Answer:
Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
F=(4i+3j)N acts on an object of mass m=2k.g and drags it by moving the object from origion to x=5m. Find the workdone on the object and the angle between the force and the displacement
Answer:
nnnjjdndbsnnshfhhgbfbdbdh
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
If a jet travels 350 m/s, how far will it travel each second?
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
Answer:
5.83 seconds
Explanation:
60 seconds in 1 minute
350 meters per second
350/60
=5.83
train starts from rest and accelerates at 1m/ s²
for 10 seconds how far does it move
Answer:
s=50m
Explanation:
you can use the formula
s=ut+1/2at²
s=0t+1/2(1)10²
=1/2(100)
=50
I hope this helps
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
Explanation:
What is measurement
Answer:
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Explanation:
A scientific method can be defined as a research method that typically involves the use of experimental and mathematical techniques which comprises of a series of steps such as systematic observation, measurement, and analysis to formulate, test and modify a hypothesis.
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Basically, the dimensions include important parameters such as width, height, length, area, volume, circumference, breadth, etc.
The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer
Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
Which of the following would most likely produce the strongest magnetic
field?
A. A single moving electron
B. A stationary electric charge
C. A current in a straight wire
D. A current in a coil
Answer:
I current in a coil,,,,,,
Answer:D?
Explanation:
Sorry if i'm wrong....
What type of potential energy is a 9 volt battery an example of?
Gravitational potential energy
Elastic potential energy
Electrical potential energy
chemical potential energy
Answer:
chemical potential energy
Explanation:
A 9v battery comes in different formats, such that the most common one is the carbon-zinc and alkaline chemistry, so these are alkaline batteries (there are also rechargeable or lithium batteries, these also depend on chemical interactions).
These batteries "draw" the energy from chemical interactions of the materials inside of it, so the type of potential energy that is stored in a battery is actually chemical (regardless of the fact that the energy can be transformed into electrical energy later) the "potential" refers to how the energy is stored.
Then the correct option is chemical potential energy
Answer:
Chemical Potential Energy
Explanation:
Hope this helps!!
Have a blessed day/night!! <33
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)
Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.
Answer:
The answer is "[tex]0.3336\ m^3[/tex]"
Explanation:
Using the Promideal gas law:
[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]
[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]
[tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]
The equilibrium value of Va is 0.3336 m³.
Ideal gas lawThe equilibrium value of Va is determine by applying ideal gas law as shown below;
Pressure of gas A = Pressure of gas B
Pa = Pb
Pa(Va - nab) = naRT----(1)
PbVb = nbRT -----(2)
Solve equation (1) and (2)
[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]
Va + Vb = 1
Vb = 1 - Va
[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]
2Va - 2b = 1 - Va
3Va = 1 + 2b
[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]
Thus, the equilibrium value of Va is 0.3336 m³.
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A lead ball is dropped into a lake from a diving board 5.20 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 s after it is released. How deep is the lake?
Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
Four identical metallic objects carry the following charges 1.08 6.74 4.61 and 9.41 C The objects are brought simultaneously into contact so that each touches the others Then they are separated a What is the final charge on each object b How many electrons or protons make up the final charge on each object
Answer:
(a) 5.46 C
(b) 3.4125 x 10^19
Explanation:
q' = 1.08 C, q'' = 6.74 C, q''' = 4.61 C, q'''' = 9.41 C
When the charges are in contact to each other.
(a) So, the net charge is
[tex]q = \frac{q' + q'' + q''' + q''''}{4}[/tex]
[tex]q = \frac{1.08+6.74+4.61+9.41}{4}\\\\q = 5.46 C[/tex]
(b) As the charge is positive in nature, so the protons are there. The number of protons is
[tex]n = \frac{q}{e}\\\\n = \frac{5.46}{1.6\times 10^{-19}}\\\\n = 3.4125\times 10^{19}[/tex]
Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a clockwise current of 16.0 A , as viewed from above, and the outer wire has a diameter of 32.0 cm.
Required:
a. What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
b. What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
Solution :
a). B at the center :
[tex]$=\frac{u\times I}{2R}$[/tex]
Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,
[tex]$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$[/tex]
So,
[tex]$\frac{I_1}{d_1}= \frac{I_2}{d_2}$[/tex]
[tex]$=\frac{16}{21}=\frac{I_2}{32}$[/tex]
[tex]$I_2=24.38 $[/tex] A
Therefore, the current in the outer wire is 24.38 ampere.
Answer:
(a) counter clockwise
(b) 24.38 A
Explanation:
inner diameter, d = 21 cm
inner radius, r = 10.5 cm
Current in inner loop, I = 16 A clock wise
Outer diameter, D = 32 cm
Outer radius, R = 16 cm
(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.
So, the direction of current in outer wire is counter clock wise in direction.
(b) Let the current in outer wire is I'.
The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.
[tex]\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A[/tex]
II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. ( ) If the buckets are at rest, what is the tension in each cord? ( ) If the two buckets are pulled upward with an acceleration of 1.25 m/s by the upper cord, calculate the tension in each cord
Answer:
Here , mass of bucket ,m = 3.2 Kg
Now , let the tension in upper rope is T1
the tension in the middle rope is T2
a)
For lower bucket, balancing forces in vertical direction
T2 - mg = 0
T2 = mg
T2 = 3.2 *9.8
T2 = 31.36 N
tension in the middle rope is 31.36 N
For the upper bucket , balancing forces in vertical direction
T1 - T2 - mg = 0
T1 = T2 + 3.2 *9.8
T1 = 62.72 N
the tension in the upper rope is 62.72 N
B)
for a = 1.25 m/s^2
Using second law of motion ,for both the buckets
Fnet = ma
T1 - 2mg = 2m*a
T1 = 2*3.2*(9.8 +1.25)
T1 = 70.72 N
the tension in the upper rope is 70.7 N
Now , the lower bucket
Using second law of motion,
T2 - mg = ma
T2 = 3.2 * (9.8 + 1.25)
T2 = 35.36 N
the tension in the lower rope is 35.36 N
HELP ME ASAP PLSSS!!
3) A lead bullet initially at 30 C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. (Specific heat of lead is 0.128 kJ/kg K and lead latent heat of fusion is 24.7 kJ/Kg and melting point of lead is 600 K).
Answer:
The speed of bullet is 354.2 m/s
Explanation:
initial temperature, T = 30 degree C
specific heat, c = 128 J/kg K
Latent heat, L = 24.7 x 1000 J/kg
melting point = 600 K = 327 degree C
Let the mass is m and the speed is v.
Kinetic energy = heat used to increase the temperature + Heat used to melt
[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]
Suppose your actual height is 5 feet and 5.2 inches. A tape measure which can be read tothe nearest 1/8 of an inch gives your height as 65 3/8 inches. The laser device at the clinic that givesreadings to the nearest hundredth of an inch says you are 65.31 inches.
Required:
a. Which measuring device is more accurate?
b. Which measuring device is more precise?
Answer:
a) The laser device
b) The tape
Explanation:
First, there is a need to understand what accuracy and precision mean.
Accuracy is the closeness of a measurement to its true (pre-determined) value.
Precision is the closeness of repeated measurements to each other.
Since 1 feet = 12 inches, then, 5 feet and 5.2 inches would be equivalent to 65.2 inches. This value represents the true value of my height.
The tape measured the height as 65 3/8, which is equivalent to 65.375 inches.
The laser device measured the height as 65.31.
Error = true value - measured value
Absolute error from the tape = 65.2 - 65.375
= -0.175 inches
Absolute error from laser device = 65.2 - 65.31
= -0.11
a) The magnitude of error from the tape is more than that of the laser device. Hence, the laser device is said to be more accurate.
b) Even though there were just single readings from both instruments, the tape can be read to the nearest 1/8 of an inch and as such, can give more precisive measurements than the laser device.
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________
Answer:
Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].
Explanation:
E inside [tex]= KQr_{1} /R^{3}[/tex]
so if r1 will be the same then
E [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
The electric field is E/8.
Explanation:
The electric field due to a solid sphere of uniform charge density inside it is given by
[tex]E =\frac{\rho r}{3}[/tex]
where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.
For case I:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]
So, electric field at a distance r is
[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]
Case II:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]
So, the electric field at a distance r is
[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?
Answer:
Explanation:
(a)Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
(a)Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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Yea, gonna need some help. Thanks
Answer:
t = 3.48 s
Explanation:
The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:
[tex]h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t[/tex]
where,
v₀ = initial speed = 110 ft/s
Therefore,
[tex]110 = 32t\\\\t = \frac{110}{32}\\\\[/tex]
t = 3.48 s
A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy
Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
Distance stretched ( Δl ) = 4 * 10^-3 m ( right value )
average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.
Answer:
always runs slower than normal.
Explanation:
The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.
According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.
In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.
widely accepted scientific principles do not change. true or false
Answer:
False
Explanation:
As technology advances and new evidence is found which either contradicts or supports accepted scientific principles, the principles are susceptible to change.
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
