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dedylja [7]
4 years ago
7

One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of

negligible weight supported by two scales. In one example the height of the man is 190 cm; when he lies on the board, the left scale reads 450 N and the right scale reads 390 N. The center of gravity of the man relative to his feet on the right is?
Physics
1 answer:
andrey2020 [161]4 years ago
7 0

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

                       = 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

 x = \dfrac{855}{840}

 x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

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Answer:

(C) Decreases by factor of 3

Explanation:

Centripetal acceleration is given by

a = \dfrac{v^2}{r}

where <em>v</em> is the linear velocity and <em>r</em> is the radius of the curve.

Let the centripetal acceleration on the curve of radius <em>R</em> be a_1.

Then

a_1 = \dfrac{v_i^2}{R}

Let the centripetal acceleration on the curve of radius 3<em>R</em> be a_2.

Then

a_2 = \dfrac{v_i^2}{3R} = \dfrac{1}{3}\dfrac{v_i^2}{R} = \dfrac{1}{3}a_1

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3 years ago
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to th
Nastasia [14]

Answer:

T =  40 +  e^{3.68t} e^{2.99}

Explanation:

The differential equation for given  is given as

\frac{dT}{dt} = - k(T-T_s)

integrating above equation we have

ln(T-T_s) = -kt + C

At t = 0 , T(0) =  60

ln(60- 40) = -k\times 0 + C

2.99 = C

 At t =1 , T(1) = 40.49887

ln(40.49787 - 40) = -k\times 1 +  2.99

- k = -3.687

So we have

T- 40  = e^{3.68t + 2.99}

T =  40 +  e^{3.68t} e^{2.99}

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3 years ago
Two masses are joined by a mass less string. A 33-N force applied vertically to the upper mass gives the system a constant upwar
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Lower mass: 1.20 kg, upper mass: 1.28 kg

Explanation:

In order to solve the problem, we consider the forces acting on the upper mass only first.

The upper mass is acted upon three forces:

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Therefore, the equation of the forces for the upper mass is:

F_a - m_u g - T = m_u a

where

a=3.5 m/s^2 is the acceleration (upward)

Solving for m_u,

m_u = \frac{F_a-T}{a+g}=\frac{33-16}{3.5+9.8}=1.28 kg

Now we can find the lower mass by considering the forces acting on it:

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A truck traveling at 80 mph runs into a bridge abutment and crumples for 0.8 m before coming to a full stop. If we estimate the
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U = 80mph = 35.76m/s

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