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3 years ago

A cyclist maintains a constant velocity of 6.7 m/s headed away from point

Physics

1 answer:

JulsSmile [24]3 years ago

3 0

His position x = V* t = 6.7*56= 375.2 m from point a

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Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s

umka2103 [35]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=2.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}$

$\\ \sf\longmapsto Acceleration=1.6m/s^2$

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2 years ago

Which statement about the physical properties of matter is true? (15)

Maslowich

Answer:

Psisical properties used to be considered the properties that can be modified by phisical means. New theories state that chemical properties of matter are interrelated to the phisical ones

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3 years ago

The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typic

tia_tia [17]

Answer:

Tension on tendon = 1669800N

Explanation:

Detailed explanation and calculation is shown in the image below

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2 years ago

Which of the following is a metal? Calcium (Ca) Iron (Fe) Sodium (Na) all of these

ANTONII [103]

Iron (Fe) or at least it can be.

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2 years ago

Read 2 more answers

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

2 years ago