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Arturiano [62]
3 years ago
15

the volume of a water tank is 5m×4m×2m. If the tank is half filled with water. calculate the pressure exerted at the bottom of t

he tank​
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer:

10⁴ Pa

Explanation:

  • Volume = 5m * 4m * 2m

When the tank is half filled ,

  • Volume = 5m * 4m * 1m = 20m³

Also we know that ,

  • Density of water = 10³ kg/ m³

\longrightarrow Density = Mass/Volume

\longrightarrow 10³ kg/m³ = m/20m³

\longrightarrow m = 2 * 10⁴ kg

So that ,

\longrightarrow Weight = mg

\longrightarrow F = 2*10⁴ × 10 N

\longrightarrow F = 2 * 10⁵ N

And ,

\longrightarrow Pressure = Force/Area

\longrightarrow P = 2 *10⁵/ 5 * 4 Pa

\longrightarrow P = 10⁴ Pa

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Kinetic energy = 1/2 * mass * velocity^2

In this case,
KE = 1/2 * 1569 kg * (15 (m/s))^2 = 176,5 kN
8 0
3 years ago
A chicken crosses a 7.50 m wide road at a constant speed of 0.367 m/s. How much time does it take to cross (in seconds)?
mars1129 [50]
<h3><u>Answer;</u></h3>

= 20.436 seconds

<h3><u>Explanation;</u></h3>

Speed = Distance × time

Therefore;

Time = Distance/speed

Distance = 7.50 m, speed = 0.367 m/s

Time = 7.50/0.367

         <u>= 20.436 seconds </u>

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3 years ago
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Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding whe
zloy xaker [14]

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

4 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
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