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svetlana [45]
3 years ago
9

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

. Find the pressure exerted on the floor by each leg of the chair, assuming the weight is evenly distributed. Express your answer using two significant figures.
Physics
1 answer:
tankabanditka [31]3 years ago
5 0

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

<em>F = Mg</em>

<em>⇒F = 84.8×9.81</em>

<em>⇒F = 831.888 N</em>

Area of the contact point

<em>A = πR²</em>

<em>⇒A = π0.006²</em>

<em>⇒A = π0.000036 m²</em>

Area of the four points is

<em>4A = 0.000144π m²</em>

Pressure

p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

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BlackZzzverrR [31]

Answer:

25J

Explanation:

power = work done ÷ time taken.

and work done = force applied × distance covered.

according to the question, the athlete lifts the weight of 100N upto 5m therefore;

100N × 5m = 500N/m

then onwards,

the work done (500 N/m) should be divided by the time taken to calculate the power,

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is your answer.

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8 0
3 years ago
Read 2 more answers
A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
nikitadnepr [17]

Answer:

The number A(t) of grams of salt in the tank at time t is A(t) = 150 - 110 e^{-\frac{t}{50} }

Explanation:

Knowing

\frac{dA}{dt} = Rin - Rout

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = (\frac{A(t)}{150} g/L) * (3 L/min) = \frac{A(t)}{50} g/min

Substituting this results

\frac{dA}{dt} = 3 - \frac{A(t)}{50} --> \frac{dA}{dt} + \frac{1}{50} A(t)  = 4

Thus, integration factors is

e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }

e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50} A(t) = 4 e^{\frac{t}{50} }

e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

A(t) = 150 - 110 e^{-\frac{t}{50} }

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zepelin [54]

The correct answer to the question is: 78.4 N.

CALCULATION:

As per the question, the mass of the object is given as m = 8 Kg.

We are asked to calculate the force of 8 Kg object pushing up from the table.

The force pushing up from the table is nothing else than the normal reaction.

The normal reaction of the object is equal to the weight of the object as the object is simply resting on the table.

Hence, normal reaction = Weight

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Here, g is known as acceleration due to gravity.

Hence, the force of the object pushing up from the table is 78.4 N.


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