Ionisation energy is the correct answer .
What is Energy ?
Energy, in physics the ability to do work. It can exist in electrical, kinetic, thermal, electrical, chemical, nuclear, or various other forms. There is also heat and work. H. After transmission, energy is always referred to according to its nature. Thus, the heat transferred can become thermal energy, while the work done can appear as mechanical energy. All forms of energy are associated with movement. For example, any object has kinetic energy when it is in motion. A taut device such as a bow or spring can create motion even when stationary. Because of its composition, it contains potential energy. Similarly, nuclear energy is potential energy as it results from the composition of subatomic particles within the nucleus .
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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)
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.
To find the answer, we need to know more about the pressure exerted by a liquid column.
How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,[tex]P=[/tex] ρgh
In our question, it is given that,[tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]
Thus, the height of the bag h will be,[tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]
Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.
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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.
We need to learn more about the pressure that a liquid column exerts in order to determine the solution.
How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.[tex]P=[/tex] ρgh
In our inquiry, it is assumed that,[tex]Density=1.308kg/L\\P=35.7mmHg[/tex]
As a result, the bag's height will be,[tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]
As a result, we may say that the I.V. bag is 2.78 m above the ground.
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State the newton's law of motion and give
application of each law.
Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Explanation:
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.
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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A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the
positive y-axis. If the magnitude of the field is 0.7 T, and the charge and speed of the proton are 1.5 x10-19 C
and 3.0 x 106 m/s respectively. Calculate the force acting on the proton
Answer: The force acting on the proton of charge 1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.
Explanation: To find the answer we need to know more about the Lorentz magnetic force.
What is the Lorentz magnetic force acting on the proton?Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,F= q (v× B)
[tex]F= qvBsin\alpha[/tex] where, [tex]\alpha[/tex] is the angle between v and B.
In the question, it is given that,[tex]B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\[/tex] because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.
Thus, we can find the force acting on the proton as,[tex]F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N[/tex]
Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.
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In response to a magnetic field of 0.7 T, a proton with a charge of 1.5 x 10-19 C travelling at a speed of 1.5 x 10-19 C will experience a force of 3.15 x 10^-13 N.
We need to learn more about the Lorentz magnetic force in order to locate the solution.
What does the proton experience as the Lorentz magnetic force?If you imagine a proton with charge q travelling at speed v in a magnetic field, you can write down the Lorentz magnetic force acting on the proton as,F= Q (v× B)
[tex]F=QvBsin\alpha[/tex]
[tex]\alpha[/tex] the angle between v and B is, where.
It is stated in the query that, the magnetic field is along the y axis and the proton is travelling along the x axis. Thus, the angle will be 90 degrees.As a result, we can identify the proton's driving force as,[tex]F=3.15*10^{-13}N[/tex]
Thus, we can infer that the proton will be subject to a 3.15 10^-13 N Lorentz force.
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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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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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what can you say about the speed of a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds
The speed of the ball at the given distance and time of motion is 4 m/s.
What is speed?The speed of an object is the rate of change of distance traveled by the object with time.
Speed is a scalar quantity because it has only magnitude and no direction. It is measured in meters per second.
Speed of the ballFor a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds, the speed of the ball at the given distance and time of motion is calculated as follows;
Speed of the ball = distance traveled by the ball / time of motion
from the question, distance = 20 mtime of motion, t = 5 secondsspeed of the ball = 20 m / 5 s
speed of the ball = 4 m/s
Thus, the speed of the ball at the given distance and time of motion is 4 m/s.
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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.
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'f or s' are possible answers in some of the cases.
The new sphere has a density of ρ > ρ0 and a mass of m = m0.
The new sphere has a mass of m > m0 and a radius of r = r0.
The new sphere has a radius of r = r0 and a density of ρ > ρ0.
The correct response for each of the condition given in the questions are,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rTo find the answer, we have to know about the Archimedes principle.
How to solve the problem for different conditions?The Archimedes principle states that the upthrust F on a body is equal to the weight W of the displaced liquid.The sum of forces must be zero for the sphere to be in equilibrium.[tex]F-W=0\\F=W\\W=mg, where.\\m=density*volume=d*V\\V=\frac{4}{3} \pi r^3[/tex]
Let's apply the idea of density to the body and water now. We are taking d instead of ρ.[tex]d_wVg = d_0(\frac{4}{3}\pi r^3 ) g \\d_wV = d_0(\frac{4}{3}\pi r^3 )[/tex] (1)
Let's examine each example for the initial condition with d₀, m₀, and r₀, where the height of the water is h.Case 1:
The new sphere's mass is m = m₀ .The new sphere's density, d > d₀.Here, the smaller, denser sphere with the same mass, If the sphere floats, the amount of water it displaces will be equal to its mass, which will be the same as the amount of water the original sphere displaces.Consequently, the water level is unchanged.But if the sphere descends, the water displaced is less than the sphere's mass, m = m0, and the level drops, f.Therefore, f or s is the appropriate response.Case 2:
The new sphere's radius, r = r₀, and mass, m > m₀.Consequently, the new sphere is denser than the old one.The right answer is r, because the mass of the water displaced where the sphere floats is m > m0, which is greater than the water displaced for the initial sphere.Case 3:
sphere with the same radius but a higher density.When the right side of equation (1) rises, the left side must also rise in order for the volume to rise and the height to rise as a result (r)Thus, we can conclude that,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rLearn more about Archimedes principle here:
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What force causes a bike to move forward?
A. Air resistance
B. Thrust
C. Friction
D. Gravity
The Answer is Option C. Friction.
Explanation:
The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.
A block of mass 2.60 kg is placed against a horizontal spring of constant k = 725 N/m and pushed so the spring compresses by 0.0500 m.
What is the elastic potential energy of the block-spring system (in J)?
_______J
If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring.
_____s/s
The elastic potential energy of the block-spring system is 0.906 J.
Velocity of the block , v = 0.83 m/s.
What is elastic potential energy?Elastic potential energy is the energy stored in a stretched or compressed elastic material.
Elastic potential energy = Ke²/2The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J
The elastic potential energy of the spring is converted to kinetic energy of the block.
1/2 mv² = 0.906 J
where v is velocity
v = √(0.906 * 2)/2.6
v = 0.83 m/s.
In conclusion, elastic potential energy is present in compressed or stretched elastic materials.
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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!
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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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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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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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
A 550-g squirrel with a surface area of 945 cm2 falls from a 4.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
m/s
The velocity of a 55.0-kg person hitting the ground, is mathematically given as
vt=39.5983m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?Generally, the equation for is mathematically given as
mass of squirrel,
[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]
Terminal velocity is given by:
[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]
where \rho is the density of fluid that is falling and it is given by
[tex]$\rho=\frac{m}{V}$[/tex]
since, volume =area * height
[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]
A is the surface area of squirrels.
C is the drag coefficient.
The surface area facing the fluid is given by:
[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]
so, terminal velocity is :
[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]
Vt=39.5983
In conclusion, the terminal velocity of the squirrel is 39.5983m/s
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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?
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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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
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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Plzzz help me with this
I’ll give brainliest
Answer:
(A) By reducing friction
Answer:
A
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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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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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
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:
Select the correct answer.
Using the statistical definition of entropy, what is the entropy of a system where W = 4?
Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.
What is entropy?Entropy is typically referred to as a measurement of a system's randomness or disorder. In 1850, a German physicist named Rudolf Clausius first proposed this idea. Entropy is a thermodynamic property used to characterize a system's behavior in terms of temperature, pressure, entropy, and heat capacity. This thermodynamic explanation took the systems' equilibrium condition into account.
Entropy can be calculated using a mathematical expression.
Entropy = Total change of heat / thermodynamic temperature
S= KblnW
S is the statistical entropy
The value of Boltzmann's constant is 1.38×10⁻²³.
S= 1.38×10⁻²³ ln(4)
S = 1.91×10⁻²³
Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.
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Answer: C.
1.91 × 10⁻²³ joules/kelvin
Explanation: edmentum
A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises by 2.4 cm. Find: (a) the work done by the gas; (b) the change in its internal energy. Atmospheric pressure is 105Pa
The work done by gas is 0.753 J and change in internal energy is 4.247J
So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm
The piston raised by 2.4cm
As we know that Work done is PΔV
Where ΔV is change in volume
Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3
Here pressure is 10^5 pa
So W = [tex]10^5\times7.53\times10^-(6)[/tex]
Therefore W = 0.753 J
Now coming to change in internal energy
Change in Internal Energy = Heat Added - Energy lost in work
∴ 5J - 0.753 J = 4.247J
Hence the change in internal energy is 4.247 J
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