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DIA [1.3K]
2 years ago
15

A soccer player attempting to steal the ball from an opponent was extending her knee at 50 deg/s when her foot struck the oppone

nt's shin pads. If the player's knee was stopped (0 deg/s) within 0.2 seconds, what angular acceleration did the knee experience?
Physics
1 answer:
denpristay [2]2 years ago
7 0

Answer:

\alpha = -4.36 rad/s^2

Explanation:

Initial angular speed of the player is given as

\omega = 50 deg/s

here we know that

1 deg = \frac{\pi}{180} rad

now we know that

\omega = 50\frac{\pi}{180}

\omega = 0.87 rad/s

now its speed comes to zero in 0.2 s

so angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0 - 0.87 rad/s}{0.2 s}

\alpha = -4.36 rad/s^2

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land
sattari [20]

Answer:

a) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s

a) Impulse

J = final momentum - initial momentum

J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s

Magnitude of impulse

|J| = 219.1 kgm/s

b) Force

F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N

4 0
2 years ago
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2 years ago
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Please re-explain the following phrases in terms of momentum
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C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>
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Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
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