This question focuses on the element shown in Figure P1.24 I, with a voltage of 10 V and current of 2e A.
The goal is to calculate the power of the circuit element and the total energy transferred between t= 0 and t= [infinity]. To determine if this energy is absorbed or supplied by the element, calculations must be done. First, the power of the circuit element can be calculated by multiplying the voltage and current together.
This yields a power of 20e W. Next, the energy transferred between t= 0 and t= [infinity] can be found by integrating the power over that time interval. This yields an energy of 20e Joules. Finally, to determine if the energy is absorbed or supplied by the element, the sign of the power must be checked. In this case, the power is positive, so the element is supplying the energy.
3. Consider a large windmill 30m in diameter. On a windy day, suppose that the windmill entrains a stream of air at a speed of 40 mph. Downstream of the windmill, the entrained stream exits over a large diameter at a speed of 20 mph. The pressure is 2atm at the inlet and equals atmospheric pressure at the outlet. Find the power (in megawatts) generated by the windmill. Density of air is 1.2 kg/m3
Answer:
The power generated by the windmill is approximately 1.364 MW
Explanation:
The diameter of the windmill, d = 30 m
The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s
The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s
The pressure at the inlet, P₁ = 2 atm
The pressure at the outlet, P₂ = 1 atm
The density of air, ρ = 1.2 kg/m³
The power obtained from the windmill, 'P', is given as follows;
[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]
Where;
[tex]g_c[/tex] = 1.0 kg/(N·s²)
A = Cross-sectional rea of the the windmill = π·D²/4 = π×(30 m)²/4 = 706.858347 m²
Plugging in the values, we get;
[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]
The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.
Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the train platform hears beats with a frequency of 2.00 beats/s when the whistles sound at the same time. What are the two possible speeds and directions that the moving train can have?
Actual answers :3.85 m/s away from the station and 3.77 m/s towards the station from the book. I just need to know how to get to the answers.
Answer:
-3.77 m/s
3.85 m/s
Explanation:
given that
Frequency at stationary = 180 Hz
Beat frequency = 2 Hz
Using Doppler effect, we know that
f' = f[(v ± v0) / (v ± vs)], where
v = speed of sound, 343 m/s
v0 = speed of the observer, 0
vs = speed of light, ?
f = stationary frequency, 180 Hz
f' = stationary ± beat frequency, 180 ± 2
Applying the formula, we have
f' = f[(v ± v0) / (v ± vs)]
182 = 180 [(343 + 0) / (343 + vs)]
182/180 = 343 / 343 + vs
343 + vs = 343 * 180/182
343 + vs = 339.23
vs = 339.23 - 343
vs = -3.77 m/s
Again, using
f' = f[(v ± v0) / (v ± vs)]
178 = 180 [(343 + 0) / (343 + vs)]
178/180 = 343 / 343 + vs
343 + vs = 343 * 180/178
343+ vs = 346.85
vs = 346.85 - 343
vs = 3.85 m/s
QUESTION 9 / 10
What is the first step you should take when you want to open a savings account?
A. Present your photo ID to the bank representative,
B. Make your initial deposit.
C. Review the different savings account options that your
bank offers.
D. Go to the bank and fill out an application.
Answer:
A
Explanation:
AWNSER:
awnser:
C
explanation:
What force causes a resistance in motion
when two surfaces are touching?
Answer:
FRICTION
Explanation:
Friction is a force, the resistance of motion when one object rubs against another.
Frictional force
Explanation:
Its the opposing force against horizontal motion
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (Figure 1). The field is changing with time, according to B(t)=(1.4T)e^−(0.057s^−1)t.
a) Find the emf induced in the loop as a function of time (assume t is in seconds).
b) When is the induced emf equal to 110 of its initial value?
c) Find the direction of the current induced in the loop, as viewed from above the loop.
The solution to the questions are given as
[tex]t=40.39 \mathrm{sec}[/tex][tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]the direction of induced current will be Counterclock vise.What is the direction of the current induced in the loop, as viewed from above the loop.?Given, $B(t)=(1.4 T) e^{-0.057 t}$
[tex]$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}[/tex]
[tex]\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$[/tex]
[tex]\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}[/tex]
[tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]
(b) [tex]Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$[/tex]
[tex]\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}[/tex]
c)
In conclusion, the direction of the induced current will be Counterclockwise.
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Galileo _____.
did not believe friction existed
believed that friction stopped objects in motion
believed that friction kept objects in motion
assumed that in a frictionless environment objects would never move
Answer:
friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.
Plzzz answer this question correctly
Answer:
changing the direction in which a force is exerted
A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10^−2 T/s )t+( 2.85×10^−5 T/s4 )t^4. The coil is connected to a 610 Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
a) Find the magnitude of the induced emf in the coil as a function of time.
b) What is the current in the resistor at time t0 = 5.20 s ?
The current is 1.13* 10^{-4}A
Given that r = radius of the coil = 4 cm = 0.04 m
Area of coil is given as
A = πr²
A = (3.14) (0.04)² = 0.005024 m²
N = Number of turns = 500
R = Resistance = 600 Ω
B = magnetic field = (0.0120)t + (3 x 10⁻⁵) t⁴
Taking derivative at both the side
[tex] \frac{dB}{dt} = (0.120 + (12 \times 10^-5)t^3)[/tex]
Induced current is given as
[tex]i= (\frac{NA}{R} )( \frac{db}{dt} )[/tex]
[tex]i \: = (\frac{NA}{R})(12 \times 10 ^{-5} )t^3[/tex]
substituting the value t = 5
[tex]i = ( \frac{(500)(0.005024)}{600}) (12 \times 10 ^{-5} )5^3[/tex]
[tex]i = 1.13 \times 10 ^{ - 4} A[/tex]
Hence the current is
[tex]1.13 \times 10^-4A[/tex]
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The Electric current is 1.11* 10^{-4}A
Given that the coil's radius is 3.55 cm (0.35 m),
The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.
R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)
t + (3 x 10⁻⁵) t⁴
The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.
The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.
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Dogs can inherit four features: fur pattern, fur
length, ear length, and tail length. The alleles are expressed as shown:
Dominant alleles
Recessive alleles
F: spotted fur
f: solid-colored fur
L: long fur
1: short fur
E: long ears
e: short ears
T: long tail
t: short tail
Two dogs mate and have puppies. Both parent dogs are heterozygous for spotted fur. Fill in the Punnett square to show the po
dropping the letters)
Answer:
This is something for biology, it shouldnt be in the physics subject
Explanation:
Determine the voltage Vab for the first circuit and also determine the voltages Vab and Vcd for the second circuit
Vab= E = 20V
because I = 0 and the voltage drop across the resistances R1 and R2 is also 0.
Second circuit:
Vab = 10V (no voltage drop across R1)
Vcd= E2-E1 = 20V
Series connection of voltage sources. But the sources are connected to the contrary and voltage drop across R1 or R2 is 0 V.
Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
What current is needed in the solenoid's wires?
A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.8 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field.
What current is needed in the solenoid's wires? Express your answer with the appropriate units.
Friction is necessary when you are on a bike to stay
Answer:
yes friction is needed hope this helps might of been to long tho
What is the weight of a 44.5 kg object?
Answer:
98.11 I think
Explanation:
I really hope this helps have a wonderful day
What is the mass of 9.11 moles of
ozone, 03?
The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
What is molecular mass?Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule. The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.
To compute the Molecular Mass of [tex]$O_{3}[/tex]
Atomic mass of oxygen(O) = 16
As [tex]$O_{3}[/tex] contains 3 atoms,
The molecular mass of [tex]$O_{3}[/tex]
= (16 x 3) = 48g/mol
Hence the mass of 9.11 moles O3
= 9.11 mol x 48g/mol
= 437.28g
= 0.43728kg
Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
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A 0.15 kg baseball collides with a 1.0 kg bat. The ball has a velocity of 40 m/s immediately before the collision. The center of mass of the bat also has a velocity of 40 m/s, but in the opposite direction, just before the collision. The coefficient of restitution between the bat and the ball is 0.50. Estimate how fast the baseball is moving as it leaves the bat following the collision.
Answer:
The final velocity of the baseball as it leaves the bat is 40 m/s
Explanation:
The given parameters of the baseball and bat are;
The mass of the baseball = 0.15 kg
The mass of the bat = 1.0 kg
The velocity of the ball before collision, v₁ = 40 m/s
The velocity of the bat before collision, v₂ = -40 m/s
The coefficient of restitution, e = 0.50
Let, 'v₃', and 'v₄' represent the final velocity of the ball and the bat respectively after collision, we have;
Taking the final velocity of the bat, v₄ = 0 m/s
According to Newton's Law of restitution
e = (v₃ - v₄)/(v₁ - v₂)
∴ 0.5 = (v₃ - v₄)/(40 - (-40))
80 × 0.5 = 40 = (v₃ - v₄)
v₃ - v₄ = 40
v₃ = 40 + v₄ = 40 + 0 = 40
The final velocity of the baseball as it leaves the bat, v₃ = 40 m/s.
Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.
1) What is Cole's acceleration?
2) What is his velocity when he reaches school?
Explanation:
this is the answer for your question. if you have any doubt.
you can send your doubt to:6369514784(what's app)
What is the difference between the reflection and refraction of light
Answer:
Reflection can simply be defined as the reflection of light when it strikes the medium on a plane. Refraction can be defined as the process of the shift of light when it passes through a medium leading to the bending of light. The light entering the medium returns to the same direction.
Answer:
reflection is your image and refraction is light
The noise from a power mower was measured at 104 dB. The noise level at a rock concert was measured at 121 dB. Find the ratio of the intensity of the rock music to that of the power mower.
Based on the calculations, the ratio of the intensity of Ir to Ip is equal to 50.12.
How to find the ratio of the intensity?Let the intensity of the power mower be Ip.Let the intensity of the rock music be Ir.Mathematically, sound intensity level can be calculated by using this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Where:
I is the intensity of the sound.
Making I the subject of formula, we have:
I = Io × [tex]10^{\frac{\beta }{10} }[/tex]
For Ip, we have:
Ip = Io × [tex]10^{\frac{104 }{10} }[/tex]
Ip = Io × [tex]10^{10.4}[/tex]
For Ir, we have:
Ir = Io × [tex]10^{\frac{121 }{10} }[/tex]
Ir = Io × [tex]10^{12.1}[/tex]
Now, we can find the ratio of the intensity:
Ir/Ip = Io × [tex]10^{12.1}[/tex]/Io × [tex]10^{10.4}[/tex]
Ir/Ip = [tex]10^{12.1}[/tex]/[tex]10^{10.4}[/tex]
Ir/Ip = [tex]10^{12.1 -10.4}[/tex]
Ir/Ip = [tex]10^{1.7}[/tex]
Ir/Ip = 50.12.
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Which of the following would have the least amount of inertia? Assume all the bags are the same size.
bag of rocks
bag of feathers
bag of bricks
bag of sand
Earth’s atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by Earth’s atmosphere?
Answer:
Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.
Explanation:
Which instrument would make rice vibrate easier, a tuba or a flute? Explain why. Hint: think about the difference between high and low notes in terms of vibrating atoms.
Answer:
I assume the higher notes would make the rice vibrate more easily, so a flute.
Plzzz help me with this
I’ll give brainliest
Answer:
(A) By reducing friction
Answer:
A
Explanation:
1. Boron has two naturally occurring isotopes, boron-10 and boron-11. Boron-10 has a mass of 10.0129 relative to carbon-12 and makes up 19.78 percent of all naturally occurring boron. Boron-11 has a mass of 11.00931 compared to carbon-12 and makes up the remaining 80.22 percent. What is the atomic weight of boron?
2. Identify the number of protons, neutrons, and electrons in the following isotopes:
a. 147N b. 3517Cl c. 4820Ca d. 6329Cu e. 23092U
3. How many outer shell electrons are found in an atom of
a. Na? b. P? c. Br? d. I? e. Te?
4. Use the periodic table to identify if the following are metals, nonmetals, or semiconductors:
a. Radon b. Francium c. Arsenic d. Phosphorus e. Hafnium
From the calculation, the atomic weight of the boron atom is 10.812.
What is the atomic weight of boron ?We can find the atomic weight of boron from;
(19.78/100 * 10.0129) + ( 80.22/100 * 11.00931)
= 1.981 + 8.831
= 10.812
The number of protons, neutrons, and electrons in the following isotopes are;
N - 7 electrons, 7 protons and 7 neutrons
Cl - 17 electrons, 17 protons and 18 neutrons
Ca - 20 protons, 20 electrons and 28 neutrons
Cu - 29 protons, 29 electrons and 3 neutrons
U - 92 protons, 92 electrons and 138 neutrons
The number of outer shell electrons in the atoms are;
Na - 1
P - 5
Br - 7
I - 7
Te - 6
The elements are classified as follows;
Radon - nonmetal
Francium - metal
Arsenic - semiconductor
phosphorus - nonmetal
Hafnium - metal
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The solar wind consists of protons from the Sun moving toward Earth (the wind actually consists of about 95% protons). The number density of protons at a distance from the Sun equal to the orbital radius of Earth is about 7.0 protons per cubic centimeter. Your research team monitors a satellite that is in orbit around the Sun at a distance from the Sun equal to Earth's orbital radius. You are in charge of the satellite's mass spectrometer, an instrument used to measure the composition and intensity of the solar wind. The aperture of your spectrometer is a circle of radius 28.4 cm. The rate of collection of protons by the spectrometer is such that they constitute a measured current of 91.0 nA. What is the speed of the protons in the solar wind
Answer:
[tex]\mathbf{V_d = 3.2 \times 10^5 \ m/s}[/tex]
Explanation:
[tex]\text{The speed of the protons can be estimated by using the formula:}[/tex]
[tex]V_d = \dfrac{I}{enA}[/tex]
[tex]where;[/tex]
[tex]\text{e = proton charge}[/tex]
[tex]\text{n = No. of protons per unit volume}[/tex]
[tex]\text{A = area of aperture}[/tex]
[tex]V_d = \dfrac{91 \times 10^{-9} \ A}{(1.602 \times 10^{-19} \ C (7.0 \times 10^6 \ m^{-3} ) (\pi) (0.284 \ m)^2}[/tex]
[tex]\mathbf{V_d = 3.2 \times 10^5 \ m/s}[/tex]
A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, of them were given multivitamin tablets daily that contained gram of vitamin C and grams of various other vitamins and minerals. The remaining volunteers were given tablets containing grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Days to recover from flu Treated with multivitamin , , , , , , , , , Treated with vitamin C , , , , , , , , , Suppose that it is known that the population standard deviation of recovery time from the flu is days when treated with multivitamins and that the population standard deviation of recovery time from the flu is days when treated with vitamin C tablets. Suppose also that both populations are approximately normally distributed. Construct a confidence interval for the difference between the mean recovery time when treated with multivitamins () and the mean recovery time when treated with vitamin C only (). Then find the lower limit and upper limit of the confidence interval.
The confidence interval is given as lower interval is -0.77 while The upper interval is 1.67
How to solve for the confidence intervalMultivitamin treatment
n1 is 10
σ1 = 1.8
x21 = 5
vitamin C treatment
n2 = 10
σ2 = 1.5
x2 = 4.55
zα/2 = 1.645
The formula for the confidence interval is given as
(x1 - x2) ± zα/2[tex]\sqrt{\frac{sd1^2}{n1}+\frac{sd2^2}{n2} }[/tex]
When we input the values we have above we would have
CI = (-0.7688550 , 1.668855 )
The the lower interval is -0.77
The upper interval is 1.67
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Complete questionA random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 10 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 10 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained:
Treated with multivitamin
Days to recover from flu
2.4, 6.4, 9.1, 4.1, 4.6, 6.4, 6.4, 3.2, 6.9, 0.5
Treated with Vitamin C
5.2, 3, 3.6, 5.5, 7.5, 6.7, 1.3, 1.9, 5.3, 5.5
Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the
population standard deviation of recovery time from the flu is 1.5 days when treated with vitamin C tablets. Suppose also that both populations are
approximately normally distributed. Construct a 90% confidence interval for the difference µµ₂ between the mean recovery time when treated with
multivitamins (μ,) and the mean recovery time when treated with vitamin C only (H2). Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?
Valence electron of the first 20 elements
Answer:
Hydrogen
1 valence electron
Helium
2 valence electrons
lithium
1 valence electrons
beryllium
2 valence electrons
boron
3 valence electrons
carbon
4 valence electrons
nitrogen
5 valence electrons
oxygen
6 valence electrons
flourine
7 valence electrons
neon
8 valence electrons
sodium
1 valence electron
magnesium
2 valence electrons
aluminum
3 valence electrons
silicon
4 valence electrons
phosphorus
5 valence electrons
Answer: 17 Chlorine -1, +1, (+2), +3, (+4), +5, +7
18 Argon 0
19 Potassium +1
20 Calcium +2
Explanation:
15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers are listed in the image. Thank you so much!
The percent difference between two numbers [tex]x[/tex] and [tex]y[/tex] is given by
[tex]\dfrac{|y-x|}x \times 100\%[/tex]
The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from [tex]x[/tex] to [tex]y[/tex] or vice versa. If we remove them, we have two possible interpretations of percent difference.
For example, the (absolute) percent difference between 3 and 6 is
[tex]\dfrac{|6-3|}3 \times 100\% = 100\%[/tex]
In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is
[tex]\dfrac{3-6}3\times100\%=-50\%[/tex]
which is to say, we take away 50% of 6 away from 6 to end up with 3.
"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take [tex]x[/tex] from the left column and [tex]y[/tex] from the right column.
[tex]\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%[/tex]
[tex]\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%[/tex]
[tex]\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%[/tex]
The percentage difference are 2.82%, 4.17%, 4.17%.
The percent difference between two numbers a and b is given by formula:
[tex]\frac{y-x}{x}*100[/tex]
We are given absolute value as we are only concerned about the absolute percent difference.
Making comparisons to object measurements determines that the differences should be computed relative to object measurements.
Here take from the left column and from the right column.
a) [tex]\frac{7.3-7.1}{7.1} *100=2.82[/tex]%
b) [tex]\frac{5.0-4.8}{4.8} *100=4.17[/tex]%
c) [tex]\frac{7.5-7.2}{7.2} *100=4.17[/tex]%
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Which part of the water cycle is where vapor from plants leaves the plants as they breath?
condensation
Transpiration
evaporation
Answer:
I think it is transpiration
Answer:
transpiration is the right answer
A car is being tested on a track. The driver approaches the test section at a speed of 28 m s−1. He then accelerates at a uniform rate between two markers separated by 100 m. The car reaches a speed of 41 m s−1.
The uniform acceleration of the car is 4.485 m/s².
Acceleration of the car
The uniform acceleration of the car is calculated as follows;
v² = u² + 2as
a = (v² - u²)/2s
where;
v is final velocity = 41 m/su is initial velocity = 28 m/ss is distance = 100 ma is acceleration = ?a = (41² - 28²)/(2 x 100)
a = 4.485 m/s²
Thus, the uniform acceleration of the car is 4.485 m/s².
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