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kolezko [41]
3 years ago
11

A person is standing on a level floor. His head, upper

Physics
1 answer:
BabaBlast [244]3 years ago
8 0

Answer:

y_{cg} = 1.03 m

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity = \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}

y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}

y_{cg} = 1.03 m

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A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
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Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

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time of motion, t = 2.0 s

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

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