All categories

3 years ago

Water flows through a pipe of

Physics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

9.38 m/s

Explanation:

Mass is conserved.

m₁ = m₂

ρ₁ Q₁ = ρ₂ Q₂

Assuming no change in density:

Q₁ = Q₂

v₁ A₁ = v₂ A₂

v₁ π r₁² = v₂ π r₂²

v₁ r₁² = v₂ r₂²

Plugging in values:

(1.50 m/s) (0.0250 m)² = v (0.0100 m)²

v = 9.38 m/s

You might be interested in

When you’re driving a golf ball, a good "follow-through" helps to increase the distance of the drive. A good follow-through mean

Natali [406]

Answer:

$v_{f}$ = F Δt / m

Explanation:

The impulse theorem and momentum is

I = Δp

F Δt = m $v_{f}$ - m v₀

Before touching the ball is without movement, the force is represented by the kinetic energy of the club when it reaches the bottom of the path, so if this force is constant, the time [or the duration of the blow is directly proportional to the speed that the ball acquires.

$v_{f}$ = F Δt / m

The higher this speed, the greater the distance traveled.

3 0

4 years ago

Which city has already hosted the Olympic Games three times?

____ [38]

London, it has hosted the 1908, 1948 and the 2012 summer olympics

8 0

4 years ago

Read 2 more answers

Arthur is testing how well various types of disinfectants can kill E. coli bacteria. He starts with a Petri dish that is covered

maw [93]

The variable is the disinfectant that is used, because there different ones being used

3 0

3 years ago

Read 2 more answers

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$