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Vanyuwa [196]
3 years ago
5

How an application of atmospheric device work?example siphon​

Physics
1 answer:
Vika [28.1K]3 years ago
6 0

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

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Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density
Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height =  1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴  Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0
3 years ago
What's the answers for all of the questions ?///
Maru [420]
More energy, colour and I think less bright
8 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
In the circuit shown below, 0.25 A of current flows through a 20-Ω resistor. How much voltage is needed to produce this current?
balu736 [363]

Answer:

D 5 V

Explanation:

Without seeing the whole circuit it is impossible to say for certain.

However the simplest circuit would produce a value of

FV = IR = 0.25(20) = 5 v

5 0
3 years ago
For horizontally-launched projectiles, which of the following describes acceleration in both directions with a = 0 and a = -9.8m
natulia [17]

Answer:

4 is the best option for it as there are some stuff in grade to ye

5 0
3 years ago
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