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3 years ago

A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.

Physics

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

Pressure = 39200 N/m²

Explanation:

Given the following data;

Mass = 60 kg

Area = 150 cm²

Conversion:

1 centimeter square = 0.0001 meter square

150 centimeter square = 150 * 0.0001 = 0.015 m²

To find the pressure exerted on the ground by the person;

Mathematically, pressure is given by the formula;

$Pressure = \frac {Force}{area}$

First of all, we would determine the force exerted by the person.

Force = mass * acceleration due to gravity.

We know that acceleration due to gravity is equal to 9.8 m/s².

Substituting into the formula, we have;

Force = 60 * 9.8

Force = 588 Newton

Next, find the pressure;

$Pressure = \frac {Force}{area}$

Substituting into the formula, we have;

$Pressure = \frac {588}{0.015}$

Pressure = 39200 N/m²

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3 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

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Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

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The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

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3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

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