Answer: The balance of care is sympathetic and loving
Explanation: According to Erikson's theory, when the balance of care is sympathetic and loving the psychological conflict of the first year is resolved on the positive side.
According to Erikson's theory of psychosocial development, the psychological conflict of the first year is resolved on the positive side when the infant develops a sense of trust in their caregivers.
Explanation:According to Erik Erikson's theory of psychosocial development, when an infant develops a sense of trust in their caregivers during the first year of life, the psychological conflict of that stage is resolved on the positive side. Erickson named this first stage trust versus mistrust. In this stage, caregivers who are responsive and sensitive to their infant's needs help the child to develop a sense of trust; thus, the child will see the world as a safe, predictable place. If the caregivers are unresponsive and do not meet the infant's needs, feelings of anxiety, fear, and mistrust emerge; the child may see the world as unpredictable.
Successful completion of each developmental task at various stages results in a sense of competence and a healthy personality, according to Erikson. Failure to master these tasks may lead to feelings of inadequacy and affect the formation of a positive self-concept.
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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.
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.
A 2.40 μC charge is subject to a 3.00 mN force due to an Electric Field. What is the magnitude of the Electric Field at the location of the charge?
The magnitude of the electric field at the location of the 2.40 μC charge is 1250 N/C.
Explanation:The magnitude of the electric field at the location of the 2.40 μC charge can be calculated using the equation:
E = F / q
Where E is the electric field, F is the force, and q is the charge.
In this case, the force is given as 3.00 mN, which is equal to 0.003 N, and the charge is 2.40 μC, which is equal to 2.40 x 10^-6 C. Plugging these values into the equation, we get:
E = (0.003 N) / (2.40 x 10^-6 C) = 1250 N/C
Therefore, the magnitude of the electric field at the location of the charge is 1250 N/C.
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!
Children in a tree house lift a small dog in a basket 4.90m up to their house. If it takes 201J of work to do this, what is the combined mass of the dog and basket?
Answer:
4.18 kg.
Explanation:
Work done: Work is said to be done when ever a force move a body through a given distance. The S.I unit of work done is Joules (J)
Mathematically, work done is expressed as
W' = F×d.................... Equation 1
Where W' = work done, F = force , d = distance.
making F the subject of the equation,
F = W'/d............................. Equation 2
Note: The force need to lift the small dog n a basket = combined weight of the dog ans the basket.
Therefore,
W = F
Where W = combined weight of the dog and the basket.
Also
W = Mg
M = W/g............................. Equation 3
Where M = combined mass of the dog and the basket, g = acceleration due to gravity.
Given: W' = 201 J, d = 4.90 m.
Substitute into equation 2
F = 201/4.9
F = 41.02 N.
Since F = W = 41.02 N and g = 9.81 m/s²
Substitute these values into equation 3
M = 41.02/9.81
M = 4.18 kg.
Thus the combined mass of the dog and the basket = 4.18 kg.
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 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force acts on the otter along the incline?
Answer:
The net force acting on the otter along the incline is 13.96 N.
Explanation:
It is given that,
Mass of the otter, m = 2 kg
Distance covered by otter, d = 85 cm = 0.85 m
It takes 0.5 seconds.
We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Here, u = 0 (at rest)
[tex]d=\dfrac{1}{2}at^2[/tex]
[tex]a=\dfrac{2d}{t^2}[/tex]
[tex]a=\dfrac{2\times 0.85}{0.5^2}[/tex]
[tex]a=6.8\ m/s^2[/tex]
The net force acting on the otter along the incline is given by :
F = ma
[tex]F=2\ kg\times 6.8\ m/s^2[/tex]
F = 13.6 N
So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.
Answer:
[tex]F=13.6\rm N[/tex] Force acts on the otter along the incline
Explanation:
Given information:
Incline length [tex]s=\rm 85cm=0.85m[/tex]
Time [tex]t=0.5\rm sec[/tex]
Mass of otter [tex]m=2\rm kg[/tex],
Initial velocity [tex]u=0[/tex] as otter is in rest
Muddy incline so we can assume surface as friction less
By use equation of motion,
[tex]s=ut+\frac{1}{2} a t^2[/tex]
[tex]s=ut+\frac{1}{2} a t^2=0\times t+\frac{1}{2}\times a\times t^2[/tex]
[tex]0.85=0+\frac{1}{2} a (0.5)^2\\\\a=\frac{2\times0.85}{0.5^2}=6.8\rm m/s^2[/tex]
The net force act on the otter is
[tex]F=ma[/tex]
[tex]F=2\rm kg\times6.8\rm m/s^2[/tex]
[tex]F=13.6\rm N[/tex]
Hence [tex]F=13.6\rm N[/tex] force acts on the otter along the incline
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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.
Compare the characteristics of an air mass to its source region
Source region: A large area of the earth's surface, where large masses of air originate with uniform temperature and humidity conditions characteristic of the region
In meteorology, an air mass is an air volume determined by its temperature and the amount of water vapour. Air masses span several hundreds or thousands of miles, and conform to the surface properties below them. They are categorized by latitude and by their areas of continental or maritime origin.
Still, from surface effects the air masses themselves are mild. The areas of the globe from which air masses are called source regions. A source area must have certain temperature and humidity properties which can stay constant for a considerable length of time to influence the air masses above it.
A molecule of DNA (deoxyribonucleic acid) is 2.33 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.01% upon becoming charged. Determine the effective spring constant of the molecule.I've seen a couple examples of this problem, including the one already solved on cramster. I tried that method but it didn't work. I got 8.114E-6, then 9.55 E-6, and webassign said that both answers were wrong. I need a fool proof method. One of my calculators said 9.544E-30 but I'm afraid to try it because I only get 10 chances to try an answer and I've tried about 5 times.
To determine the effective spring constant of a molecule of DNA, we can use Hooke's Law and Coulomb's Law. The effective spring constant of the DNA molecule is approximately -1.967 x 10-4 N/m.
To determine the effective spring constant of a molecule of DNA, we can use Hooke's Law, which states that the force required to compress or extend a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, we are given that the DNA molecule compresses 1.01% when it becomes charged, so we can set up the equation:
F = -kx
Where F is the force, k is the spring constant, and x is the displacement. The negative sign indicates that the force and displacement are in opposite directions. We can rearrange this equation to solve for k:
k = -F/x
Since we are given the percentage compression, we can calculate the displacement as a fraction of the original length:
x = (1.01/100) * 2.33 µm = 0.023533 µm
Now, we need to calculate the force. Since the ends of the molecule become singly ionized, one end becomes negatively charged and the other end becomes positively charged. This creates an electric field between the ends, and the molecule experiences an electric force. The magnitude of this force can be calculated using Coulomb's Law:
F = (ke * q1 * q2) / r2
Where F is the force, ke is the electrostatic constant (9.0 x 109 N m2 / C2), q1 and q2 are the charges on the ends of the molecule, and r is the distance between the charges. Since the ends are singly ionized, we can assume equal and opposite charges:
F = (ke * q2) / r2
Now we can substitute the values into the equation:
F = (9.0 x 109 N m2 / C2) * (1.6 x 10-19 C)2 / (0.023533 x 10-6 m)2 = 4.6307 x 10-12 N
Finally, we can substitute the values for force and displacement into the equation for the spring constant:
k = - (4.6307 x 10-12 N) / (0.023533 x 10-6 m) = -1.967 x 10-4 N/m
Therefore, the effective spring constant of the DNA molecule is approximately -1.967 x 10-4 N/m.
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The effective spring constant of the DNA molecule is approximately [tex]\( 9.54 \times 10^{-6} \) N/m.[/tex]
To determine the effective spring constant [tex]\( k \)[/tex] of the DNA molecule, we can use Hooke's Law, which states that the force [tex]\( F \)[/tex] exerted by a spring is proportional to the displacement [tex]\( x \)[/tex] from its equilibrium position, i.e.,[tex]\( F = -kx \)[/tex].
[tex]\[ x = \frac{1.01}{100} \times L = \frac{1.01}{100} \times 2.33 \times 10^{-6} \text{ m} \] \[ x = 2.3533 \times 10^{-8} \text{ m} \][/tex]
Next, we need to calculate the force [tex]\( F \)[/tex] that causes this compression. Since the molecule is ionized, the force can be calculated using Coulomb's Law, which states that the force between two point charges is:
[tex]\[ F = \frac{k_e \cdot q_1 \cdot q_2}{r^2} \][/tex]
Using Hooke's Law, we can express the force as:[tex]\( F \)[/tex]
[tex]\[ F = k \cdot x \][/tex]
We can now solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{F}{x} \][/tex]
Therefore, we can write:
[tex]\[ k = \frac{F}{x} = \frac{k_e \cdot q^2}{x \cdot r^2} \][/tex]
Thus, we have:
[tex]\[ k = \frac{k_e \cdot q^2}{x \cdot L^2} \][/tex]
Since we do not have the values for [tex]\( q \)[/tex], we can assume that the force is such that it causes a 1.01% compression, and we can use the percentage compression to represent the force. This means we can write:
[tex]\[ k = \frac{1.01 \cdot L}{x \cdot L} \] \[ k = \frac{1.01}{x} \][/tex]
Now we can plug in the value for [tex]\( x \)[/tex]:
[tex]\[ k = \frac{1.01}{2.3533 \times 10^{-8} \text{ m}} \] \[ k \approx 9.54 \times 10^{-6} \text{ N/m} \][/tex]
Therefore, the effective spring constant of the DNA molecule is approximately [tex]\( 9.54 \times 10^{-6} \)[/tex] N/m.
How much work, in kJ, can a spring whose spring constant is 4 kN/cm produce after it has been compressed 3 cm from its unloaded length?
Answer:
work done on compressing spring will be 135 j
Explanation:
We have given spring constant [tex]K=3kN/cm=3\times \frac{1000N}{10^{-2}m}=3\times 10^5N/m[/tex] ( As 1 kN = 1000N and 1 m = 100 cm )
Spring is compressed by 3 cm
As 1 m = 100 cm
So [tex]3cm=3\times 10^{-2}m[/tex]
Work done on compressing spring is given by [tex]W=\frac{1}{2}kx^2[/tex]
So [tex]W=\frac{1}{2}\times 3\times 10^{5}\times (3\times 10^{-2})^2=13.5\times 10=135J[/tex]
So work done on compressing spring will be 135 j
The work a spring can produce after being compressed from its unloaded length can be calculated using the formula for potential energy stored in a spring. In this case, given the spring constant of 4 kN/cm and a compression of 3 cm, the spring can produce 18 kJ of work.
Explanation:The work done by a spring, in this case, can be calculated using the formula for the potential energy stored in a compressed or stretched spring, which is given by PE = 1/2 * k * x². Here, k is the spring constant and x is the displacement or compression of the spring from its original length.
Given that the spring constant k = 4 kN/cm = 4000 N/cm and displacement x = 3 cm, we can substitute these values into the formula: PE = 0.5 * 4000 * (3)² = 18,000 N.cm = 18 kJ. The spring can therefore produce 18 kJ of work when it has been compressed 3 cm from its unloaded length.
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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 bullet is fired straight up from a gun with amuzzle velocity of 208 m/s.Neglecting air resistance, what will be itsdisplacement after 8.9 s?
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 208 m/s
Acceleration, a = -9.81 m/s²
Time, t = 8.9 s
Substituting
s = ut + 0.5 at²
s = 208 x 8.9 + 0.5 x -9.81 x 8.9²
s = 1462.68 m
Displacement after 8.9 seconds is 1462.68 m
The wavelength of a wave is the distance between A. the amplitude and the normal position. B. two consecutive crests. C. the crest and the following trough. D. the rarefaction and the following compression.
Answer: two consecutive crests.
Explanation: A wave length is the distance from crest to crest or from a trough to another. The crest represent the highest maximum point and the trough represent the lowest point on the wave.
All waves undergoes some properties such as refraction, distraction,reflection and interference.
Pacman the arcade game went out of order at fat max's arcade. Max is dragging it 5 meters across the ground and out of the store with a rope at a 45 degree angle. What is the force be exerts if he performs 3,500 joules of work
The magnitude of the force is 990 N
Explanation:
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem, we have the following:
W = 3500 J (work done by Max)
d = 5 m (displacement)
[tex]\theta=45^{\circ}[/tex] (direction at which the force is applied)
Solving for F, we find the magnitude of the force:
[tex]F=\frac{W}{dcos \theta}=\frac{3500}{(5)(cos 45^{\circ})}=990 N[/tex]
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To find the force Max exerts while moving the Pacman arcade game, use the formula Work = Force x Distance x cos(θ). Solve for force, giving Force = Work / (Distance x cos(θ)). Substituting the given values into the formula will provide the answer.
Explanation:The work done can be used in the equation that relates work, force, and displacement to identify the force exerted by Max. The relationship between these concepts is expressed as: Work = Force x Distance x cos(θ), where 'θ' is the angle at which the force is applied.
In this scenario, the work done (W) is 3,500 joules, the distance (d) is 5 meters, the angle (θ) is 45 degrees, and the force (F) is what we're trying to find.
Rearranging the formula to solve for Force, we get: Force = Work / (Distance x cos(θ)).
Substituting the given values and doing the math, Max's exerted force will be calculated. The cos(45) is approximately 0.7071, and substituting this in our formula might aid us in accurately finding the force exerted by Max while moving the Pacman arcade game.
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Consider the indicated events in the history of the universe that have helped make human life possible. Rank the events based on when they occurred, from longest ago to most recent. To rank items as equivalent, overlap them. Note: If two events occurred within seconds of each other, rank them as equivalent.a- the Big Bang & the universe begins to expandb- elements such as carbon and oxygen first existc- nuclear fusion begins in the Sund- earliest life on Earthe- dinosaurs go extinctf- earliest humans
Answer:
- the Big Bang=the universe begins to expand.
-elements such as carbon and oxygen first exist.
-nuclear fusion begins in the Sun.
-earliest life on Earth.
-dinosaurs go extinct.
-earliest humans.
Explanation:
There are several events in history that changed the universe and human life. The events led to the most famous scientific discovery, evolution, technological development among others. The universe and human life have drastically changed due to the occurrence of the events such as the Big Bang and the presence of elements.
Answer:
A. The big bang and the universe begins to expand.
B. elements such as oxygen and carbon first exist
C .nuclear fusion begins in the sun
D. Earliest life on earth
E. Dinosaurs go extinct.
F. Earliest humans
Explanation:
-The big bang suggests that our universe was born 13.7 billion years ago in a massive expansion that exploded the space like a gigantic balloon.
- carbon and oxygen were as a result of nuclear fusion that we call stars. They consume their hydrogen, helium and lithium to produce heavy element when they're with a bang to spread element of life around the universe.
-when the central temprature of the sun reached 10 million degrees, nuclear burning of hydrogen into helium commenced after series of protosuns collapsed indirectly raising the central temperature.
- The cretaceous paleogene(k- Pg) extinction event occurrence about 66 million years ago which was sudden mass extinction of ¾ of plant and animals on the planet.
- The earliest member of the genus Homo which evolved around 2.8 million years ago.
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
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.
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.
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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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
A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) north has the sailboat traveled in 35 min?
Explanation:
Let east be positive x axis and north be positive y axis
Velocity of boat = 2.8 m/s in a direction 52° north of west.
Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s
Time taken = 35 min = 35 x 60 = 2100 s
Displacement = Velocity x Time
Displacement = (-1.72 i + 2.21 j) x 2100
Displacement = -3612 i + 4633.50 j m
Displacement along west = 3612 m
Displacement along north = 4633.50 m
Final answer:
To find the westward and northward distances traveled by the sailboat, we decompose the boat's velocity into westward and northward components and then multiply each component by the travel time of 2100 seconds (35 minutes). Calculations involving trigonometry such as cosine for the westward component and sine for the northward component will yield the respective distances.
Explanation:
The question asks for the distance traveled westward and northward by a sailboat that runs before the wind with a constant speed of 2.8 m/s, heading 52° north of west. To solve this, we can decompose the sailboat's velocity into its westward and northward components using trigonometry. The total time traveled is 35 minutes, which is equivalent to 2100 seconds (35 min x 60 s/min).
Westward Component (a)
The westward component of the velocity can be found using the cosine function:
Vwest = V * cos(θ)
Where V is the speed of the boat, and θ is the angle north of west. Plugging in the given values:
Vwest = 2.8 m/s * cos(52°)
The distance traveled westward is the westward component of the velocity multiplied by the time:
Distancewest = Vwest * time
Northward Component (b)
Similarly, the northward component of the velocity is:
Vnorth = V * sin(θ)
The distance traveled northward is:
Distancenorth = Vnorth * time
By calculating these components, we can determine how far the sailboat has traveled in both the westward and northward directions.
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
Each day, Ted can wax 8 cars or wash 10 cars, and Tom can wax 3 cars or wash 5 cars. What is each man's opportunity cost of washing a car?
Answer:
1/2
Explanation:
This will lead us to simultaneous equation since there are two person involved doing the same job each day.
For Ted;
Since Ted can wax 8 cars or wash 10 cars in a day. Mathematically, we have
8x + 10w = 1... (1)
Where x is for waxing cars, w is for washing cars
Similarly for Tom,
Tom can wax 3 cars or wash 5 cars in a day as well, this gives us;
3x + 5w = 1... (2)
Equating 1 and 2, we have;
8x + 10w = 1... (1)
3x + 5w = 1... (2)
Using elimination method, we will multiply eqn 1 by 3 and eqn 2 by 8 to have;
24x + 30w = 3
24x + 40w = 8
Subtracting eqn 4 from 5 we have;
30w - 40w = 3 - 8
-10w = -5
w = 5/10
w = 1/2
Therefore each man's opportunity cost of washing a car is 1/2
Almost every beginning physics course will teach you that the force on an object due to gravity is F = mg, where g is just the acceleration due to gravity at the surface of the Earth. In other words, this equation is just a specific example of Newton’s second law. The mass of the Earth is M⊕ and the radius of the Earth is R⊕. Write down an expression for g in terms of M⊕ and R⊕.
Answer: g = GM0/R0
Explanation:
Second law of Newton (gravitational law) postulates that
The gravitational force on a body on the earth is
i) directly proportional to the mass of earth M0
ii) directly proportional to the mass of the object m
iii) inversely proportional to the raduis R0 of earth.
The gravitational constant G is the proportional constant linking all of these parameters
F = GM0m/R0
But F = mg
Where F is the weight (i.e gravitational force on the object)
m is the mass of the object and
g is the acceleration due to gravity
Hence mg = GM0m/R0
m is cancelled as it exists on
both sides
GM0/R0 = g
Therefore
g = GM0/R0
A 20 g bullet is fired horizontally with a velocity of 250 m/s from a 1.5 kg rifle. If the rifle was held loosely, what would be the recoil velocity?
Answer:
3.33 m/s
Explanation:
given,
mass of bullet, m = 20 g = 0.02 Kg
speed of bullet, v = 250 m/s
mass of rifle, M = 1.5 kg
speed of recoil, u = ?
initial speed of the bullet and the rifle is zero.
using conservation of momentum
(M + m) V = m v + M u
(M + m) x 0 = 0.02 x 250 + 1.5 x u
1.5 u = -5
u = -3.33 m/s
negative sign represent that the recoil velocity is opposite to bullet velocity.
Hence, the recoil velocity is equal to 3.33 m/s
A technique in MRA where signal intensity depends on the direction of flow and thus requires gradient application in all three planes for proper signal acquisition:___________
Answer:
A technique in MRA where signal intensity depends on the direction of flow and thus requires gradient application in all three planes for proper signal acquisition is: Phase-contrast (PC-MRA)
Phase-contrast magnetic resonance imaging MRA is a technique used in moving blood flow, where its speed is encoded in the magnetic resonance signal's phase after the applying bipolar gradient along any axis and the measurement point.
Explanation:
In this technique the bipolar gradient is manipulated varying its magnetic fields to be preset to a maximum expected flow velocity and it´s applied along any axis or axes depending on the direction along which flow is to be measured to get a reversed image of the bipolar gradient and the difference of the two images is calculated. The unaffected phase accrued during the application of the gradient, is 0 for stationary spins. Since phase-contrast can only acquire flow in one direction at a time, 3 separate image acquisitions in all three directions must be computed to give a complete quantitative measurements of blood flow.
Although this technique is slow,its strength lays in the possibility of calculating spins moving with a constant velocity of the applied bipolar gradient. The accrued phase is proportional to both and the 1st moment of the bipolar gradient, thus providing a means to estimate gamma is the Larmor frequency of the imaged spins on moving tissues such as blood, which acquire a different phase since it moves constantly through the gradient, thus also giving its speed of the flow.
A traveling wave is described by the following function y=0.12 cos (4x +2t). Here y(x,t) is the displacement of the particle at location x at time t. All quantities are in SI units. Determine the speed and the direction of the wave.
a. 2 m/s along+x direction.
b. 0.5 m/s along +x direction.
c. 2 m/s along -x direction.
d. 0.5 m/s along - direction.
e. 2 m/s along +y direction.
f. 0.5 m/s along -y direction.
Answer:
d. 0.5 m/s along -x direction
Explanation:
Wave: A wave is a disturbance, that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself.
The general equation of a traveling wave can be expressed as
y = Acos(2πft-2πx/λ).................................. Equation 1
Where A = amplitude of the wave, f = frequency of the wave, λ = wavelength of the wave, x = linear distance, t = time, π = pie.
From the question,
the equation of the moving wave is
y = 0.12cos(4x+2t) ................................... equation 2
Comparing equation 1 and 2
-2πx/λ = 4x
λ = -2π/4
λ = -2(3.14)/4
λ = -1.57 m.
Also,
2πft = 2t
f = 2t/2πft
f = 1/π
f = 1/3.14
f = 0.3185 Hz.
Recall that
v = λf.......................... Equation 3
Substitute the value of f and λ into equation 3
v = -1.57(0.3185)
v = - 0.5 m/s.
Note: v is negative because - x direction
Hence the right option is d. 0.5 m/s along -x direction
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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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.