Answer:
D1
Explanation:
Higher diffusion constants mean that the substance diffuses more in a given time. Because D1>D2, D1 should diffuse more.
Each year 500 runners run up the stairs to the 86th floor of the Empire State Building in New York City. There are 1576 steps and each step is 0.241 m high. In 2003, Australian Paul Crake (20-29 age group) set the overall record by reaching the 86th floor in 9:33. His mass was 70.0 kg. Question 2 HomeworkUnanswered What was Paul Crake's power output during this climb
Answer:
The power is 465.44 W.
Explanation:
mass, m = 70 kg
number of steps, n = 1576
height of each step, h = 0.241 m
time taken, t = 9.33 min= 9.33 x 60 s
The power is given by the rate of doing work.
W = n m g h
W = 1576 x 70 x 9.8 x 0.241
W = 260553.776 J
The power is given by
[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]
A set of data was collected measuring the length of a sheet of steel using three different instruments. The actual length of the sheet is 2.00 m. Which instrument is the most accurate in its measurements
Accuracy of measurement is the degree of how close the measurement is, to the actual measurement.
For a measurement of 2.00 m, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m.
The set of data is not given. So, I will give a general explanation.
For a measurement to be the accurate, the measurement must be close to the actual measurement, and in some cases, the measurement should be approximated to the actual measurement.
Given that:
[tex]Length =2.00m[/tex]
An example of an accurate measurement is:
[tex]Length =1.95m[/tex]
Why? The reasons are
1.95 m is close to 2.00 m1.95 m is approximately 2.00 mAnother example is:
[tex]Length = 1.99m[/tex]
1.99 m is also an accurate measurement because of the reasons above.
However, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m
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When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
A jogger travels a route that has two parts. The first is a displacement ->A of 2.20 km due south, and the second involves a displacement ->B that points due east.
(a) The resultant displacement ->A + ->B has a magnitude of 3.81 km.
What is the magnitude of B?
______ km
What is the direction of A + B relative to due south?
_____° west of south or east of south?
(b) Suppose that A - B had a magnitude of 3.81 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?
Answer:
a) B = 3.11 km. θ= 54.7º E of S
b) B = 3.11 km θ= 54.7º W of S
Explanation:
a)
Since we know the value of the total displacement, and the value of the displacement A and its direction, we can find the magnitude of B just applying the Pythagorean Theorem, as follows:[tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]
Solving for B, the only unknown, we get:[tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]
Now, applying some simple trig, we can find the angle that (A+B) makes with the S axis, as follows:[tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]
Since it's a positive number, applying the convention that the positive angles are measured counterclockwise, this means that this angle is measured East of South.b)
If the magnitude of A-B is the same than the one for A+B, this means that the magnitude of B remains the same, i.e. 3.11 Km.But if we do graphically A-B, as it is the same as adding A + (-B), we find that the angle of A-B is different to the one in A+B, even the magnitudes of both displacements are the same.In this case, B is a negative number, because it's a displacement due west.So, applying the same trig that for a) we can find the angle that (A-B) makes with the S Axis, as follows:[tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]
So, since it is negative, it's measured clockwise from the S axis, so it's 54.7º W of S.Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2 < r1) and whose turns per unit length are n1 and n2.
Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
a.
70. All the following statements are true. Which one can be explained by Kepler's Second Law?
The Sun is not in the precise center of Saturn's orbit.
b. Earth is sightly closer to the Sun in January than in July
All the planets orbit the Sun in nearly the same plane.
d. Mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun.
e. Venus orbits the Sun at a faster orbital speed than Earth.
The statement that can be explained using Kepler's Second Law is "Mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun."
According to Encyclopedia Britannica, Kepler's second law states that; "a planet moves in its ellipse so that the line between it and the Sun placed at a focus sweeps out equal areas in equal times."
This law is sometimes referred to as the law of equal areas. It describes the fact that a planet moves faster when it is closer to the sun and slower when it farther from the sun.
An implication of this law is that mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun.
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Two long straight wires lie parallel to each other 2.00 cm apart, and carry equal and opposite currents of 145 A. What is the strength of the magnetic field they produce, halfway between them
Answer:
B = 2.9 mT
Explanation:
Given that,
The distance between the wires, r = 2 cm
Distance halfway between them d = 1 cm = 0.01 m
The current in the wires, I = 145 A
We need to find the strength of the magnetic field they produce, halfway between them. The formula for the magnetic field in the wire is given by :
[tex]B=\dfrac{\mu_o I}{2\pi d}\\\\B=\dfrac{4\pi \times 10^{-7}\times 145}{2\pi \times 0.01}\\\\B=2.9\ mT[/tex]
So, the required magnetic field is 2.9 mT.
An ideal heat engine operates between two
temperatures 367.7 K and 865.1 K. What is
the efficiency of the engine?
0.787
0.42504
None
0.213
0.5750
Answer:
As it is the most efficient heat engine, it's efficiency is [T1 - T2]/T1. It can be measured for every Carnot cycle. From the formula and diagram, we can understand that the efficiency of an ideal heat engine also depends on the difference between the hot and cold reservoirs.
Explanation:
5. How much heat is generated when an
electrical coil rated 200W is used to heat a
given mass of water for 1hour
Answer:
I HOPE THIS IS CORRECT
Explanation:
Power of water =2 kw=2000w
Mass of water =200kg
difference in temperature ΔT=70−10=60oC
Concept
energy required to heat the water = energy given by water in time t=pt
energy required to increase tempeature of water by 60oC,Q=msΔT
S= specific heat =4200J/kgoC
pt=msΔT
2000×t=200×4200×60
t=25200
or t=25.2×103sec.
The quantity of heat generated is 720,000 Joules or 720 kiloJoules.
Given the following data:
Time = 1 hourPower = 200 WattsTo find the quantity of heat generated when the electrical coil is used:
Note: the quantity of heat generated is energy.
First of all, we would convert the time in hour to seconds.
Conversion:
1 hour = 3600 seconds
Mathematically, the energy of a device is given by the formula;
[tex]Energy = power[/tex] × [tex]time[/tex]
Substituting the values into the formula, we have;
[tex]Energy = 200[/tex] × [tex]3600[/tex]
Energy = 720,000 Joules or 720 kiloJoules.
Therefore, the quantity of heat generated is 720,000 Joules or 720 kiloJoules.
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who is the biggest man in the world
Answer:
Sultan Kösen
here is a pic
A 1540-kg truck has a wheel base of 3.13 m (this is the distance between the front and rear axles). The center of mass of the truck is 1.09 m behind the front axle.a) What is the force exerted by the ground on each of the front wheels?b) What is the force exerted by the ground on each of the back wheels? Hint: Remember that the truck has four wheels.
a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.
b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.
Given the following data:
Mass of truck = 1540 kgDistance between the front and rear axles = 3.13 meters.Center of mass of the truck = 1.09 meters.a. To determine the force exerted by the ground on each of the front wheels:
First of all, we would take moment about the rear wheels.
[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]
Force, F = 9836.32 Newton
For each front wheel:
[tex]Force = \frac{9836.32}{2}[/tex]
Force = 4918.16 Newton.
b. To determine the force exerted by the ground on each of the back wheels:
We would determine the sum of the vertical forces acting on the wheels.
[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]
B = 5255.68 Newton.
For each back wheel:
[tex]Force = \frac{5255.68}{2}[/tex]
Force = 2627.84 Newton.
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Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
three people are standing by maintaining social distancing i.e 6 feet from each other. A doctor checks one person first then go to the other standing towards the east and then third person towards north. what will be the resultant displacement covered the doctor?
Answer:
8.5 feet.
Explanation:
A sketch of the position of the three people gives a right angled triangle. The hypotenuse of the triangle gives the resultant displacement, while the two other sides are 6 feet each.
The resultant displacement, R, is the overall displacement covered by the doctor. This can be determined by;
R = [tex]\sqrt{d_{1} ^{2} + d_{2} ^{2} }[/tex]
where: [tex]d_{1}[/tex] = 6 feet and [tex]d_{2}[/tex] = 6 feet.
Thus,
R = [tex]\sqrt{6^{2} + 6^{2} }[/tex]
= [tex]\sqrt{72}[/tex]
R = 8.49
The resultant displacement covered by the doctor is 8.5 feet.
Assume you are going to race the three objects (hollow sphere, disk and ring used in Experiment 8.2) by releasing them from rest at the top of an inclined plane.Which object do you expect to reach the bottom of the inclined plan first? Why?
Answer:
a. The disk
b. Because it has the smallest rotational inertia
Explanation:
a. Which object do you expect to reach the bottom of the inclined plan first?
I would expect the disk to reach the bottom first.
b. Why?
This is because the disk has the smallest rotational inertia.
The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.
Since the three objects are rolling from the same height, they have the same mechanical energy.
But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.
The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
What is a moment of inertia?Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
There are three objects, hollow sphere, disk and ring.
The moment of inertia of the hollow sphere object is given as,[tex]I=\dfrac{2}{3}mr^2[/tex]
The moment of inertia of the ring is,[tex]I=mr^2[/tex]
The moment of inertia of the disk is,[tex]I=\dfrac{1}{2}mr^2[/tex]Here, (m) is the mass and (r) is the radius of the object.
These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.
As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.
Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
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what is entrapersonal environment
Answer:
The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization
Explanation:
. Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/s, and the mass of rain per second striking the roof is 0.071 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof.
Answer:
the average force exerted by the rain on the roof is 1.065 N
Explanation:
Given;
speed of the rainfall, v = 15 m/s
the mass of the rate of the rainfall, m' (m/t) = 0.071 kg
Let the average force exerted by the rain on the roof = F
The average force exerted by the rain on the roof is calculated by Newton's second law of motion;
[tex]F = ma = m \frac{v}{t} = \frac{m}{t} \times v \\\\Recall, \ m' = \frac{m}{t} \\\\F = m'\times v\\\\F = 0.071 \times 15\\\\F = 1.065 \ N[/tex]
Therefore, the average force exerted by the rain on the roof is 1.065 N
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
It's nighttime, and you've dropped your goggles into a 3m deep swimming pool. If you hold a laser pointer 1.0m abovve the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0m from the edge. How far are the goggles from the edge of the pool?
Answer:
x_total = 3.07 m
Explanation:
Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.
Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry
tan θ₁ = 1.0 /2.0
θ₁ = tan⁻¹ 0.50
θ₁ = 26.57º
now let's use the law of refraction to find the angle of refraction (in water)
n₁ sin θ₁ = n₂ sin θ₂
the refractive index of air is n₁ = 1 and that of water n₂ = 1.33
sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]
sin θ₂ = [tex]\frac{1}{1.33} \ sin \ 26.57[/tex]
θ₂ = sin⁻¹ 0.3363
θ₂ = 19.65º
now we can find the distance from the entry point to the water to the lenses
tan θ₂ = x₂ / 3
x₂ = 3 tan θ₂
x₂ = 3 tan 19.65
x₂ = 1.07 m
the total distance from the edge of the pool is
x_total = 2 + x₂
x_total = 2 + 1.07
x_total = 3.07 m
Help please help I don’t want to fail please
Answer:
B
Explanation:
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if the system it is rotated about (a) one end of the rod and (b) the center of the rod?
Explanation:
a) [tex]I=\displaystyle \sum_{i}m_ir_i^2[/tex]
where [tex]r_i[/tex] is the distance of the mass [tex]m_i[/tex] from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,
[tex]I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2[/tex]
b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,
[tex]\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2[/tex]
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 2.12 m/s and observes that it takes 1.5 s to reach the water.
(a) List the knowns in this problem.
(b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave
Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.
Answer:
O The environment did work on an object
Explanation:
A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V battery. Draw a labeled circuit diagram for the circuit described using correct symbols. Calculate the equivalent resistance. Calculate the voltage drop across each resistor in the circuit.
(a) Attached to the response as Figure 1.
(b) 35.0Ω
(c) Across 5.0Ω = 1.3V
Across 10.0Ω = 2.6Ω
Across 20.0Ω = 5.2Ω
Explanation:(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.
(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e
R = R₁ + R₂ + R₃ -------------------(i)
Where;
R₁ = 5.0 Ω
R₂ = 10.0 Ω
R₃ = 20.0 Ω
Substitute these values into equation (i) as follows;
∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω
∴ R = 35.0 Ω
Therefore, the equivalent resistance is ∴ R = 35.0Ω
(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;
i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.
ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;
V = I R ------------------(ii)
Where;
V = Voltage supplied to the circuit = 9.0V
I = Current through the circuit
R = Resistance of the equivalent resistor = 35.0Ω
Substitute these values into equation (ii)
9.0 = I x 35.0
I = [tex]\frac{9.0}{35.0}[/tex]
I = 0.26A
This is also the current flowing through each of the resistors separately.
iii. Calculate the voltage drop across
1. 5.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 5.0Ω resistor
I = current through the 5.0Ω resistor = 0.26A
R = resistance of the 5.0Ω resistor = 5.0Ω
=> V = 0.26 x 5.0
=> V = 1.3V
2. 10.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 10.0Ω resistor
I = current through the 10.0Ω resistor = 0.26A
R = resistance of the 10.0Ω resistor = 10.0Ω
=> V = 0.26 x 10.0
=> V = 2.6V
3. 20.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 20.0Ω resistor
I = current through the 20.0Ω resistor = 0.26A
R = resistance of the 20.0Ω resistor = 10.0Ω
=> V = 0.26 x 20.0
=> V = 5.2V
During which radioactive decay process does a neutron change into a proton?
A. Alpha decay
B. Gamma decay
C. Beta decay (positron)
D. Beta decay (electron)
A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.
What is beta decay?A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.
A positron is an electron with a positive charge.
During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.
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In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
[tex]2gh = v_f^2 - v_i^2[/tex]
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]
h = 0.82 m
Now, for the time in air during upward motion we use first equation of motion:
[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]
(c)
Now we will consider the downward motion and use the third equation of motion:
[tex]2gh = v_f^2-v_i^2[/tex]
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]
vf = 7.17 m/s
Now, for the time in air during downward motion we use the first equation of motion:
[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
t = 1.14 s
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
