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Lyrx [107]
2 years ago
15

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

um speed can the child walk?
Physics
1 answer:
AnnZ [28]2 years ago
8 0

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

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A = h / n => h = a*n

a = 0.290 hit / time
n = 300 times

=> h = 0.290 hit / time * 300 time = 87 hits

Answer: 87 hits
4 0
2 years ago
A ray diagram without the produced image is shown.
Goryan [66]

Answer:

B) inverted and real

Explanation:

6 0
2 years ago
Water floats on a liquid called "carbon tetrachloride." The two liquids do not mix. A light ray passing from water into carbon t
Oduvanchick [21]

Answer:

1.26

Explanation:

index  \: of  \: refraction \:  =  \:  \ \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }

\frac{ \sin(45.0) }{ \sin(40.1) }

= 1.26

5 0
2 years ago
A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

\to \sigma y  =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\

Calculating Shear Transverse:

\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}

        = \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi

= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\  in

\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\

       = [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi

8 0
2 years ago
To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?
koban [17]

<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>

<u>Answer:</u> <em>sine</em>

<u>Explanation:</u>

Thus a_x=0,a_y=g

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion v= u+at

Thus velocity along x direction v_x=u cos θ

Velocity along y direction v_y= u sinθ -gt

Sign of g is negative.

3 0
2 years ago
Read 2 more answers
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