A particular organic molecule forms an electric dipole, the magnitude of the molecule's electric dipole moment
3.152 x 10^-29 C4.45458171*10^{-24} NmThe max torque that the electric field can exert on this molecule is the same as the torque found in question 2.What is an electric field vector?Generally, The magnitude of the electric dipole moment (p) is given by the product of the charge separation (d) and the magnitude of the charges (q):
p = dq
Therefore, the magnitude of the electric dipole moment of the molecule is:
p = (0.197 nm)(1.60 x 10^-19 C)
= 3.152*10^{-20}Cm
2) The magnitude of the torque (τ) that acts on the molecule in a uniform electric field (E) is given by the vector cross product of the electric dipole moment vector (p) and the electric field vector (E):
τ = p x E
The magnitude of the torque is given by the formula:
τ = pE sin(θ)
where
θ is the angle between the electric dipole moment vector and the electric field vector
Therefore, the magnitude of the torque that acts on the molecule in the given electric field is:
τ = (3.152 x 10^-29 Cm)(2.87 x 10^5 N/C) sin(29.5°)
= 4.45458171*10^{-24} Nm
3)The torque exerted on a dipole*in an electric field is dependent on the angle between the direction of the electric field and the dipole moment.
The torque is maximum when the angle between the direction of the electric field and the dipole moment is 90 degrees and the torque will be zero when it's parallel with the field.
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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
Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet is 7.34 km/s.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
10,378.82 m/s is the second cosmic speed.
69,801 km is the radius of the synchronous orbit of a satellite.
Given
Mass of planet = 4.74 × [tex]10^{24}[/tex] kg
Radius of planet = 5870 km = 5870000m
First Cosmic speed = 7.34 km/sec
1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.
It can be calculated by
v = √2GM/r where,
v= Escape velocity of the satellite
G = Gravitational constant
M = Mass of planet
r = Radius of planet
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
2) Speed of the satellite at the given period
v = 2πr/T where,
T= Time period of rotation = 16.6 × 3600 seconds
r = v×T/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Hence
The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.
And the radius of the synchronous orbit of a satellite is 69,801 km.
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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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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:
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.
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
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.
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.
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:
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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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.
A 1100-N crate rests on the floor.
1) How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 310 N .
Express your answer to two significant figures and include the appropriate units.
2)
How much work is required to move it at constant speed 5.0 m vertically.
Express your answer to two significant figures and include the appropriate units.
(a) The work done in moving the crate along the floor is 7,050 J.
(b) The work done in moving the crate vertically is 0 J.
Work done in moving the crate horizontally
W = Fdcosθ
where;
F is total force to be appliedd is displacement of the crateθ is the horizontal angle = 0W = (1100 + 310) x 5
W = 7,050 J
Work done in moving the crate verticallyW = Fd cos(90)
W = 0
Thus, the work done in moving the crate along the floor is 7,050 J.
the work done in moving the crate vertically is 0 J.
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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:
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)
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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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?
Which element is a metalloid?
Answer:
The metalloids are located on the right side of the periodic table in a "step-like" arrangement.
All of the possible metalloids are:
boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)
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 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
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:
Even if all stars were the same distance from Earth, their absolute magnitude and
apparent magnitude would be very different.
True
False
Answer: True
hope this helps!
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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Plzzz help me with this
I’ll give brainliest
Answer:
(A) By reducing friction
Answer:
A
Explanation:
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.
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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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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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)
In order for the refrigerator not to tip over, the maximum acceleration of 1.86 m/s² must not be exceeded.
What is acceleration?The term acceleration has to do with the rate at which velocity changes with time.
We have to take the moments at the tipping point of rotation as follows;
Clockwise moment = Anticlockwise moment
Hence;
F₂ * 1.58 m = F₁ * 0.67 m
The weight at half the width= 30 cm or 0.3 m
Height of refrigerator = 158 cm 0r 1.58 cm
Then;
m * a * 1.58 = m * 9.81 * 0.30
a = 1.86 m/s²
In order for the refrigerator not to tip over, the maximum acceleration of 1.86 m/s² must not be exceeded.
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What is the Difference between accuracy and precision ?
Answer:
Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.
Accuracy means the state of being accurate, without any mistakes and the results should be 100% true.
Whereas, precision means, approximately true or almost true.
Hope it help you
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]
Select the correct answer. Which graphs show the correct relationship between kinetic energy and mass? A. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases upwards B. Graph representing the relationship between mass on the x-axis and kinetic energy on the y-axis. the curve starts at the origin and keeps on increasing C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis D. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases as it goes with a small bend Reset Next
C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
What is kinetic energy?
Kinetic energy is the energy possessed by a body due to its motion.
K.E = ¹/₂mv²
where;
m is mass of the objectThe kinetic energy of a body is directly proportional to the mass of the object.
Thus, the correct relationship between kinetic energy and mass is Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
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