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IrinaK [193]
3 years ago
14

A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5

cm in diameter and the outlet flow is 50 L/s.
Calculate the propulsion force at the moment of departure, that is, with the boat at rest. The pressure at the inlets and outlets is assumed to be practically atmospheric (water density = 1000 Kg /m3).
Physics
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

The propulsion force at the moment of departure, is 49 N  

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N      

Therefore, the propulsion force at the moment of departure, is 49 N  

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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
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elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

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V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
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zheka24 [161]
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kkurt [141]

Answer:

I font now but I think its 2.0 + 78 w -60

8 0
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