Answer:
A) E = 278925.62 N/C with direction; radially out.
B) E = 43048.47 N/C with direction radially out.
C) E = -3214.29 N/C with direction radially in.
Explanation:
From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;
E = kQ/r²
where;
Q is the net charge within the distance r.
We are given the charge Q = 15-nC and
spherical shell of radius 10cm
A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
E = 278925.62 N/C
This will be radially out ,since the net charge is positive.
B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
E = 43048.47 N/C
This will be radially out ,since the net charge is positive.
C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Q = 15 nC - 22 nC
Q = -7 nC = -7 x 10^(-9) C
and;
E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
E = -3214.29 N/C
This will be radially in, since the net charge is negative. You can indicate this with a negative answer.
A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C
Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²Then E = 278925.62 N/CThen This will be radially out since the net charge is positive.
B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C
then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²Then E = 43048.47 N/CAfter that This will be radially out since the net charge is positive.
C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Then Q = 15 nC - 22 nCAfter that Q = -7 nC = -7 x 10^(-9) CWhen E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²Then E = -3214.29 N/C Thus, This will be radially in, since the net charge is negative.Find out more information about magnitude here:
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A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area
of contact is 2000 cm2.
Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?
Answer:
17.85°
Explanation:
To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
n1: index of refraction of Sophia's cornea = 1.387
n2: index of refraction of aqueous fluid = 1.36
θ1: angle to normal in the first medium = 17.5°
θ2: angle to normal in the second medium
You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:
[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]
hence, the angle to normal in the aqueous medium is 17.85°
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer to three significant figures and include the appropriate units.
Answer:
ΔE = ‒0.271 kJ
Explanation:
Let's begin by listing out the given variables:
q = -0.653 kJ, w = 0.389 kJ
Using the formula ΔE = q + w
ΔE = -0.653 + 0.388
ΔE = (‒0.655 + 0.382) kJ
ΔE = ‒0.271 kJ
Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
Which of the following is analogous to the pipes in an electrical circuit?
A. capacitors storing the incoming charge from the battery
B. large resistors causing restrictions to the flow of charge
C. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery electric current being forced uphill by the battery
D. electric current being forced uphill by the battery back to the positive terminal
The correct answer is D. electric current being forced uphill by the battery back to the positive terminal.
What is Electric Current?
Electric current is the flow of electric charge through a conducting medium, such as a wire, due to the movement of electrons or ions. The flow of charge is typically caused by the presence of an electric field that creates a potential difference (voltage) between two points in a circuit
In an electrical circuit, pipes are analogous to wires or conductive paths that allow the flow of electric current. The flow of electric current is from the positive terminal of the battery to the negative terminal, which is opposite to the direction of conventional current flow. Therefore, option C is incorrect.
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Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..
Answer:
* air resistance.
*the direction of the rotation of the Earth
rotation of the thrown body
Explanation:
The projectile launch is described by the expressions
x-axis x = v₀ₓ t
y-axis y = [tex]v_{oy}[/tex] t - ½ gt²
When the things that affect this movement are analyzed, in order of importance we have:
* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.
* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.
* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption
* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height
Which formation is one feature of karst topography?
Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.
What is karst topography?Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.
The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.
When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.
Hence, sinkholes formation is one feature of karst topography
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Answer: A) Caves
Explanation:
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex] ............1
put here value and we will get value
normal force = [tex]\frac{0.400}{0.04}[/tex]
solve it we get
normal force = 10 N
PLS HELP,WILL GIVE BRAINLIEST + 30 POINTS
Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.
Answer:
☟
Explanation:
Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.
Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:
heated to 600-700°C
passed over a catalyst of silica or alumina
These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:
hexane → butane + ethene
C6H14 → C4H10 + C2H4
Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.
Fractional distillation of crude oil
Fractional distillation separates a mixture into a number of different parts, called fractions.
A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.
Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.
Oil fractions
The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.
As you go up the fractionating column, the hydrocarbons have:
lower boiling points
lower viscosity (they flow more easily)
higher flammability (they ignite more easily).
Other fossil fuels
Crude oil is not the only fossil fuel.
Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.
Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.
In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.
What is fractional distillation?Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.
Generally, the components have boiling points that differ by less than 25 °C from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.
The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.
Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.
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4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
A 1000-kg car is moving down the highway at 14m/s. What is the momentum?
Explanation:
Momentum = mass × velocity
p = mv
p = (1000 kg) (14 m/s)
p = 14000 kg m/s
The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
Given the data in the question
Mass of the car; [tex]g = 1000kg[/tex]Velocity of the car; [tex]v = 14m/s[/tex]Momentum; [tex]p = ?[/tex]Momentum is the product of the mass of a particle and its velocity.
Momentum = Mass × Velocity
[tex]P = m \ * \ v[/tex]
We substitute our given values into the equation
[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]
Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
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A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
1. An object with a mass of 15 kilograms is pushed by a force of 30 Newtons. How much does
it accelerate?
Answer: [tex]2m/s^2[/tex]
Explanation:
[tex]Formula: F=ma[/tex]
Where;
F = force
m = mass
a = acceleration
Solve for a;
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Answer:
Explanation:
The work required to push will be equal to work done by friction . Let d be the displacement required .
force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction
work done by force of friction
mg x μ x d = 660
80 x 9.8 x .272 x d = 660
d = 3 .1 m .
A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to
Answer:
The time when the ball strikes the ground is closest to [tex]t_t = 9.4 \ s[/tex]
Explanation:
From the question we are told that
The time of projection is t = 0.0 s
The distance of the point from the ground is [tex]d = 90 \ m[/tex]
The initial velocity of the ball is [tex]v _i = 36 .2 \ m/s[/tex]
generally the time required to reach maximum height is
[tex]t_r = \frac{g}{v}[/tex]
Where is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
Substituting values
[tex]t_r = \frac{36.2}{9.8}[/tex]
[tex]t_r = 3.69 s[/tex]
when returning the time and velocity at the roof level is t = 3.69 s and u = 36.2 m/s this due to the fact that air resistance is negligible
The final velocity at which it hit the ground is
[tex]v_f^2 = u^2 + 2ag[/tex]
So
[tex]v_f = \sqrt{ u^2 + 2gs}[/tex]
substituting values
[tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]
[tex]v_f = 55.45 \ m/s[/tex]
The time taken for the ball to move from the roof level to the ground is
[tex]t_g = \frac{v-u}{a}[/tex]
substituting values
[tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]
[tex]t_g = 1.96 \ s[/tex]
The total time for this travel is
[tex]t_t = t_g + 2 t_r[/tex]
[tex]t_t = 1.96 + 2(3.69)[/tex]
[tex]t_t = 9.4 \ s[/tex]
Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a price ceiling for a bottle of water equal to the market equilibrium price during normal times. After all, it seems unfair that sellers of water gain because of a hurricane.
Answer:idk
Explanation:idk
Answer:
shortage of 50 water bottles
$2
30
Explanation:
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.
Before she jumped from the cliff, her gravitational potential energy was
GPE = (her weight) x (height of the cliff) .
That's exactly the GPE that she lost on the way down to the water. So we can write
24,500 J = (her weight) x (44.0 m)
Divide each side by 44.0 m:
Her weight = 24,500 J / 44 m
Her weight = 556.8 Newtons
(about 125 pounds)
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m
Answer:
Letter C) and D) is the correct answer.
Explanation:
We know that the a is an acceleration's magnitude, so the units of a are m/s².
Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.
[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]
[tex][vt]=[m][/tex]
[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]
Therefore, the term F/m is the correct answer.
I hope it helps you!
We can see that a and F/M are dimensionally homogeneous.
In solving dimensions, we try to express a quantity in terms of the fundamental quantities;
MassLengthTimeFor the term a, its dimension is LT^-2
For the term F/m, its dimension is LT^-2
Hence, it follows that a and F/M are dimensionally homogeneous.
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If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
What's a line of best fit? Will give BRAINLIEST
A line of best fit expresses the relationship between the points.
Explanation:
It does not go through all the points but goes through most of them and it is like a hardrawn curve
A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane
Answer:
5.1 rad/s
Explanation:
Mechanical energy of the system is conserved since no external work is done on the sphere.
[tex]mgh = mv^2/2 + I\omega^2/2[/tex]
Substituting v = ωr and I = 2 m r^2/5, we get,
=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]
=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]
=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]
=> [tex]gh = 7\omega^2 r^2/10[/tex]
=> [tex]\omega r = (10gh/7)^{1/2}[/tex]
=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.
Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,
An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?
Answer:
Ff = 33.4N
Explanation:
To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.
The frictional force is given by:
[tex]F_f=\mu_kN[/tex] (1)
Ff: frictional force = ?
µk: coefficient of kinetic friction = 0.167
N: normal force of the object = 200N
You replace the values of the parameters in the equation (1):
[tex]F_f=(0.167)(200N)=33.4N[/tex]
The frictional force, while the objects is moving, is 33.4N
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
[tex]^{\circ}C=K-273[/tex]
Put T = 100 K
[tex]T=100-273=-173^{\circ} C[/tex]
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4
Complete Question
The complete question is shown on the first uploaded image
Answer:
The uncertainty in inverse frequency is [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
Explanation:
From the question we are told that
The value of the proportionality constant is [tex]k = 5 \frac{Hz }{T}[/tex]
The strength of the magnetic field is [tex]B = 20 \ T[/tex]
The change in this strength of magnetic field is [tex]\Delta B = 3 \ T[/tex]
The magnetic field is given as
[tex]B = \frac{k}{\frac{1}{w} }[/tex]
Where [tex]w[/tex] is frequency
The uncertainty or error of the field is given as
[tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]
The uncertainty in inverse frequency is given as
[tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]
substituting values
[tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
