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zlopas [31]
3 years ago
6

If a truck weighs 18,000 N and it’s tires are inflated to a pressure of 190 kPa how large is the area of the trucks tires that a

re in contact with the ground
Physics
1 answer:
Mamont248 [21]3 years ago
4 0

0.095m²

Explanation:

Given parameters;

Weight of truck = 18000N

Pressure = 190kPa or 190000Pa

Unknown:

Area of truck tires = ?

Solution:

Pressure is given as the force per unit area of a substance:

            Pressure = \frac{force }{Area}

  Since Area is the unknown, we make it the subject of the expression;

            Area = \frac{force }{pressure}

                     =\frac{18000}{190000}

                     =  0.095m²

Learn more:

Pressure brainly.com/question/7882867

#learnwithBrainly

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A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i
Luden [163]

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

5 0
2 years ago
The asteroid, Ida, has a small moon, Dactyl, that orbits at a speed of 5.66 m/s in an orbit of radius 90, 000m . What is Ida's m
cricket20 [7]

Answer:

4.3 x 10^16 kg

Explanation:

M = rv^2/G =[90,000 x 5.66^2] / [6.67 x 10^-11]

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4 0
3 years ago
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