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ryzh [129]
2 years ago
12

A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the

weight of the person is evenly distributed, determine the pressure exerted on the floor by each leg.
Physics
1 answer:
kkurt [141]2 years ago
8 0

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

F = (m_{person}+m_{stool})g\\\\F =  (4.5 + 89)*9.8\\\\F = 916.3 N

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2

Pressure due to weight,

P = Weight/A

P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa

P = 4.96 × 10⁵ Pa

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8 0
3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
How does a higher heat capacity affect the amount of time it takes for an object’s temperature to change?
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8 0
3 years ago
If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen
amm1812

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

           d sin θ = m λ

let's use trigonometry for the angle

           tan θ = y / L

how the angles are small

          tan θ = sin θ / cos tea = sin θ

 

we substitute

         sin θ = y / L

        d y / L = m λ

         y = m λ L / d

the light fulfills the relation of the waves

       c = λ f

       λ = c / f

       λ = 3 10⁸ /375

       λ = 8 10⁵ m

first order m = 1

let's calculate

        y = 1  8 10⁵  4030 10-9 / 950 10-9

        y = 33.93 10⁵ m

6 0
2 years ago
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