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A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq
Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>
  • The following expression can be used to relate the fundamental frequency to the driving frequency;

                                        f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

  • The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

                                          f(4) = 4× f (1)

                                          480 = 4× f(1)

                                         f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

  • To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
  • For, n = 5,

                      f(n) = n 120Hz,

                      f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

#SPJ4

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A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear
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Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
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3 years ago
What is #6<br><br> IM GIVING 40 POINTS
frosja888 [35]

Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity. Its standard units are meters per second per second, or meters per second squared. (this is for number 3)

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3 years ago
Where is this earthquake's epicenter?
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The correct answer is C
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3 years ago
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A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
nikitadnepr [17]

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

6 0
2 years ago
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