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yawa3891 [41]
3 years ago
9

5. If you walk 70 meters north in 5 minutes and 20 meters south in 15 seconds. What was the average

Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

kjfvkuzfvnv

Explanation:

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larisa86 [58]

Answer: (a) t1 = omega1/alpha

(b) theta1 = 1/2 * alpha*theta1^2

(c) t2 = omega2/5*alpha

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How many kisses is 50 hectokisses?<br><br> What would you be if you earn a megabuck year?
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1.) 6 hectokisses 50 times

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3 years ago
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A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

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3 years ago
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nika2105 [10]

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6 0
2 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

6 0
3 years ago
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