Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of Q is located at one of the corners of the square. What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of Q
Answer:
12.0 V
Explanation:
Data :
Potential difference due to a single charge (+Q), E = 3.0 V
The Electric potential for the system of charges is given as:
[tex]E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}][/tex]
for single charge, E = 3.0 V = [tex]\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}][/tex] ->eq(1)
And for 4 charges:
[tex]E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}][/tex] -eq(2)
from eq(1) and (2) we have
E = 4 × 3.0 V = 12 V
URGENT : Which of the following is the most stable isotope? Explain.
Answer:
Plutonium–238
Explanation:
The stability of isotopes is largely dependent on their half-life.
Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.
From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.
We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.
Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.
Which statement best describes one way that the molecules differ from atoms? a. A molecule can contain a nucleus about which its electrons orbit b. A molecule can contain two atoms of the same element. C. Only a molecule can be the smallest particle of a certain element. d. Only a molecule can be broken down into two or more different elements.
B and D are both true statements. I'm not comfortable saying that either one is better than the other one.
The statement that best describes one way that molecules differ from atoms is a molecule can contain two atoms of the same element, and only a molecule can be broken down into two or more different elements. The correct options are b and d.
What are atoms and molecules?According to science, an atom is the smallest component of an element that can exist freely or not. A molecule, on the other hand, is the smallest component of a chemical and is made up of a group of atoms linked together by a bond.
A molecule is the smallest component of a substance that has the chemical properties of the compound.
The term "independent molecule" is not commonly used to refer to atoms and complexes linked by non-covalent interactions such as hydrogen or ionic bonds. Molecules are common constituents of matter.
Therefore, the correct options are b and d.
To learn more about atoms, refer to the link:
https://brainly.com/question/25617532
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A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?
Explanation:
We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of the pendulum
g is acceleration due to gravity on which it is placed
It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.
A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Explanation:
Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.
5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343
d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.
.0160349 = d / 3 x 10⁸ + .0078717
d / 3 x 10⁸ = .0081632
d = 2448960 m
= 2448.96 km
= 2449 km .
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time
Answer:
Explanation:
The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force
impulse by net force = change in momentum
Net force = R - mg , mg is weight of the man .
( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0
R - mg = 227.81 x 10³
Average net force = 227.81 x 10³ N .
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimetre high hydroelectric dam would be required to charge battery
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
a body with v=20m/s changes its speed to 28m/s in 2sec. its acceleration will be
Answer:
Explanation:
Givens
vi = 20 m/s
vf = 28 m/s
t = 2 seconds
Formula
a = (vf - vi) / t
Solution
a = (28 - 20)/2
a = 8/2
a = 4 m/s^2
The air flowing over the top of the wing travels
in the same amount of time than the air
flowing beneath the wing.
Answer: Short Answer: NO ( In Most Cases)
Explanation:
If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is -0.26s^2. Find its speed at the next signal.
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
You are helping your friend prepare for the next skateboard exhibition by determining if the planned program will work. Your friend will take a running start and then jump onto a heavy-duty 13-lb stationary skateboard. The skateboard will glide in a straight line along a short, level section of track, then up a curved concrete wall. The goal is to reach a height of at least 10 feet above the starting point before coming back down the slope. Your friend's maximum running speed to safely jump on the skateboard is 24 feet/second. Your friend weighs 155-lbs. What is the height hf that your friend will reach according to his plan?
Answer:
8.3 feet
Explanation:
The kinetic energy of the system on the ground is ...
KE = Σ(1/2)(mv^2) = (1/2)(155)(24^2) +(1/2)(13)(0^2) = 44640 lb·ft²/s²
The potential energy at the highest point is the same:
PE = mgh
44640 = (155 +13)(32)h
h = 44640/5376 = 8.30 . . . . feet
_____
We haven't worried too much about the conversion between pounds mass and pounds force. Whatever factor may be involved will divide out when computing the maximum height. We have used g=32 ft/s².
__
To achieve a 10 ft height, the running speed would need to be 26.34 ft/s, about 10% higher.
As you get ready for bed, you roll up one of your socks into a tight ball and toss it into the laundry basket across the room. Then, you try to toss the other sock without rolling it up.. What effects whether or not your socks land in the basket?
Answer:
The drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, the size of the sock (weight), and the speed with which the socks is tossed.
Explanation:
The socks, like every other particle or body travelling through air is met by a resistance that impedes its motion. This resistance is due to the air molecules around, that collide with the body as it travels through them. The resistance offered by this force is proportional to the surface area of the body that collides with the air molecule, so, rolling the socks into a ball reduces the effect of air resistance on the socks, compared to the one tossed without rolling. Air resistance is also largely dependent on the relative motion of the body and the air molecules, the density of the fluid (air), and the size of the body (weight).
Therefore, whether the socks lands in the basket or not is affected by the drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, size of the socks (weight), and the speed with which the socks is tossed.
Drag force opposes motion of objects through fluid with its magnitude depending on the velocity of the object in the fluid
The single parameter that effects whether or not the socks lands in the basket is the drag force, [tex]\mathbf{F_D}[/tex] acting on the socks
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
The reason that drag force is the parameter that effects the landing point of the socks is as follows:
The parameters that effects whether or not the socks land in the basket or not are;
The distance of the basket away from the thrower = The range, RThe velocity with which the socks are thrown, uThe angle of elevation with which each socks is thrown, θThe amount of drag experienced by each socks, [tex]\mathbf{F_D}[/tex]The parameters, R, u, and θ depends on the thrower, that parameter that effects the whether or not the socks lands in the basket that is independent of the thrower, is the drag, [tex]\mathbf{F_D}[/tex]
Drag is the force opposing (slows) the motion of an object in a fluid.
The drag force, [tex]\mathbf{F_D}[/tex], slowing down motion, is given by the following formula;
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
Where;
[tex]v_r[/tex] = The velocity of flow of the fluid, relative to the object
ρ = The density of the fluid
[tex]C_D[/tex] = The drag coefficient
A = The cross sectional area of the fluid
Therefore, the independent parameter that effects whether or not the socks lands in the basket is the drag force on the socks
Learn more about drag force here:
https://brainly.com/question/17074446
Which of the following is NOT true about main sequence stars?
A.) The forces of gravity and nuclear fusion balance so the star does not collapse or explode.
B.) Temperature is directly related to brightness.
C.) The forces of gravity and nuclear fusion are not in balance so the star's core collapses while the outer layers expand.
D.) Temperature is related to size.
Answer:
the statements the B is not true
Explanation:
In the stars the force of gravity tends to collapse them, by joining the atoms nuclear reactions that create an outward force until the two forces reach an equilibrium
The temperature of a star is a reflection of the energy within it and it is related to the intensity of the nuclear rations and not to the size of the stars
In examining the statements the B is not true
What is the goal of the Standing Waves lab? Group of answer choices To determine how frequency changes with mode number. To determine the velocity of a wave traveling on string. To determine wavelength of a wave on a string. To be the very best like no one ever was.
Answer:
To determine wavelength of a wave on a string.
Explanation:
The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.
When you take your 1900-kg car out for a spin, you go around a corner of radius 56 m with a speed of 14 m/s. The coefficient of static friction between the car and the road is 0.88. Part A Assuming your car doesn't skid, what is the force exerted on it by static friction
Answer:
6,650 newtons
Explanation:
The computation of the force exerted on it by static friction is shown below:
Data provided in the question
Mass of car = m = 1,900 kg
speed = v = 14 m/s
radius = r = 56 m
Let us assume friction force be f
And, the Coefficient of friction = [tex]\mu[/tex]= 0.88
As we know that
[tex]f = \frac{mv^2}{r}[/tex]
[tex]= \frac{1,900 \times 14^2}{56}[/tex]
= 6,650 newtons
We simply applied the above formula so that the force exerted could come
Blocks of mass 10, 30, and 90 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 32 N is applied to the left-most block. 1) What is the magnitude of the force that the left block exerts on the middle one
Answer:
32N
Explanation:
The Left force exerts an opposite horizontal force of 32N on the middle object
The Nardo ring is a circular test track for cars. It has a circumference of 12.5km. Cars travel around the track at a constant speed of 100km/h. A car starts at the easternmost point of the ring and drives for 15 minutes at this speed.
1. What distance, in km, does the car travel?
2. What is the magnitude of the car's displacement, in km, from its initial position?
3. What is the speed of the car in m/s?
Answer:
1. 25 Km
2. zero
3. 27.7 m/s
Explanation:
Data provided in the question:
Circumference of the track = 12.5 km
Speed of the car = 100 Km/h
Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr
Now,
1. Distance traveled = Speed × Time
= 100 × [tex]\frac{15}{60}[/tex]
= 25 Km
2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)
Which means that the car is again at the initial position
Therefore, The displacement is zero.
3. Speed of car in Km/hr = 100 Km/h
now,
1 Km = 1000 m
1 hr = 3600 seconds
therefore,
100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s
= 27.7 m/s
Hence, the speed of car in m/s = 27.7
5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities
Answer:
D
Explanation:
Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.
Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.
With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.
Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.
Dacia asks Katarina why it is important to learn a new coordinate system, because they have been using the Cartesian coordinate system and it seems to Dacia that it works fine. Which of Katarina's replies to Dacia are correct?
Answer: A and D
a. Many objects move in arcs of circles or complete circles at times, and polar coordinates allowthe motion of such objects to be comprehended more easily. Some of Newton's laws in certaincases, such as calculation ofg from first principles, are much easier to calculate using polar coordinates.
b. "These coordinates are just used to confuse students.
c. ""Physics teachers are helping math teachers by getting students to practice theirtrigonometry.
d. ""In some cases, such as addition of forces, where a force magnitude is specified, it is simpler todescribe the forces in polar coordinates and be able to convert to the xy representation.
Explanation:
(A). Any force possessing a fixed magnitude and direction, it's always important to describe the for exactly as it is and also be able to convert it from one coordinate representation to another.
(D). Anything which involves radial motion(this is a motion along a radius) or motion along an arc of a circle or ellipse, this kind of motion is best explained and easily understood when in polar coordinates.
How much work is done by 0.30 m of gas if its pressure increases by 8.0 x105 Pa and the volume remains constant
Salerno
Answer:
0
Explanation:
if the volume remains constans, the works is 0 because the equation
W = P . ∆V
P = pressure
∆V = change in volume
Organ pipe A, with both ends open, has a fundamental frequency of 475 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. Use 343 m/s for the speed of sound in air. How long are (a) pipe A and (b) pipe B?
Answer:
The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]
The length of organ pipe B is [tex]L_b = 0.2708 \ m[/tex]
Explanation:
From the question we are told that
The fundamental frequency is [tex]f = 475 Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The fundamental frequency of the organ pipe A is mathematically represented as
[tex]f= \frac{v_s}{2 L}[/tex]
Where L is the length of organ pipe
Now making L the subject
[tex]L = \frac{v_s}{2f}[/tex]
substituting values
[tex]L = \frac{343}{2 *475}[/tex]
[tex]L = 0.3611 \ m[/tex]
The second harmonic frequency of the organ pipe A is mathematically represented as
[tex]f_2 = \frac{v_2}{L}[/tex]
The third harmonic frequency of the organ pipe B is mathematically represented as
[tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]
So from the question
[tex]f_2 = f_3[/tex]
So
[tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]
Making [tex]L_b[/tex] the subject
[tex]L_b = \frac{3}{4} L[/tex]
substituting values
[tex]L_b = \frac{3}{4} (0.3611)[/tex]
[tex]L_b = 0.2708 \ m[/tex]
A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Calculate the state of plane stress at point C located 50 mm below the top of the beam and 0.5 m to the right of point A. Also find the principal stresses and the maximum shear stress at C. Neglect the weight of the beam.
Answer:
Explanation:
Given that:
width b=100mm
depth h=150 mm
length L=2 m =200mm
point load P =500 N
Calculate moment of inertia
[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]
Point C is subjected to bending moment
Calculate the bending moment of point C
M = P x 1.5
= 500 x 1.5
= 750 N.m
M = 750 × 10³ N.mm
Calculate bending stress at point C
[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]
Calculate the first moment of area below point C
[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]
Now calculate shear stress at point C
[tex]=\frac{FQ}{It}[/tex]
[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]
Calculate the principal stress at point C
[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]
Calculate the maximum shear stress at piont C
[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]
An electric point charge of Q = 22.5 nC is placed at the center of a cube with a side length of a = 16.3 cm. The cube in this question is only a mathematical object, it is not made out of any physical material. What is the electric flux through all six sides of the cube?
Answer:
The electric flux is [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
Explanation:
From the question we are told that
The magnitude of the electric point charge [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]
The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]
The number of sides is [tex]N= 6[/tex]
The electric flux according to Gauss law is mathematically evaluated as
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]
[tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
HELPP ?Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volume of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?
Answer:
a. 0.16240664737515434 moles
b. 67.5 degrees Celcius
Explanation:
a. Use Ideal Gas Equation
PV=nRT
Where P = pressure in pascals, V=Volume in cubic meters, n=number of moles, R is a constant=8.314 J/mol.K and T is temperature in Kelvin.
27C = 273+27=300Kelvin
volume 4L = 0.004m^3
Pressure = 1atm = 101325 Pascal
PV=nRT
101325Pa*0.004m^3=n*8.314J/mol.K*300K
Solving for n from the above you get n=0.16240664737515434 moles
b.Use combined gas law equation
P1*V1/T1=P2*V2/T2
P1= 1atm
V1=4L
T1=27C
P2= 4/16 L =0.25L
P=1*40 atm = 40atm
We do not know T2
USING THE FORMULA
(1atm*4L)/27C = (40atm*0.25L)/T2
(1*4)/27=(40*0.25)/T2
IF you simplify for T2, you get 67.5
Hence final temperature = 67.5 degrees Celcius
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as measured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?
A.
inverse plot, 0.45 kilograms/meter3
B.
inverse plot, 0.51 kilograms/meter3
C.
quadratic plot, 1.05 kilograms/meter3
D.
inverse plot, 1.23 kilograms/meter3
E.
inverse plot, 0.95 kilograms/meter3
' A ' looks like the best choice.
Answer:
B. inverse plot, 0.51 kilograms/meter3
Explanation:
Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
An organism has 20 chromosomes after fertilization.after meiosis, how many chromosomes would each sex cell have?
Answer:
EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n
means haploid parent cells join or fuse to form diploid zygote
Answer:
10
Explanation:
When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir
(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Answer:
a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.
Explanation:
a) The change in entropy of the hot reservoir is:
[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]
[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]
b) The change in entropy of the cold reservoir is:
[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]
[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]
c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:
[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]
[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]
[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]
[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]
d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 m on a side, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. Part A How much thermal energy is added to the air by the drag force
Answer:
13.9 kJ
Explanation:
Given that
Length of the side, l = 1.8 m
Drag coefficient, C(d) = 1.4
Distance of run, d = 200 m
Velocity of run, v = 5 m/s
Density, ρ = 1.23
Using the Aerodynamics Drag Force formula. We have
F(d) = 1/2.ρ.A.C(d).v²
The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,
F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²
F(d) = 139.482/2
F(d) = 69.74
recall that, energy =
W = F * d
W = 69.74 * 200
W= 13948
W = 13.9kJ
Therefore, the thermal energy added to the air by the drag force is 13.9kJ
