Answer:
The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"
Explanation:
For point a:
Energy balance equation:
[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]
[tex]W=0\\\\Q=0\\\\m_e=0[/tex]
From the above equation:
[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]
because the rate of air entering the tank that is [tex]h_i[/tex] constant.
[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]
Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]
Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air
temperature:
[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]
Substituting the value from ideal gas:
[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]
Follow the ideal gas table.
The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]
Interpolate
[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]
Substitute values from the table.
[tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]
For point b:
Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.
[tex]n=\frac{pV}{TR}\\\\[/tex]
[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]
For point c:
Entropy is given by the following formula:
[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]
MULTIPLE CHOICE
What is the role of the Sun in a forest ecosystem?
What is the Role of the sun in a forest ecosystem?
Answer:
It is to produce sunlight in the forest plants
A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.
∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a
==> a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²
To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².
The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:
Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)
F[tex]_{horizontal}[/tex] = 150 N × cos(60°)
F[tex]_{horizontal}[/tex] = 150 N × 0.5
F[tex]_{horizontal}[/tex] = 75 N
Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)
F[tex]_{vertical}[/tex] = 150 N × sin(60°)
F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)
F[tex]_{vertical}[/tex] ≈ 129.9 N
Now, let's calculate the net force in the horizontal direction:
Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]
F[tex]_{net horizontal}[/tex] = 75 N - 15 N
F[tex]_{net horizontal}[/tex] = 60 N
Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:
F[tex]_{net horizontal}[/tex] = m × a
60 N = 20 kg × a
Now, solve for acceleration (a):
a = 60 N / 20 kg
a = 3 m/s²
So, the acceleration of the box is 3 m/s².
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The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle
Answer:
4.38 m/s
Explanation:
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
two objects A and B vertically thrown up with velocities 80m/s and 100m/s at two sec interval.where and when will they meet each other?
Answer:
hcbvdgsyyvjusvbxjxu usbsbhsi
Explanation:
ysggsghxuxgscsixigdvgsibxhdhshshjf
Why we use semiconductor instead of metal in thermopile.
Answer:
Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.
why do we go to hospital
Answer:
bcz we want to have fun there lolol
Answer:
for emergency, treatment, medicines,etc.....
Can Some1 help??
When using the "ball and stick" drawing method of drawing a compound, which element usually goes in the center of the model?
the rate of cooling determines ....... and ......
Answer:
freezing point and melting point
b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?
Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]
t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
What is a measure of how hard it is to stop a moving object?
25.
A. gravity
B. weight
C.
inertia
D. momentum
Answer:
C. inertia
Explanation:
inertia describes an object’s resistance to change in motion (or to get in motion due to a lack of motion), and momentum describes how much motion it has.
both are connected, as inertia depends on the object's momentum, but the answer here is inertia.
why the stone moves away when the string is broken rotation
Answer:
When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path
Answer:
when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped
Proper physical exercise makes bones _[blank 1]_.
People with stronger muscles and bones have better _[blank 2]_.
Which option shows the words that correctly fill in blank 1 and blank 2, in that order?
longer, flexibilitylonger, flexibility , ,
stronger, posturestronger, posture , ,
longer, posturelonger, posture , ,
stronger, flexibility
stronger, posturestronger, posture
hope that helped
Difference between scissors and nut cracker
ASK YOUR TEACHER An oil slick on water is 99.8 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength, in nanometers), given its index of refraction is 1.38
Answer:
There will be a phase change at the 1-1.38 interface and no phase change at the 1.38-1.33 interface.
At a thickness of lambda / 4 (y/4) one should get constructive interference for the reflected light.
y = 4 * 99.8 * 10E-9 m = 400 nm (about) = 4 * 10E-7 m
The color of this light will be violet or blue
4. A diver is 20 m underwater and they are startled by a shark. They are tempted to take a big breath of air, drop their gear, and swim to the surface while holding their breath. Explain why this is dangerous g
Answer:
Explanation:
The air enters their lungs at the same pressure as the water at that depth.
If they hold their breath as they rise to atmospheric pressure, the expanding volume of air (due to decreasing pressure) trapped in their lungs will hyperextend the alveoli in their lungs, likely tearing blood lines and risking death by drowning in their own blood.
What is the maximum speed at which a car can round a curve of 25-m radius on a level road if the coefficient of static friction between the tires and road is 0.80?
I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:
• its weight, mg, pulling it downward
• the normal force from contact with the road, n, pushing upward
• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)
By Newton's second law, the net forces on the car in either the vertical or horizontal directions are
∑ F (vertical) = n - mg = 0
∑ F (horizontal) = f = ma
where a is the car's centripetal acceleration, given by
a = v ²/r
and where v is the maximum speed you want to find and r = 25 m.
From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have
µmg = mv ²/r ==> v ² = µgr ==> v = √(µgr )
So the maximum speed at which the car can make the turn without sliding off the road is
v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire
Answer:
the weight of the wire + 25kg
Explanation:
A kind of variable that a researcher purposely changes in investigation is
Answer:
independent variable
Explanation:
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
Partial tides _______. Question 7 options: represent various components of local tides that are resolved mathematically are predicted individually are added together to predict the height and timing of astronomical tides All of the above are correct. Only a and c are correct.
Complete Question
Partial tides __________.
Question 7 options:
a. represent various components of local tides that are resolved mathematically
b. are added together to predict the height and timing of astronomical tides
c. consist of 4 components due to the influence of celestial bodies
d. consist of up to 60 components due to astronomical and non-astronomical factors
e. All of the above except c are correct.
Answer:
Option E
Explanation:
Generally
Partial tides represent various components of local tides that are resolved mathematically
Partial tides are added together to predict the height and timing of astronomical tides
Partial tides consist of up to 60 components due to astronomical and non-astronomical factors
But Partial tides do not consist of 4 components due to the influence of celestial bodies
Therefore
All of the above except c are correct.
Option E
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Answer:
Explanation:
Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is
F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes
[tex]F_n-w=ma[/tex] where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:
[tex]F_n=ma+w[/tex] . m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:
w = mg so
w = 28(9.8) and
w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually
w = 270 N.
Filling in the elevator equation:
[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:
[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:
[tex]F_n=280N[/tex] So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?
The coefficients of friction between a race cars tyres and the track surface are
the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).
Explanation:
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The first law of motion describes the principle of __________
Answer:
The first law of motion describes the principle of law of inertia.
Ashley, a psychology major, remarks that she has become interested in the study of intelligence. In other words, Ashley is interested in?
Group of answer choices.
a) the capacity to learn from experience, solve problems, and to adapt to new situations.
b) how behavior changes as a result of experience.
c) the factors directing behavior toward a goal.
d) the ability to generate novel
Answer:
a) the capacity to understand the world, think rationally, and use resources effectively.
Explanation:
Psychology can be defined as the scientific study of both the consciousness and unconsciousness of the human mind such as feelings, emotions and thoughts, so as to understand how it functions and affect human behaviors in contextual terms.
This ultimately implies that, psychology focuses on studying behaviors and the mind that controls it.
In this scenario, Ashley who is a psychology major, stated that she's interested in the study of intelligence.
Intelligence can be defined as a measure of the ability of an individual to think, learn, proffer solutions to day-to-day life problems and effectively make informed decisions.
In other words, Ashley is interested in the capacity of humans to understand the world, think rationally, and use resources effectively to produce goods and services that meet the unending requirements, needs or wants of the people (consumers or end users) living around the world.
The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the propeller SLOWS at a steady rate of 0.9 revolutions per second per second. After 17 revolutions, how fast is the propeller spinning in revolutions per second
Answer: [tex]5.77\ rps[/tex]
Explanation:
Given
Initial angular velocity is [tex]\omega_i=8\ rps[/tex]
rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]
after 17 revolution i.e. [tex]\theta =17\ rev[/tex]
using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]
Insert the values
[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]
