Consider the cylindrical weir of diameter 3 m and length 6m. If the fluid on the left has a specific gravity of 0.8, find the magnitude and direction of the resultant force

Answers

Answer 1

This question is incomplete, the complete question is;

Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.

Answer:

- the magnitude of the resultant force is 557.32 kN

- the direction of resultant force is  48.29°

Explanation:

Given the data in the question and the diagram below,

First we work on the force on the left hand side.

Left Horizontal

[tex]F_{LH[/tex] = βgAr

here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²

we substitute

[tex]F_{LH[/tex] = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N

Left Vertical

[tex]F_{LV[/tex] = ( βgπh² / 2 ) × W

we substitute

[tex]F_{LV[/tex] = [ ( ( 1.6 × 1000 ) × 9.81  × π(1.5)² ) / 2 ] × 6 = 332845.458 N

Now we go to the right hand side

Right Horizontal

[tex]F_{RH[/tex] = βgAh

here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²

we substitute

[tex]F_{RH[/tex] = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N

Right Vertical

[tex]F_{RV[/tex] = ( βgπh² / 4 ) × W

we substitute

[tex]F_{RV[/tex] = [ ( ( 0.8 × 1000 ) × 9.81  × π(1.5)² ) / 4 ] × 6 =  83211.36 N

Hence

Fx = [tex]F_{LH[/tex] - [tex]F_{RH[/tex] = 52974 N - 423792 N =  370818 N

Fy = [tex]F_{LV[/tex] + [tex]F_{RV[/tex] = 332845.458 N + 83211.36 N = 416056.818 N

R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N

R = 557.32 kN

Therefore, the magnitude of the resultant force is 557.32 kN

Direction of resultant force;

tanθ = Fy / Fx

we substitute

tanθ = 416056.818 N / 370818 N

tanθ = 1.121997

θ = tan⁻¹( 1.121997 )

θ = 48.29°

Therefore, the direction of resultant force is  48.29°

Consider The Cylindrical Weir Of Diameter 3 M And Length 6m. If The Fluid On The Left Has A Specific

Related Questions

9) An electrical appliance has a resistance of 25 N. When this electrical ap-
pliance is connected to a 230 V supply line, the current passing through it
will be:
d) 92 A
a) 0.92 A
c) 9.2 A
b) 2.9 A​

Answers

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

A spherical light bulb dissipates 100W and is of 5cm diameter. Assume the emissivity is 0.8 and the irradiation is negligible. What is the surface temperature of this spherical light bulb

Answers

Answer:

[tex]T=728.9K[/tex]

Explanation:

Power [tex]P=100W[/tex]

Diameter [tex]d=5[/tex]

Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]

Emissivity [tex]e=0.8[/tex]

Generally the equation for Area of Spherical bulb is mathematically given by

[tex]A=4\pi r^2[/tex]

[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]

[tex]A=7.85*10^{-3}m^2[/tex]

Generally the equation for Emissive Power bulb is mathematically given by

[tex]E=e\mu AT^4[/tex]

Where

[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]

Therefore

[tex]T^4=\frac{E}{e\mu A}[/tex]

[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]

[tex]T=^4\sqrt{2.80*10^{11}}[/tex]

[tex]T=728.9K[/tex]

Giúp em câu này với ạ,em cảm ơn

Answers

Ảnh lỗi hả bạn ơi, gửi lại ảnh đi

(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find the first six positive frequencies that will be present in the replicated spectrum.

Answers

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

[tex]$f_{m_1}= 65 \ Hz$[/tex] ,  and

[tex]$f_{m_2}= 95 \ Hz$[/tex]

Sampling rate [tex]f_s = \ 245 \ Hz[/tex]

The positive frequencies at the output of the sampling system are :

[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]

When n = 0,

[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]

when n  = 1,

[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]

[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]

[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]

When n = 2,

[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]

[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz


14. Name the 3 primary causes for the expected mass migration due to Climate
Change.

Answers

Answer: sea-level rise, extreme weather events, and drought, and water scarcity.

Explanation:

The plum pudding model of the atom states that

Answers

Answer:

According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

Explanation:

Hope this helps you

Answer:

The plum pudding model of the atom states that  had negatively-charged electrons embedded within a positively-charged "soup."

Explanation:

Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.

A proton moves across a magnetic field and feels a force. If an electron were to
move at the same speed in the same direction across the same magnetic field, the
force
O 1) would be smaller in the opposite direction
O2) would be the same magnitude in the same direction
3) would be the same magnitude in the opposite direction
4) would be larger in the opposite direction
5) would be smaller in the same direction

Answers

potang ina mooooooo bubu hayop kaaaaapestrng yawa

Explanation:

peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa

At an amusement park, a 7.00 kg swimmer uses a water slide to enter the main pool. The swimmer starts at rest, slides without friction, and descends through a vertical height of 0.40 m. Find the total energy present when the swimmer is at the top of the slide.

Answers

Answer:

27.44 J

Explanation:

We can find the energy at the top of the slide by using the potential energy equation:

PE = mgh

At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.

The swimmer's mass is given as 7.00 kg.

The acceleration due to gravity is 9.8 m/s².

The (vertical) height of the water slide is 0.40 m.

Substitute these values into the potential energy equation:

PE = (7.00)(9.8)(0.40) PE = 27.44

Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.

Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.

A thermodynamic system consists of an ideal gas at a volume of 3.50 L and initial pressure of 6.2 × 104 Pa. As the volume is held constant, the pressure is increased to 8.2 × 104 Pa. What work is involved in this process?

Answers

Answer:

0 J

Explanation:

Since work done W = PΔV where P = pressure and ΔV = change in volume.

Since the volume is constant, ΔV = 0

So, Work done, W = PΔV = P × 0 = 0 J

So, the work done is 0 J.

When a person sits erect, increasing the vertical position of their brain by 38.6 cm, the heart must continue to pump blood to the brain at the same rate. (a) What is the gain in gravitational potential energy (in J) for 110 mL of blood raised 38.6 cm

Answers

Answer:

the gain in gravitational potential energy is 0.4369 J

Explanation:

Given the data in the question;

from the definition of density, we know that;

ρ = m / V

where ρ is density, m is the mass and V is the volume.

Now, lets make mass the subject of the formula

m = ρV  ------ let this be equation 1

Now, we know that potential energy PE = mgh ------- let this be equation 2

where m is mass, g is acceleration due gravity, and h is the height.

substitute equation 1 into 2

PE = ρVgh

given that; V =  110 mL = 110 × 10⁻⁶ m³, h = 38.6 cm = 38.6 × 10⁻² m, g = 9.8 m/s², ρ = 1.05 × 10³ kg/m³

we substitute

PE = (1.05 × 10³ kg/m³) × (110 × 10⁻⁶ m³) × 9.8 m/s² × 38.6 × 10⁻² m

PE =  0.4369 J

Therefore,  the gain in gravitational potential energy is 0.4369 J

When the bell in a clock tower rings with a sound of 474 Hz, a pigeon roosting in the belfry flies directly away from the bell.

If the pigeon hears a frequency of 453Hz, what is its speed?

Answers

Answer:

15.44 m/s

Explanation:

Below is the given values:

The bell in rings with a sound = 474 Hz

If the frequency of pigeon hears = 453 Hz

Speed = ?

Use below formula:

Frequency = [(Vs - Vo) / Vs ] x fo

Vs = Speed of sound

Vo = Speed of observer

fo = Sound frequency

Frequency = [(Vs - Vo) / Vs ] x fo

453 = [(343 - Vo) / 343 ] x 474

453 / 474 =  [(343 - Vo) / 343 ]  

0.955 = (343 - Vo) / 343

0.955 x 343 = 343 - Vo

327.56 = 343 - Vo

Vo = 343 - 327.56

Vo = 15.44 m/s

A meterstick of negligible mass is placed on a fulcrum at the 0.4 m mark, with a 1 kg mass hung at the zero mark and a 0.5 kg mass hung at the 1.0 m mark. The meterstick is held horizontal and released. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly:________

a. 1 Nm
b. 2 Nm.
c. 2.5 Nm.
d. 7 Nm
e. 7.5 Nm

Answers

Answer:

The net torque is 0.98 Nm.

Explanation:

The torque is given by

Torque = force x perpendicular distance

The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.

Take the torques about the fulcrum.

Torque =  1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6

Torque = 3.92 - 2.94 = 0.98 Nm

What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?

Answers

Answer:

,hgfghjhytreftgyhujijjuhygtfrderftgyh

Explanation:

g Select True or False: Boyle's Law states that the volume of a fixed amount of gas maintained at constant temperature is directly proportional to the gas pressure

Answers

Answer: The given statement is False.

Explanation:

Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature and the number of moles.

Mathematically,

[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)

OR

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are the initial pressure and volume of the gas

[tex]P_2\text{ and }V_2[/tex] are the final pressure and volume of the gas

Hence, the given statement is False.

Which of the following absorbs the energy required by photosynthesis?

Answers

Answer:

There are no options, so....

Explanation:

Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.

Suppose a proton is moving with a speed of 10 m/s in a direction parallel to a uniform magnetic field of 3.0 T. What is the magnitude and direction of the magnetic force on the proton

Answers

Answer:

the magnetic force on the proton is zero.

Explanation:

Given;

speed of the proton, v = 10 m/s

magnitude of the magnetic field, B = 3 T

The magnitude of the magnetic force on the particle is calculated as;

F = qvBsinθ

where;

θ is the angle between the velocity of the particle and the magnetic field

Since the particle is moving parallel to the magnetic field, θ = 0

F = qvBsin(0)

F = 0

Therefore, the magnetic force on the proton is zero.

Answer:

This is a trick question, in that the numbers do not matter.

Study the dependence of the magnetic force on the direction of the magnetic field and the direction of motion of the particle. When is it a maximum? When is it a minimum?

Hint: In vector notation, this is often expressed as q v x B, where q is the electric charge of the particle, v is its velocity (a vector), and B is the magnetic field vector.

A gauge is attached to a pressurized nitrogen tank reads a gauge pressure of 28 in of mercury. If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank

Answers

Answer:

The absolute pressure is 28.15 psi.

Explanation:

Gauge pressure = 28 inch of Mercury

Absolute pressure, Po = 14.4 psi

The absolute pressure is the sum of the gauge pressure and the absolute pressure.  

gauge pressure = 28 inch = 0.7112 m of Mercury

= 0.7112 x 13.6 x 1000 x 9.8 = 94788.736 Pa

= 13.75 psi

The absolute pressure is

P = 14.4 + 13.75 = 28.15 psi

What are moana's hobbies

Answers

The movie? Or the book

Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values (rather than two as it does in our universe). If this were the case, and the Pauli Exclusion Principle still applies, how many electrons would be allowed in each orbital

Answers

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

Calculate the energy in electron volts of X-rays that have a frequency of 6.00 x 1016 Hz.

207 eV
228 eV
249 eV

Answers

Answer:

228

Explanation:

The 2nd maximum of a double slit diffraction pattern makes an angle 20 degrees when a wavelength of 600 nm is used. What is the angle for the 5th maximum when the slit separation d is increased to 1.5d keeping the wavelength the same. g

Answers

Answer:

θ = 34.77°

Explanation:

From diffraction equation:

[tex]m\lambda = dSin\theta[/tex]

where,

m = order of diffraction

λ = wavelength of light used

d = slit separation

θ = angle

Therefore, for initial case:

m = 2

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = ?

θ = angle 20°

Therefore,

[tex](2)(6\ x\ 10^{-7}\ m)=d(Sin\ 20^o)\\\\d = \frac{12\ x 10^{-7}\ m}{0.342}\\\\d = 3.5\ x\ 10^{-6}\ m[/tex]

Now, for the second case:

m = 5

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m

θ = angle = ?

Therefore,

[tex](5)(6\ x\ 10^{-6}\ m) = (5.26\ x\ 10^{-6}\ m)Sin\theta\\\\Sin\theta = \frac{(5)(6\ x\ 10^{-7}\ m)}{(5.26\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.5703)[/tex]

θ = 34.77°

what mass of water will release 60000 J of energy while cooling from 40.0 degrees celsius to 20.0 degrees celsius?

Answers

Answer:

714.286g

Explanation:

Q=mcT

60000=m×4.2×(40-20)

m=714.285....g

3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
at 25°C. If the rate of degradation of the drug is 0.05 day-1,
calculate the time required for the initial concentration to
drop to (a) 50% (half-life) and (b) 90% (shelf-life) of its initial
value.

Answers

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

Assume the following vehicles are all moving at the Sam speed .it would be harder to change the velocity of which vehicle . What law is it

Answers

Answer:what the choices

Explanation:

Describe 3 Levers of Power and how they work.

DESCRIBE

Answers

First class lever – the fulcrum is in the middle of the effort and the load.
Second class lever – the load is in the middle between the fulcrum and the effort.
Third class lever – the effort is in the middle between the fulcrum and the load.

If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?

50 W
90 W

Answers

the 50 W bulb would have a greater current than the 90 W bulb

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (4i - 9 j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:

Answers

Answer: [tex]107.8\ J[/tex]

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

[tex]\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J[/tex]

Initial kinetic energy

[tex]K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J[/tex]

Change in kinetic energy is equal to work done by object

[tex]\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J[/tex]

A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:____.
a. 0.75 N.
b. 1.96 N.
c. 0.03 N.
d. 30 N.
e. 0.2 N.

Answers

Answer:

the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

Explanation:

Given the data in the question;

mass m = 0.2 kg

radius r = 0.6 m

θ = 150 revolutions = 300π rad

time t = 60 seconds

we know that; Angular speed ω = θ / t

we substitute

ω = 300π / 60

ω = 5π rad

Linear speed of stone u = ω × r

we substitute

u = 5π × 0.6

u = 3π m/s

The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;

so

T = mv² / r

we substitute

T = [ 0.2 × (3π)² ] / 0.6

T = 17.7652879 / 0.6

T = 29.6 ≈ 30 N

Therefore, the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

A car (m=1200kg) accelerates at 3m/s/s for d=10m. How much work has the engine done? *
A) 5400J
B) 36000J
C) 352800J
show your work please

Answers

This is the answer hope it helps

mấy bạn việt nam giúp mình với. cần gấp quá

Answers

Saying english so we can help u
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