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-Dominant- [34]
4 years ago
9

Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the ai

r outside the tank. Assume that the density of water is 1.00 g/cm3
Physics
1 answer:
Dovator [93]4 years ago
8 0

Answer:

The net downward force on the tank is 1.85\times10^{5}\ N

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

F=(P+\rho\times g\times h-P_{out})A

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

F=(150\times10^3+1.00\times10^3\times3.71\times14.4-88.0\times10^{3})\times1.60

F=1.85\times10^{5}\ N

Hence, The net downward force on the tank is 1.85\times10^{5}\ N

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Distance for which the bike is ridden = 30 km
Speed at which the bike is driven = 0.75 km/minute
Let us assume the number of minutes taken to travel the distance of 30 km = x
Now we already know the formula of speed can be written as
Speed = Distance traveled/ Time taken
0.75 = 30/x
0.75x = 30
x = 30/0.75
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So the time taken for riding a distance of 30 km will be 40 minutes. I hope this procedure is simple enough for you to understand.
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Which vector is the sum of vectors a and b
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A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T
mart [117]
<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

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

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

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But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

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but E =  4.32 × 10^-13 Joules

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Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

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Can someone answer and explain these questions?
Andru [333]

Answer:

oh that looks hard but sorry im not good at math ether so sorry i couldnt help

Explanation:

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