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Gennadij [26K]
3 years ago
13

PLEASE HELP ME THANK YOUUU

Physics
2 answers:
givi [52]3 years ago
7 0
The correct answer is B #markasbrainleist
Taya2010 [7]3 years ago
3 0

Answer:

A) or B) which ever you choose in order

Explanation

Hope it helps:)

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A weightlifter raises a 200-kg barbell through a height of 2 m in 2.2 s. the average power he develops during the lift is
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Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.
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Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the
Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate

\rho = Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3

V_1 = 20m/s

A_1 = 60*10^{-4}m^2

\rho_2 = 0.762kg/m^3

V_2 = 160m/s

PART A) Applying the flow equation we have to

\dot{m} = \rho_1 A_1 V_1

\dot{m} = (2.21)(60*10^{-4})(20)

\dot{m} = 0.2652kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}

A_2 = \frac{0.2652}{(0.762)(160)}

A_2 = 2.175*10^{-3}m^2

Therefore the Area at the end is 21.75cm^2

3 0
3 years ago
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3 years ago
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Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
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