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Yuki888 [10]
3 years ago
5

A 6.00-kg box is sliding to the right across the horizontal floor of an elevator. The coefficient of kinetic friction between th

e box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2.

Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer: Kinetic frictional force = 23.76N

Explanation: Please see the attachments below

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<span>a cell eliminates endocytosis.

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What services do plants provide? Select the three that apply.
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6 0
3 years ago
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

8 0
1 year ago
Two thousand two hundred frequent business travelers are asked which midwestern city they prefer: Indianapolis, Saint Louis, Chi
drek231 [11]
Frequency Table
City                       Frequency
Indianapolis              124
Saint Louis                416
Chicago                  1,225
Milwaukee                 435

Relative Frequency Table
City                            Relative Frequency
Indianapolis              124/2200 = 31/550
Saint Louis                416/2200 = 52/275
Chicago                  1,225/2200 = 49/88
Milwaukee                 435/2200 = 87/440
5 0
3 years ago
The space velocity of a star is 120 km/s and its radial velocity is 72 km/s. what is its tangential velocity?
Marina CMI [18]
You divide 120 and 72 and get 1.67 repeated i think
3 0
3 years ago
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