A fine-grained sedimentary rock is known as ____

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

6 0

3 years ago

Read 2 more answers

Write the factors of potential energy

IgorLugansk [536]

It depends on two factors height and mass

6 0

3 years ago

A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t

jonny [76]

Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

(E_m) = (m × Lf)

b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

(Total heat) = (m × Lf) + mc (T2 - T1)

c) 3,646,458 J = 3646.46 kJ

Explanation:

a) When a pure body changes its phase at meltimgbor boiling point, it does so at a constant temperature. When a pure body melts, the amount of heat responsible for this change is just given by a product od the mass of the body and the body's heat of fusion.

(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

(Total heat) = (m × Lf) + mc (T2 - T1)

c) What is the energy in Joules?

(Total heat) = (m × Lf) + mc (T2 - T1)

m = mass of ice = resulting mass of water = 9 kg

Lf = latent heat of fusion = 334000 J/kg

c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

(Total heat) = (m × Lf) + mc (T2 - T1)

Q = (9 × 334000) + [9 × 4186 × (17 - 0)]

Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

7 0

3 years ago

The weight of a body of certain mass becomes zero in space.why?write with reasons

Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

A fine-grained sedimentary rock is known as _____.