Answer:
Range, [tex]R = MV²/2QE[/tex]
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
What is the boiling point (in °C) of a solution of 7.94 g of I2 in 69.2 g of toluene, assuming the I2 is nonvolatile? (For toluene, Tb = 110.63°C and Kb = 3.40°C·kg/mol.)
Answer: The boiling point of solution is [tex]112.16^0C[/tex]
Explanation:
Elevation in boiling point:
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of toluene = [tex]110.63^oC[/tex]
[tex]k_b[/tex] = boiling point constant of toluene =[tex]3.40^oC/m[/tex]
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]w_2[/tex] = mass of solute [tex](I_2)[/tex] = 7.94 g
[tex]w_1[/tex] = mass of solvent (toluene) = 69.2 g
[tex]M_2[/tex] = molar mass of solute [tex](I_2)[/tex]= 254g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-110.63)^oC=1\times (3.40^oC/m)\times \frac{(7.94g)\times 1000}{254\times (69.2g)}[/tex]
[tex]T_b=112.16^0C[/tex]
Therefore, the boiling point (in °C) of a solution is 112.16
A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘ below the horizontal.
a. Find the work done by the force F⃗ P on the cart if the ramp is 6.5 m long.
b. Find the work done by the force mg⃗ on the cart.
Answer: a. 198.6J b. - 198.6J
Explanation: Parameters given:
m = 15kg
g = 9.8m/s²
∅ = 12°
a. Work done by the force Fp on the cart if the ramp is 6.5m long.
Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N
Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J
b. The work done by the force mg on the cart.
Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance
= 15kg x -9.8m/s² x Sin12° x 6.5m
= - 198.6J
Answer:
(a)The work done by the force Fp is 198.6J
(b) The work done by the weight mg is -198.6J
The net force acting on the cart is zero. This is because the cart is moving with a constant velocity and by newton's first law the the net force on the cart is equal to zero.
Fp was calculated to be equal to 34.94 N.
Explanation:
In order to solve this kind of problems successfully, the best approach is to resolve all forces acting on the cart parallel and perpendicular to the ramp surface. So that the x-axis is parallel to the ramp surface and the y axis is perpendicular to the ramp surface.
Fpx = FpCos 29°
Fpy = FpSin 29°
Wx = mg sin12°
Wy = mg cos 12°
Summation Fx = 0
And Summation Fy = 0
The full solution can be found below in the attachment.
Thank you for reading and I hope this is helpful to you.
A 0.500 kg mass is oscillating on a spring with k=330 N/m.The total energy of its oscillation is 3.24 J. What is speed of the mass when it is 0.100 m from the EP?
(Unit=m/s)
The speed is 2.5 m/s
Explanation:
The total (mechanical) energy of the mass-spring system at any point during the motion is the sum of the kinetic energy (KE) and the potential energy (PE):
[tex]E=KE+PE=3.24 J[/tex]
and it is constant.
The kinetic energy can be written as
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m = 0.500 kg is the mass
v is the speed
While the potential energy is
[tex]PE=\frac{1}{2}kx^2[/tex]
where
k = 330 N/m is the spring constant
x is the elongation
So the first equation becomes
[tex]E=\frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
Therefore, if we substitute
x = 0.100 m
We can find the speed when the elongation is x = 0.100 m:
[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(3.24)-(330)(0.100)^2}{0.500}}=2.5 m/s[/tex]
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Correct Answer:
2.5 m/s
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. At what angle (in degrees) must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s?
Answer:
The answer to your question is angle = 18.46°
Explanation:
Data
d = 75 m
v₀ = 35 m/s
α = ?
Formula
[tex]d = \frac{vo^{2}sin2\alpha}{g}[/tex]
solve for sin2α
sin 2α = [tex]\frac{dg}{vo^{2}}[/tex]
Substitution
sin 2α = [tex]\frac{75(9.81)}{35^{2}}[/tex]
Simplify
sin 2α = [tex]\frac{735.75}{1225}[/tex]
Divide
sin 2α = 0.600
Get sin⁻¹
2α = 36.9°
Divide by 2
α = 18.5°
To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should you choose to throw a rubber ball or a beanbag of equal size and weight?
Answer:
Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.
For Rubber Ball
Upon elastic collision it will reverses the direction and move with velocity equal or less then original
change in momentum = P
[tex]P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}[/tex]
For Beanbag
value of impulse will large if velocity is zero.
[tex]v_{f}=0\\ P = -mv_{i}[/tex]
Explanation:
The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.
A collision between two objects can be elastic or inelastic.
In elastic collision both momentum and kinetic energy are conserved.In inelastic collision only momentum is conserved.The impulse experienced by each throwing object is equal to change in the momentum of the object.
[tex]J = \Delta P[/tex]
The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.
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At 20°C, the resistance of a sample of nickel is 525 Ω. What is the resistance when the sample is heated to 70°C? Let α = 0.005866 at 20°C. Explain please.
The final resistance is [tex]679\Omega[/tex]
Explanation:
The relationship between the resistance of a metal and the temperature is
[tex]R(T) = R_0(1+\alpha (T-T_0))[/tex]
where
[tex]R_0[/tex] is the resistance at a temperature of [tex]T_0[/tex]
R is the resistance at temperature T
[tex]\alpha[/tex] is the temperature coefficient of resistance
In this problem, we have:
[tex]R_0 = 525 \Omega[/tex]
[tex]T_0 = 20^{\circ}C[/tex]
[tex]\alpha = 0.005866 \Omega/^{\circ}C[/tex]
Therefore, the resistance when [tex]T=70^{\circ}C[/tex] is
[tex]R=(525)(1+0.005866(70-20))=679\Omega[/tex]
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Answer:
Therefore the new resistance would be 679 Ω
Explanation:
Resistance is the opposition to the flow of electric current. The resistance of an object given its coefficient of resistance can be obtained with the expression bellow;
R =R_ref [1+ α (T - T_ref)]
Where R is the new resistance
R_ref is the base resistance = 525 Ω
α is the coefficient of resistance at 20°C = 5
T is the new temperature = 70°C
T_ref is the base temperature = 20°C
Substituting the values into the equation we have;
R = 525 x [ 1 + 0.005866 (70-20)]
R = 525 x [ 1 + 0.005866 (50)]
R = 525 x 1.2933
R = 678.98
R≈ 679 Ω
Therefore the new resistance is 679 Ω
__________ brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction. Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.
"Anti-Lock" brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
Explanation:
The safety anti-skid braking system is known as "anti-lock braking system" having huge application on land vehicles like one, two and multiple wheeler vehicles and aircraft. During braking, it avoids wheels to get locked by building tractive contacts to the road's surface.
This seems to be an automated system work on the principles of techniques - threshold and cadence braking. The wheel velocity sensors are utilized by ABS to find whether one or more than one wheels chose to get lock while braking.
Answer:
Anti- Lock
Explanation:
Anti- lock brake systems(ABS) is a safety system used in hindering the wheels from locking up when the brake is applied.
Braking release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.
Which of the following ingredients, if added to foods in equal amounts, would have the LEAST impact on the energy density of the food?
a. fruit juice
b. table sugar
c. sausage
d. butter
Answer:
a. fruit juice
Answer:
a. fruit juice
Explanation:
have a great day everyone! also merry christmas!
What do we expect to happen to Earth’s air temperature if the amount of energy from the sun that reaches the Earth’s surface is reduced by half?
Explanation:
The Natural Greenhouse Effect However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.
What do you call the gravitational attraction between you and the earth?
A.weight
B. mass
Answer:
A. Weight
Explanation:
Mass And Weight
Mass is a fundamental property of objects and is defined as a measure of the amount of matter in the object. The symbol for mass is m and its SI unit is the kilogram (kg). At normal speeds, i.e. much smaller than the speed of light, the mass is considered as a constant property.
The weight W is defined as the force of gravity on the object, that is, the gravitational attraction between the object and the Earth. Its SI unit is the Newton (Nw or N).
Compared with cool air, warm air is:________
a. more dense, and tends
b. to rise less dense, and tends
c. to rise more dense, and tends
d. to sink less dense, and tends to sink
Answer:
option (b)
Explanation:
The density of cool air is more so always falls downwards.
Compared to the cool air, the density of air is less and it tends to rise.
Thus, option (b) is correct.
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well.Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kgm2 and for arms and legs in is 0.90 kgm2 . If she starts out spinning at 4.5rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s
Explanation:
given information:
moment inertia of arm and leg when in, I₁ = 0.9 kgm²
moment inertia of arm and leg when extended, I₂ = 2.9 kgm²
angular speed when in, ω₁ = 4.5 rev/s
so, her angular speed (in rev/s) when her arms and one leg open outward is
L₁ = L₂
I₁ω₁ = I₂ω₂
ω₂ = I₁ω₁/I₂
= 0.9 x 4.5/2,9
= 1.4 rev/s
A block of mass 4 kg, which has an initialspeed of 3 m/s at timet= 0, slides on ahorizontal surface.Find the magnitude of the work that mustbe done on the block to bring it to rest?
Answer:
18 J
Explanation:
Work done: This can be defined as the product of force and distance acting on a body. The S. I unit of work is Joules (J)
From the question, the work that must be done in bringing the block to rest is equal to the kinetic energy of the block.
Ek = 1/2mv².................... Equation 1
Where Ek = kinetic eenrgy m = mass of the block, v = velocity of the block.
Given: m = 4 kg, v = 3 m/s.
Substituting into equation 1
Ek = 1/2(4)(3²)
Ek = 2(9)
Ek = 18 J.
Thus the work that must me done on the block to bring it to rest = 18 J.
How star properties affect star formation? Provided following are the spectral types of four different main-sequence stars. Rank the stars based on the strength of the radiation pressure that pushes outward as they are forming, from highest pressure to lowest pressure.A) O9.B) A5.C) G2.D) M6.
Answer:
-M6
-G2
-A5
-O9
Explanation:
A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running north with a speed of 3.6 m/s. Assuming momentum conservation, determine the speed and direction of the two players immediately after the tackle.Give the direction as an angle, in degrees, south of west.
They travel at a speed of 0.68 m/s south ([tex]90^{\circ}[/tex] south of west)
Explanation:
We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two players must be conserved before and after they collide.
Mathematically:
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]
where, taking north as positive direction:
[tex]m_1 = 92 kg[/tex] is the mass of the first player
[tex]u_1 = -5.8 m/s[/tex] is the initial velocity of the first player (south, so negative)
[tex]m_2 = 110 kg[/tex] is the mass of the second player
[tex]u_2 = 3.6 m/s[/tex] is the initial velocity of the second player (north, so positive)
[tex]v[/tex] is the final combined velocity of the two players
Solving for v, we find the final velocity of the two players combined:
[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(92)(-5.8)+(110)(3.6)}{92+110}=-0.68 m/s[/tex]
where the negative sign indicates their final direction is south.
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To find the final speed and direction after the tackle, we use the conservation of momentum. The fullback and lineman's initial momenta are calculated and combined to find the total initial momentum, which equals the final momentum because momentum is conserved. The direction is either south (90 degrees) or north (270 degrees) depending upon the sign of the final velocity.
Explanation:To solve this physics problem, we will use the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant. The formula for momentum (p) is the product of mass (m) and velocity (v), or p = mv.
Before the tackle, the fullback's momentum is southward and the lineman's is northward. We treat south as the positive direction. Thus, the initial momentums of the fullback and lineman are:
Fullback momentum: pFB = 92 kg × 5.8 m/sLineman momentum: pL = 110 kg × (-3.6 m/s) (negative due to the northward direction)The total initial momentum (pinitial) is the sum of these:
pinitial = pFB + pL
After the tackle, the players stick together and move as one mass (M = 92 kg + 110 kg), with a final velocity (V), so:
pfinal = MV
Since momentum is conserved, pinitial = pfinal, and we can solve for V:
V = pinitial / M
Once we calculate the final velocity, the direction is south if V > 0, north if V < 0. To convert the direction into an angle south of west, note that if they move southward, the angle is simply 90 degrees (directly south), while any northward movement would result in a 270-degree angle.
A car traveling at 7 m/s accelerates uniformly at 2.5 m/s to reach a speed of 12 m/s. How long does it take for this acceleration to occur?
There is one mistake in the question as unit of acceleration is not written correctly.The correct question is here
A car traveling at 7 m/s accelerates uniformly at 2.5 m/s² to reach a speed of 12 m/s. How long does it take for this acceleration to occur?
Answer:
time taken =2 seconds
Explanation:
Given Data
Initial Speed Vi= 7 m/s
Final Speed Vf=12 m/s
Acceleration a= 2.5 m/s²
To find
Time taken for this acceleration
Solution
As we know that
Final velocity=Initial velocity + acceleration×time
[tex]V_{f}=V_{i}+at\\ t=\frac{V_{f}-V_{t}}{a}\\ t=\frac{12m/s-7m/s}{2.5m/s^{2} }\\ t=2 seconds[/tex]
So car takes 2 seconds for this acceleration to occur
Satellite 1 revolves around a planet at the altitude equal to one-half the radius of the planet. The period of revolution of satellite 1 is . What is the period of revolution of an identical satellite 2 that revolves around the same planet at the altitude equal to the radius of the planet?
Answer:
To calculate the period of satellite orbiting around a planet, we use Kepler's third law;
Square of T = [(4π)/(G*m)] * R^3.
Therefore,
T = sqrt{[(4π)/(G*m)]*R^3}.
T is the period, m is mass orbiting satellite, G is gravitational constant, R is the radius of of the planet, r is the radius of the orbiting satellite.
For Satellite 1, r is one-half of the planet, that is r = (3/2) * R
For satellite 2, r = R
Explanation:
Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of mass m1, in meters per square second.
Answer:
a1 = 3.68m/s²
Explanation:
Given values:
Mass of the block placed on the table, m1 = 12.25 kg
Mass of the block hanging vertically, m2 = 7.5 kg
Acceleration due to gravity, g = 9.8 m/s2
Tension in the string is T
Let the acceleration of mass 1 and mass 2 be a1 and a2
a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,
a = m2*g/(m1 + m2)
a = 7.5 x 9.8 / (12.5 + 7.5)
a = 3.68 m/s²
a1 = a2 = 3.68m/s²
a1 is directed to the right and a2 is directed downwards
Below is a diamonds to show the geometrical arrangements of both masses
What type of nuclear radiation is emitted when carbon-14 decays
Answer:
Beta radiation
Explanation:
Beta radiation is a radioactive phenomenon of nuclear decay in which an unstable atom or isotop, by transforming a neutron into a proton, or by transforming a proton into a neutron, becomes stable. For example, the decay of carbon 14 produces beta radiation.
According to the Belmont Report, the moral requirement that there be fair outcomes in the selection of research subjects, expresses the principle of:________________.
Answer:
The correct answer is: justice
Explanation:
The Belmont Report refers to a report that was published 25 year ago, focusing on the ethical treatment and protection of participants in medical and behavioral research. This report centers around 3 principles:
1. Beneficence- striving to maximize benefits for participants of the research study and minimizing any harms/ risks that might occur.
2. Justice- The fair selection of potential participants for a study. This ensures equitable and fair distribution of risks/ benefits to all potential participants of a research study. Subjects of a study must not be chosen merely out of convenience or easy access. The inclusion/ exclusion criteria should be chosen according to the nature of the study and steps/ treatments that it will involve.
3. Respect for persons- Each participant of a research study should be able to provide informed consent prior to their participation, protected from controllable harm and treated with respect.
Therefore, moral requirement that there be fair outcomes in the selection of research subjects, expresses the principle of justice.
A child has been diagnosed with an autoimmune disease that predominantly affects epithelial tissue. Histologic examination should include which body parts as potential sites of involvement? Select all that apply.a) Heartb) Bone marrowc) Bladderd) Epidermise) Blood vessel lining
Answer:C) BLADDER
D) EPIDERMIS (E) BLOOD VESSEL LINING
Explanation: Autoimmune diseases are diseases caused by the body fighting against its self. In autoimmune diseases the body's defense system tends to attack the body cells, tissues, Organs and the entire system mistaking it for a foreign body such as Viruses or Bacteria,Fungi etc.
Epithelial tissues are tissues which form the surface covering of most of the hollow Organs,all body Organs and are present inside the glands. The body parts that will be involved are the BLADDER, EPIDERMIS AND BLOOD VESSEL LINING.
A plastic boat with a 25 cm² square cross section floats in a liquid. One by one, you place 50-g masses inside the boat and measure how far the boat extends below the surface.
Your data are as follows:
Mass added, m(g) - Depth, d(cm)
50 - 2.9
100 - 5.0
150 - 6.6
200 - 8.6
Graphing either m versus d or d versus m gives a straight line. In the graph shown above, we chose to plot d on the vertical axis and m on the horizontal axis. From the equation for the line of best fit given, determine the density rho of the liquid.
Please Explain.
Answer:
Explanation:
Archimedes principle states that the upward buoyant foce exrted on a body is equal to th wight o the liquid displaced.
Now, the buoyant force on the boat is given by:
[tex](m+m_b)g=V\rho g[/tex]
[tex]V[/tex] is the volum [tex]\rho[/tex] is the density [tex]m_b[/tex] is the mass of the boat and [tex]m[/tex] is the mass added to the boat.
[tex](m+m_b)g=(Sd)\rho[/tex]
[tex]S[/tex] is the surface area and [tex]d[/tex] is the depth.
[tex]m=Sd\rho - m_b...(1)[/tex]
The equation for thebest fit linis,
[tex]d=(0.374m/kg)m+0.11m[/tex]
Re-arrangethis equati for [tex]m[/tex]
[tex]m=\frac{d}{(0.374m/kg)}-\frac{0.11m}{0.374m/kg}...(2)[/tex]
From equations(1) and (2),
[tex]Sd\rho=\frac{d}{0.374m/kg}[/tex]
the density is,
[tex]\rho=\frac{1}{S(0.0374m/kg)}=\frac{1}{(25cm^2)(\frac{1m^2}{10^4cm^2})(0.374m/kg)}=1.069\times 10^3 kg/m^3[/tex]
Therefore, the density of the liquid is
[tex]\rho=1.07\times 10^3 kg/m^3[/tex]
To determine the density of the liquid, one needs to get the slope of the graph from the data given. Using this slope in the formula ρ = k/g, where 'g' is the gravity, gives the density of the liquid.
Explanation:
Let's first understand this concept with the help of Archimedes' Principle, which states that, the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. To determine the density of the liquid, we need the slope of the line from the graph.
Let's assume that the slope of the line is 'k' (which you will obtain from your graph). The slope of the line of best fit in the graph of m versus d will be m/d = ρVg/(Ag), where 'm' is the mass added, 'd' is the depth, 'ρ' is the density of the fluid, 'V' is the volume of the fluid displaced, 'A' is the cross-sectional area of the boat, and 'g' is the acceleration due to gravity.
We can write Volume 'V' = Ad, so the equation simplifies to k = ρg and hence the density ρ = k/g, where 'k' is the obtained slope and 'g' (assuming you are on the earth) is 9.81 m/s². Therefore, once you obtain the value of 'k', you can easily calculate the density of the liquid.
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Fill in the terms that accurately complete the statements. The nucleus contains positively charged particles called and neutral particles called , which are bound together by the strong nuclear force. The electron cloud contains negatively charged particles, which participate in reactions.
Explanation:
Nucleus of every atom contains both protons and neutrons. Protons are positively charged species whereas neutrons are neutral species, that is, neutrons do not contain any charge.
And, when an electron cloud contains a negatively charged ion then it is able to participate in a chemical reaction as it needs to gain stability.
Therefore, we can conclude that the nucleus contains positively charged particles called protons and neutral particles called neutrons, which are bound together by the strong nuclear force. The electron cloud contains negatively charged particles, which participate in chemical reactions.
Answer:
Protons, neutrons,and chemicals.
Explanation:
on edge
Sally is pushing a shopping cart with a force of 20 N. Because the wheels are stuck, the friction caused by the ground is exerting a force of 8 N in the opposite direction. Calculate the net force on the shopping cart including an arrow indicating what direction the shopping cart is going
The net force on the shopping cart is 12 N to the right.
This is a question related to Newton's Laws. It can be solved by using a Free body diagram, which shows all the forces acting on the object. There are 4 forces acting on the object:
Gravity (g) pointing down.Normal (N) pointing up. Push force by Sally (Fp) pointing right.Friction force (Ff) pointing left.Forces (1) and (2) cancel each other because they have the same magnitude, and the overall force is given by the addition of forces (3) and (4)
[tex]F=Fp+Ff=20N+ (-8N)=12N[/tex]
The positive value indicates that the shopping cart is moving to the right.
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Which are true?
1) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.
2) Electric field lines near positive point charges radiate outward.
3) The electric force acting on a point charge is proportional to the magnitude of the point charge.
4) Electric field lines near negative point charges circle clockwise.
5) The electric field created by a point charge is constant throughout space.
Answer:
Following are the correct options.
1. In a uniform electric field, the field lines are straight, parallel and uniformly spaced. ( As the field strength does not change in the uniform electric field that is why the field lines are parallel, equally spaced etc).
2. Electric field lines near positive point charges radiate outward. (Due to repulsive forces field lines radiate in outward direction)
3.The electric force acting on a point charge is proportional to the magnitude of the point charge. ( As columb law states that the electrostatic forces between the point charges is proportional to the product of their magnitude and inversely proportional to square of distance between them).
Explanation:
In a uniform electric field, electric field lines are straight, parallel, and uniformly spaced while near positive point charge it emits outwards. Also, the electric force on a point charge is proportional to its magnitude. But, lines near negative point charges radiate inward, not circle clockwise. The electric field created by a point charge isn't constant across space, but diminishes with distance.
Explanation:Of the five statements you provided:
In a uniform electric field, the field lines are indeed straight, parallel, and uniformly spaced. Electric field lines near positive point charges do radiate outward. The electric force acting on a point charge is proportional to the magnitude of the point charge.However, electric field lines near negative point charges don't circle clockwise, they radiate inward towards the negative charge.Additionally, the electric field created by a point charge isn't constant throughout space; it diminishes with distance from the point charge according to Coulomb's Law.Therefore, statements 1, 2 and 3 are true, while 4 and 5 are false.
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A proton is released from rest inside a region of constant, uniform electric field E 1 pointing due north. 27.3 s after it is released, the electric field instantaneously changes to a constant, uniform electric field E 2 pointing due south. 3.03 s after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E 2 to the magnitude of E 1 ? You may neglect the effects of gravity on the proton.
Answer:
E2/E1 =99.2
Explanation:
time after release of E1 (t) = 27.3 s
time after release of E2 (t') = 3.03 s
acceleration (a) = [tex]\frac{QE1}{M}[/tex]
where
Q is the protons charge M is the massafter 27.3 s
velocity (V) = a x t = [tex]\frac{QE1}{M}[/tex] x 27.3 = [tex]\frac{27.3QE1}{M}[/tex]
distance to turning point (s) = 0.5a[tex]t^{2}[/tex] = 0.5 x [tex]\frac{QE1}{M}[/tex]x [tex]27.3^{2}[/tex] = [tex]\frac{372.65QE1}{M}[/tex]
now for its return back to its starting point
acceleration (a') = [tex]-\frac{QE2}{M}[/tex]
total distance S' = distance to turning point + distance from turning point to starting point
S' = S + vt' + 0.5 a'[tex]t'^{2}[/tex]
S' = [tex]\frac{372.65QE1}{M}[/tex] + ([tex]\frac{27.3QE1}{M}[/tex] x 3.03) + (0.5 x [tex]-\frac{QE2}{M}[/tex]x [tex]3.03^{2}[/tex])
S' is the distance at the starting point and = 0
0 = [tex]\frac{372.65QE1}{M} + \frac{82.72QE1}{M}-\frac{4.59QE2}{M}[/tex]
[tex]\frac{4.59QE2}{M}=\frac{372.65QE1}{M} + \frac{82.72QE1}{M}[/tex]
multiplying both side by M/Q we have
4.59.E2 = 372.65E1 + 82.72E1
4.59.E2 = 455.37E1
E2/E1 = 455.37 / 4.59
E2/E1 =99.2
The ratio of the magnitude of electric field E2 to E1, when a proton returns to its initial point after the direction of a uniform electric field changes, is approximately 0.0123.
Explanation:A proton is initially at rest in an electric field that points due north. After 27.3 seconds, the electric field changes direction and points due south, and after 3.03 seconds the proton returns to its starting point. To find the ratio of the magnitudes of E2 to E1, we must consider the distances covered by the proton under the influence of both fields given that it starts and ends at the same position.
The distance covered under E1 can be calculated using the formula s = 0.5 × a × t^2, where 'a' is the acceleration and 't' is the time. Since the proton moves for 27.3 s under E1, the distance s1 is s1 = 0.5 × a1 × (27.3)^2. The proton then moves in the opposite direction under E2 for 3.03 s, covering the same distance in opposing direction, thus s2 = 0.5 × a2 × (3.03)^2. Since s1 = s2, we can equate them: 0.5 × a1 × (27.3)^2 = 0.5 × a2 × (3.03)^2.
Furthermore, the acceleration of the proton is directly proportional to the electric field (since a = F/m and F = qE), giving us a1 ∝ E1 and a2 ∝ E2. By simplifying the above equation, we find that a1/a2 = (3.03/27.3)^2, and therefore the ratio of the magnitudes of electric fields is also E2/E1 = (3.03/27.3)^2. Calculating this, we get the ratio E2/E1 approximately equal to 0.0123.
What is the average useful power output (in W) of a person who does 6.60 ✕ 106 J of useful work in 8.40 h?
Answer:
218.25 W.
Explanation:
Power: This is defined as the rate at which work is done. The S.I unit is Watt (W). Mathematically, it can be expressed as
P = W/t ............................... Equation 1.
Where P = useful power output, W = useful work, t = time taken to do the work.
Given: W = 6.6×10⁶ J, t = 8.4 h = 8.4×60×60 = 30240 s.
Substitute into equation 1
P = 6.6×10⁶/30240
P = 218.25 W.
Hence the useful power output = 218.25 W.
Two graduate students measure the strength of an MRI's magnetic field to have the following values: Student A: 4.2 ± 0.8 T Student B: 5.6 ± 0.5 T Do the two students agree for the strength of the magnetic field?
Answer:
do not agree
Explanation:
For student A
[tex]B=4.2\pm 0.8\ T\\\Rightarrow B=4.2+0.8\ or\ B=4.2-0.8\\\Rightarrow B=5\ T\ or\ 3.4\ T[/tex]
The magnetic field measured by student A = 5 T or 3.4 T
For student B
[tex]B=5.6\pm 0.5\ T\\\Rightarrow B=5.6+0.5\ or\ B=5.6-0.5\\\Rightarrow B=6.1\ T\ or\ 5.1\ T[/tex]
The magnetic field measured by student B = 6.1 T or 5.1 T
The magnetic fields measured are not equal hence they do not agree.
If an insulated (power side) wire rubbed through a part of the insulation and the wire conductor touched the steel body of a vehicle, the type of failure would be called a(an)
Answer:
a short-to-ground
Explanation:
Many failures that occur in vehicles can be diagnosed as electrical problems. The reason for these usually relate mainly to the electricity generated in the battery or alternator, in the ignition system as well as those produced by blown wires or fuses.
Cables or fuses: a blown fuse or a damaged electrical wire or one that is making ground or earth effect can be the cause of an electrical problem. Keep in mind that everyone is connected to each other through the wires of the electrical system. Fuses protect car components from power surges. If you have a problem with one, check it to see if you will have to replace it.
A failure where an insulated wire rubs through and touches the steel body of a vehicle is known as a short circuit.
Explanation:If an insulated (power side) wire rubs through a part of the insulation and the wire conductor touches the steel body of a vehicle, this type of failure is known as a short circuit. This happens when a wire carrying current touches the vehicle's body, which is grounded. As the body acts like a path of least resistance, the current prefers flowing through it rather than its intended path, creating a short circuit. This can result in various problems like damage to components, blowing a fuse, or even triggering a fire.
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melvin pulls a sled across level snow with a force of 317 N along a rope that is 33 degrees above the horizont?
Answer: work Melvin did=9000J
Explanation:
Given to complete the question: If the sled moved 33.9m,how much work did Melvin do? Answer in unit of J and round to the nearest thousandth.
W = F ×S
W = 317 × cos 33°×33.9
W=9012.6055J
W=9000J to the nearest thousandth