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Veronika [31]
2 years ago
7

A brick 25cm long ,9cm wide 5cm thick has a mass of 1000g determine the greatest pressure that can be exerted by the brick on a

flat surface
​
Physics
1 answer:
lana66690 [7]2 years ago
8 0

Answer:

<em>The greatest pressure is 2178 Pa</em>

Explanation:

<u>Pressure</u>

It's the force applied perpendicular to the surface of an object per unit area over which the force is distributed.

If F is the force applied and A is the surface area, the pressure is calculated as:

\displaystyle P=\frac{F}{A}

The brick has dimensions of 25 cm by 9 cm by 5 cm. There are three possible areas of contact:

A_1=25*9 = 225 cm^2=0.0225\ m^2

A_2=25*5= 125 cm^2=0.0125\ m^2

A_3=9*5= 45 cm^2=0.0045\ m^2

The  brick has a mass of m=1000 g = 1 Kg and exerts a force equal to its weight when placed on a flat surface, thus:

F = m.g = 1 * 9.8 = 9.8 N

The greatest possible pressure will occur when the area is the least possible (0.0045\ m^2), since the pressure and the area are inversely proportional, thus:

\displaystyle P_{max}=\frac{9.8}{0.0045}

\displaystyle P_{max}=2178 \ Pa

The greatest pressure is 2178 Pa

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Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a
victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

  1. F=\frac{GMm}{R^{2} }

Where:

  • F => Gravitational force or force of attraction between two masses
  • M => Mass of asteroid 1
  • m => Mass of asteroid 2
  • R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

  • M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

  • R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

  • m=ρV

This is going to be helpful because the volume of a sphere is:

  • \frac{4}{3}\pi r^{3}

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

  • F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }
  • F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }
  • F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2}  }{36}

Now let´s plug in the values we know:

  1. F = 1 lb     mutual gravitational attraction force
  2. G = 6.67(10)^{-11}     gravitational constant
  3. ρ_{iron} =491.5 \frac{lb}{ft^{3} }

  • 1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}

Solve for r and multiply by 2 because 2r = diameter

  • d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }

Result is d = 21.81 Feet

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pshichka [43]

Answer:

The average speed for the entire run is 12 km/h.

Explanation:      

The average speed is given by the following equation:

\overline{v} = \frac{d_{T}}{t_{T}}

Where:

d_{T}: is the total distance

t_{T}: is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

d_{T} = 13 km + 5 km = 18 km

t_{T} = 1 h + \frac{1}{2} h = 1.5 h

Hence, the average speed is:

\overline{v} = \frac{d_{T}}{t_{T}} = \frac{18 km}{1.5 h} = 12 km/h

Therefore, the average speed for the entire run is 12 km/h.

I hope it helps you!                                                                                      

3 0
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