Complete Question
Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule like [tex]N_2[/tex]?
Answer:
The average volume per molecule is
[tex]\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule[/tex]
The average distance between molecules
[tex]d = 3.45 *10^{-9} \ m[/tex]
The size of [tex]N_2[/tex] is 100 times smaller than the obtained value
Explanation:
From the question we can deduce that we are considering an ideal
Generally the ideal gas equation is mathematically represented as
[tex]PV = NkT[/tex]
Here T is the room temperature with value T = 300 \ K
k is the Boltzmann constant with value [tex]k = 1.38 *10^{-23} \ J/K[/tex]
P is the atmospheric pressure with value [tex]P = 1.0 *10^{5} \ N/m^2[/tex]
N is the number of molecules
Now the volume per molecule is mathematically deduced from the above equation as
[tex]\frac{V}{N} = \frac{kT}{P}[/tex]
=> [tex]\frac{V}{N} = \frac{ 1.381 *10^{-23} * 300}{ 1.0*10^{5}}[/tex]
=> [tex]\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule[/tex]
Now the distance is mathematically evaluated as
[tex]d = \sqrt{\frac{V}{N} }[/tex]
=> [tex]d = \sqrt[3]{4.09*10^{-26}}[/tex]
=> [tex]d = 3.45 *10^{-9} \ m[/tex]
Generally the size of [tex]N_2[/tex] is 115 pm which is 100 times smaller than the obtained value
Generally the size of [tex]H_2O[/tex] is [tex]95.84 \ pm[/tex] which is 10 times smaller than the obtained value
Based on the ideal gas equation, the calculated values are as follows:
the average volume per molecule is 4.09 × 10^-23 m^3/moleculethe average distance between molecules is 3.45 nmthe N2 molecule is about 30 times smaller than the average distance between molecules.How can the average volume per molecule be calculated?The average volume per molecule is calculated using the given formula derived from the ideal gas equation:
V/n = kT/Pwhere:
V = gas volumen = number of molesk is Boltzmann constant = 1.38 × 10^-23 J/KT is temperature, and P is pressureAt room temperature and atmospheric pressure;
T = 300 k
P = 1.0 × 10^5 N/m^2
Substituting the values in the equation above to calculate the average volume per molecule, V/n:
V/n = 1.38 × 10^-23 × 300/1.0 × 10^5 N/m^2
V/n = 4.09 × 10^-23 m^3/molecule
Thus, the average volume per molecule is 4.09 × 10^-23 m^3/molecule
How can the average distance between molecules be determined?The average distance between molecules can be determined by using the formula:
[tex]d = \sqrt[3]{ \frac{v}{n} } [/tex]
Substituting the values for V/n
[tex]d = \sqrt[3]{4.09 \times {10}^{ - 23} } [/tex]
d = 3.45 × 10^-9 m or 3.45 nm
Therefore, the average distance between molecules is 3.45 nm
A molecule of N2 has an average size of 115 pm or 0.115 nm.
Comparing the two values:
3.45/0.115 = 30
Therefore, the N2 molecule is about 30 times smaller than the average distance between molecules.
Learn more about distance between molecules ans Ideal gas equation at: https://brainly.com/question/14375674
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A hollow conductor carries a net charge of ++3QQ. A small charge of −−2QQ is placed inside the cavity in such a way that it is isolated from the conductor. How much charge is on the outer surface of the conductor?
Answer:
The value is [tex]q_o = Q[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = + 3Q[/tex]
The charge place on the inside of the cavity is [tex]q = -2Q[/tex]
Since we are told from the question that small charge placed inside the cavity is isolated from the conductor
Then it implies that the electric flux is Zero
Which mean that the charge place within the conductor + the charge on the inner region of the conductor = 0
i.e
[tex]q + q_i = 0[/tex]
=> [tex]-2Q + q_i = 0[/tex]
=> [tex]q_i = 2Q[/tex]
Now the net charge on the conductor is mathematically represented as
[tex]Q = q_o + q_i[/tex]
Here [tex]q_o[/tex] is the charge on the outer surface
So
[tex]3Q = q_o + 2Q[/tex]
=> [tex]q_o = Q[/tex]
A hockey puck moves 26 meters northward, then 12 meters southward, and finally 6 meters
northward
For this motion, what is the distance moved?
What is the magnitude and direction of the displacement?
Answer:
The distance moved is 44 metres.
The magnitude of displacement is 20 metres with northward direction.
Answer:
44 m.
North 20 m.
Explanation:
Distance moved = 26 + 12 + 6
= 44 m.
Magnitude of the displacement = 26 - 12 + 6
= 20m
Direction is Northward.
The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?
Answer:
The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems
A beam of helium-3 atoms (m = 3.016 u) is incident on a target of nitrogen-14 atoms (m = 14.003 u) at rest. During the collision, a proton from the helium-3 nucleus passes to the nitrogen nucleus, so that following the collision there are two atoms: an atom of "heavy hydrogen" (deuterium, m = 2.014 u) and an atom of oxygen-15 (m = 15.003 u). The incident helium atoms are moving at a velocity of 6.346 x 10° m/s. After the collision, the deuterium atoms are observed to be moving forward (in the same direction as the initial helium atoms) with a velocity of 1.531 x 107 m/s.A) What is the final velocity of the oxygen-15 atoms? B) Compare the total kinetic energies before and after the collision.
Answer:
Explanation:
We shall apply law of conservation of momentum to solve the problem .
Helium-3 collides with nitrogen-14 at rest . After the collision the newly formed deuterium atom and oxygen-15 atom moves .
momentum before the collision
= 3.016 x 6.346 x 10⁶ + 14.003 x 0 = 19.14 x 10⁶ unit
momentum after collision
2.014 x 1.531 x 10⁷ + 15.003 V
3.083 x 10⁷ + 15.003 V units
Applying the law of conservation of momentum ,
19.14 x 10⁶ = 3.083 x 10⁷ + 15.003 V
1.914 x 10⁷ = 3.083 x 10⁷ + 15.003 V
15.003 V = - 1.169 x 10⁷
V = .077917 x 10⁷
= 7.79 x 10⁵ m /s
= .0779 x 10⁷ m /s
mass of helium atom = 3.016 u = 3.016 x 1.67 x 10⁻²⁷ kg
velocity = 6.346 x 10⁶ m /s
kinetic energy = 1 /2 x 3.016 x 1.67 x 10⁻²⁷ x (6.346 x 10⁶ )²
= 101.42 x 10⁻¹⁵ J
kinetic energy of nitrogen atoms = 0
Total energy before collision = 101.42 x 10⁻¹⁵ J
Similarly kinetic energy after collision
= 1 /2 x [ 2.014 x 1.531² + 15.003 x .0779² ] x 1.67 x 10⁻²⁷ x 10¹⁴
= .835 x [ 4.72 + .09 ] x 10⁻¹³ J
= 4.016 x 10⁻¹³ J
= 401.6 x 10⁻¹⁵ J
value of kinetic energy is increased .
What is the average power output (in W) of a heart defibrillator that dissipates 435 J of energy in 10.5 ms?
Answer:
The power is [tex]P = 41429 \ W[/tex]
Explanation:
From the question we are told that
The energy is [tex]E = 435 \ J[/tex]
The time taken is [tex]t = 10.5 \ ms = 10.5 *10^{-3} \ s[/tex]
Generally the the average power is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
=> [tex]P = \frac{ 435}{ 10.5*10^{-3}}[/tex]
=> [tex]P = 41429 \ W[/tex]
Which observation is the most objective? a My frog died after 3 days in the aquarium. I will miss him. b My frog died after three days in the aquarium. We will test the temperature and water conditions to find out why. c Frogs tend to die in captivity. Ours did after three days.
Answer:
b
Explanation:
a looks so silly
(a) A narrow beam of light containing yellow (580 nm) and green (550 nm) wavelengths goes from polystyrene to air, striking the surface at a 30.0°30.0° incident angle. What is the angle between the colors when they emerge?
Answer:
0.043°
Explanation:
Snell's law States that the ratio of the angle of incidence to angle of refraction is a constant for a given pair of media.
n = sin(i)/sin(r)
n1/n2 = sin(i)/sin(r)
For the green polystyrene
n2 = refractive index of green polystyrene = 1.493
n1 = refractive index of air = 1
1/1.493= sin30°/sin(r1)
sin(r1) = sin30°×1.493
sin(r1) = 0.5×1.493
sin(r1) = 0.7465
r1 = sin^-1(0.7465)
r1 = 48.288°
For the yellow polystyrene
n2 = refractive index of yellow polystyrene = 1.492
n1 = refractive index of air = 1
1.492/1= sin30°/sin(r2)
sin(r2) = sin30°×1.492
sin(r2) = 0.5×1.492
sin(r2) = 0.746
r2 = sin^-1(0.746)
r2 = 48.245°
The angle between the colors when they emerge = r1-r2
angle between the colors when they emerge = 48.288°-48.245°
angle between the colors when they emerge = 0.043°
give me an example of orderliness In nature
Answer:
Here are some examples of orderliness In nature
1. The proposal that the order of nature showed evidence of having its own human-like "intelligence" goes back to the origins of Greek natural philosophy and science, and its attention to the orderliness of nature, often with special reference to the revolving of the heavens.
2. For Stillman, the orderliness of Astaire-like dance is an actual cure for destructive emotions.
3. Tells about the orderliness of the crowds, and of the dispatch with which the trains were being filled and emptied.
en un experimento se retiro a un pardela de su nido se le llevo a 5150 km de distancia y luego fue liberada su regreso a su nido fue 13.5 dias despues de haberse soltado si el origen es el nido y extendemos el eje +x al punto de liberacion ¿cual fue la velocidad media del ave?a)en el vuelo de regreso b) desde que se retirodel nido hasta que regreso
Answer:
a) v = - 4.4168 m / s , b) V = 0
Explanation:
The average speed is the variation of the displacement in time used.
v = Δx /Δt = (x₂-x₁) / (t₂-t₁)
let's apply this equation to our case
Let's reduce the magnitudes of the system Yes
Δx = 5150 km (1000 m / 1km) = 5.15 106 m
Δt = 13.5 day (86,400 s / 1 day) = 1,166 10 6 s
a) return trip
the vector is negative because it points long towards the center of the system
v = - 5.15 106 / 1.166 106
v = - 4.4168 m / s
the negative sign indicates that he is coming back, to the lair
b) In a complete trip the distance is zero, because it is a vector, consequently the mean fickleness is also zero
V(j_ = 0
1st block
Worksheet: Metric Prefixes
A. Circle the larger unit:
1.millimeter, centimeter
2. kilogram, megagram
3. microsecond, millisecond
4.dL, mL
5.mg, kg
Answer:
2
Explanation:
kilogram and megagram is the larger unit
Which term describes a quantity that has both magnitude and direction?
Vector explanation is apex and I got 100 on the test and it’s the definition
A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change. A) True B) False
Answer:
False
Explanation:
Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.
[tex]\epsilon=\dfrac{-d\phi}{dt}[/tex]
The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.
Hence, it is false.
ame
2. A train moving at 15 m/s slows down, and eventually stops after 5
seconds. What is the acceleration of the train?
G: Vi =
Vf=
U:
E: Formula
S: Substitute
S: Solve
Answer:
Acceleration, a = -3 m/s²
Explanation:
Given that,
Initial speed, u = 15 m/s
Final speed, v = 0
Time, t = 5 s
We need to find the acceleration of the train. It is equal to the change in velocity divided by time taken. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-15\ m/s}{5\ s}\\\\a=-3\ m/s^2[/tex]
So, the acceleration of the train is 3 m/s² and it is deaccelerating.
Which of the following are subjects of scientific laws?a. gravityb. motionc. thermodynamics
A 50 kg child is riding on a carousel (merry-go-round) at a constant speed of 5 m/s. What is the magnitude of the change in the child's momentum ∣Δp⃗ ∣∣ in going all the way around (360∘)?In going halfway around (180∘)? It is very helpful to draw a diagram, and to do the vector subtraction graphically.
Answer:
a) [tex]\mathbf{|\Delta p^{\to}_{11}| = 0 \ kg.m/s}[/tex]
b) [tex]\mathbf{|\Delta p^{\to}_{12}| = 500 \ kg.m/s}[/tex]
Explanation:
From the image attached below.
Suppose the child goes all the way around, i.e., 360, the child will execute a movement of 1 complete revolution and be at his starting point. At that point, the velocity vector is towards the y-direction.
Thus, the velocity of the child is:
[tex]v_1^{\to} = v \hat _v_1} \\ \\ v_1^{\to} = (5)(0,1,0)\\ \\ v_1^{\to} = (0,5,0) \ m/s[/tex]
the momentum will be:
[tex]p_1^{\to} = m v_1^{\to} \\ \\ p_1^{\to} = (50)(0,5,0) \\ \\ p_1^{\to} = (0,250,0) \ kg.m/s[/tex]
the change in momentum now is [tex]\Delta p = p_1^{\to} -p_1^{\to}[/tex] since that is the child's momentum initially.
∴ [tex]\Delta p =(0,250,0) - (0,250,0)[/tex]
[tex]\mathbf{\Delta p =(0,0,0) \ kg.m/s}[/tex]
By subtracting the two vector graphically as being asked in the question, we have :
[tex]|\Delta p^{\to}_{11}| = \sqrt{(0)^2+(0)^2 +(0)^2 }[/tex]
[tex]\mathbf{|\Delta p^{\to}_{11}| = 0 \ kg.m/s}[/tex]
b) In going halfway around (180°), the child will be opposite with respect to the starting point. Hence, the velocity vector will be in the negative y-direction.
Thus, the velocity of the child is:
[tex]v_2^{\to} = v \hat _v_2} \\ \\ v_2^{\to} = (5)(0,-1,0)\\ \\ v_2^{\to} = (0,-5,0) \ m/s[/tex]
the momentum will be:
[tex]p_2^{\to} = m v_2^{\to} \\ \\ p_2^{\to} = (50)(0,-5,0) \\ \\ p_2^{\to} = (0,-250,0) \ kg.m/s[/tex]
the change in momentum now is [tex]\Delta p = p_2^{\to} -p_1^{\to}[/tex] since that is the child's momentum initially.
∴ [tex]\Delta p =(0,-250,0) - (0,250,0)[/tex]
[tex]{\Delta p =(0,-500,0) \ kg.m/s[/tex]
By subtracting the two vector graphically as being asked in the question, we have :
[tex]|\Delta p^{\to}_{12}| = \sqrt{(0)^2+(-500)^2 +(0)^2 }[/tex]
[tex]|\Delta p^{\to}_{12}| = \sqrt{250000}[/tex]
[tex]\mathbf{|\Delta p^{\to}_{12}| = 500 \ kg.m/s}[/tex]
At t = 0, one toy car is set rolling on a straight track with initial position 13.0 cm, initial velocity -3.6 cm/s, and constant acceleration 2.20 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 11.5 cm, initial velocity 5.40 cm/s, and constant zero acceleration. (a) At what time, if any, do the two cars have equal speeds? (Enter NA if the cars never have equal speeds.) s (b) What are their speeds at that time? (Enter NA if the cars never have equal speeds.) cm/s (c) At what time(s), if any, do the cars pass each other? (If there is only one time, enter NA in the second blank. If there are two times, enter the smaller time first. If they never pass, enter NA in both blanks.) s s (d) What are their locations at that time? (If there is only one position, enter NA in the second blank. If there are two positions, enter the smaller position first. If they never pass, enter NA in both blanks.) cm cm (e) Explain the difference between question (a) and question (c) as clearly as possible.
Answer:
that's too much to read
A NASA spacecraft measures the rate of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is:
Complete Question
A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is: [tex]R = 0.0498 \ kPAkm^{-1}[/tex] Convert R to [tex]kJ*m^{-4}[/tex]
Answer:
The value is [tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
Explanation:
From the question we are told that
The altitude is [tex]R = 0.0498 \ kPAkm^{-1}[/tex]
Generally
[tex]1 k PA = 1000 PA[/tex]
So
[tex]R = 0.0498 \frac{1000PA}{ km}[/tex]
Also
1 km = 1000 m
So
[tex]R = 0.0498 \frac{1000PA}{ 1000m}[/tex]
=> [tex]R = 0.0498 \frac{1 PA}{ 1 m}[/tex]
Now PA is Pascal which is mathematically represented as
[tex]PA = \frac{N}{m^2 }[/tex]
So
[tex]R = 0.0498 \frac{\frac{N}{m^2} }{m}[/tex]
[tex]R = 0.0498 \frac{N}{m^3}[/tex]
Looking the unit we are arrive at we see that it contains J which is mathematically represented as
[tex]J = N * m[/tex]
So
[tex]R = 0.0498 \frac{ N \frac{m}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{\frac{J}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{J}{m^4}[/tex]
Generally
[tex]1 J \to 1.0*10^{-3} kJ[/tex]
[tex]0.0498 J \to x kJ[/tex]
=> [tex]x = \frac{0.0498 * 1.0*10^{-3}}{1}[/tex]
=> [tex]0.0498 *10^{-3} kJ[/tex]
So
[tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 4.7-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.5 μC .
What is the energy of this system?
Express your answer using two significant figures.
What is the tension in the string?
Express your answer using two significant figures.
The string is cut. What is the speed of each sphere when they are far apart?
Hint: There are two conserved quantities. Make use of both.
Express your answers using two significant figures. Enter your answers numerically separated by a comma.
(v2g,v4g)=_____m/s
Answer:
1.2J
26 N
(-28, 14) m/s
Explanation:
energy
U = kQq / d = 8.99*10^9 * (2.5*10^-6C)² / 0.047m
U = 0.0562/0.047
U = 1.20 J
to two significant figures
tension
T = kQq / d²
T = U / d
T = 1.2 / 0.047
T = 25.53 N = 26 N to 2 sf
Momentum is conserved, and the initial momentum is zero:
0 = 0.0020 * V2 + 0.0040 * V4
so
V2 = -2 * V4
Energy is also conserved:
½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J
-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J
-0.0040V4² + 0.002V4² = 1.2 J
0.0060V4² = 1.2 J
V4² = 1.2/0.0060
V4² = 200
V4 = √200
V4 = 14 m/s
and since V2 = -2 * V4
V2 = -28 m/s
(V2, V4) = (-28, 14)
The energy of this system is 1.2J
The tension in the string is 26 N
The speed of each sphere when they are far apart is (-28, 14) m/s
Calculation of energy, tension, and speed:The energy should be
U = kQq / d
= 8.99*10^9 * (2.5*10^-6C)² / 0.047m
U = 0.0562/0.047
U = 1.20 J
The tension should be
T = kQq / d²
T = U / d
T = 1.2 / 0.047
T = 25.53 N
= 26 N
The speed should be
Since Momentum should be conserved, and the initial momentum is zero:
So,
0 = 0.0020 * V2 + 0.0040 * V4
Now
V2 = -2 * V4
Due to this, Energy is also conserved:
½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J
-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J
-0.0040V4² + 0.002V4² = 1.2 J
0.0060V4² = 1.2 J
V4² = 1.2/0.0060
V4² = 200
V4 = √200
V4 = 14 m/s
and now V2 = -2 * V4
V2 = -28 m/s
(V2, V4) = (-28, 14)
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Why is it harder pushing a car than pushing a bike?
Answer:
When inertia increases, it's because the mass increased, which increases the normal force, which ultimately increases friction.
Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?
Complete Question
Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?
A His results are accurate but not precise.
B His results are neither accurate nor precise.
C His results are both accurate and precise
D His results are precise but not accurate.
Answer:
Correct option is A
Explanation:
From the question we are told that
The yield of the first test [tex]k = 5.2 \ g[/tex]
The yield of the second is [tex]u = 1.3 \ g[/tex]
The third yield is [tex]p = 8.5 \ g[/tex]
The average yield [tex]A = 5.0 \ g[/tex]
The know yield is [tex]A_S = 5.1 \ g[/tex]
From the data given we see that
[tex]A_S \ne A[/tex]
Since his average yield is closer to the known yield then the answer is accurate
But since the yield for each test are not repeated the answer is not precise
So the answer is accurate but not precise
His results are accurate but not precise.
Your question is not complete, it seems to be missing the following information;
"A His results are accurate but not precise.
B His results are neither accurate nor precise.
C His results are both accurate and precise.
D His results are precise but not accurate."
The given parameters;
first measurement = 5.2 gsecond measurement = 1.3 gthird measurement = 8.5 gaverage measurement = 5.0known substance average yield = 5.1 gEach of the measurement is far from each other. That is 5.2 grams, 1.3 grams and 8.5 grams are all far apart. So this measurement is not precise.
The known average (5.1 g) and the measured average (5.0 g) are close to each other, so the measurement is accurate.
Thus, we can conclude that his results are accurate but not precise.
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Density is calculated by dividing
Answer:
Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a volume (the same volume as 1 milliliter).
Explanation:
Hey, there!
Density is defined as mass per unit volume. so, when we keep is as a formula we get like,
[tex]density = \frac{mass}{volume} [/tex]
So, you can state that density is calculated by dividing mass by its unit volume.
Hope it helps...
A truck took 4 hours to complete a journey. At the first 1 h 45
min, it travelled at an average speed of 70 km/h. For the rest of
the journey, it travelled at an average speed of 80 km/h. What
was the total distance of the journey?
Answer:302.5
Explanation:
43 Points For Answering & +22 for Brainliest
Question 1
A student conducts an experiment to test how the temperature of a ball affects its bounce height. The same ball is used for each test, and the ball is dropped from the same height each time. What is the dependent variable?
A.The type of ball
B.The temperature of the ball
C.The drop height of the ball
D.The bounce height of the ball
Question 2
A student conducts an experiment to test how the temperature affects the amount of sugar that can dissolve in water. In the experiment, she uses 100 millilitres of water in each trial and stirs for five minutes each time. What is the independent variable?
A.The amount of water
B.The temperature of the water
C.The amount of sugar
D The time stirred
Question 3
Which of the following is a way for scientists to limit the amount of errors in their experimentation?
A.Using controls
B.Only completing an experiment once
C.Using equipment to measure the experiment that has been damaged
D.There is no need to record data from an experiment
Answer:
Question 1: D because the height the ball bounces depends on all the other factors in the experiment.
Question 2: B because the the temperature of the water is not affected by the other variables.
Question 3: A because the more that they can control in the experiment, the more accurate the results will be.
Hopefully this helps :)
1:D 2:B 3:A
just took the test
The siren of a fire engine that is driving northward at 31.0 m/s emits a sound of frequency 2020 Hz. A truck in front of this fire engine is moving northward at 19.0 m/s.
a) What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck?
b) What wavelength would this driver measure for these reflected sound waves?
Answer:
A. Using
Fl= ( v+vl/v+vz)fz
= (340+19/340+31) x 2020
= 1954.7Hz
Then to find the frequency of sound when reflected from the truck such that the driver becomes the listener
we use
F"= ( v+vz/v+vl) fz
= 340+31/340+19 x 2020
2087.5Hz
B to find the wavelength of sound we use
Wavelength= V+vl/ F"
= 340+31/2087.5= 0.18m
In the drawing, what is the vector sum of forces A→+B→+C→ if each grid square is 7.00 N on a side? If the resultant is eastward, enter a positive value and if the resultant is westward, enter a negative value.
Answer:
resultant force = 14 N ( East direction)
Explanation:
A = [tex]\sqrt{(4*7)^2 + (4*7)^2}[/tex]
A = 39.6 N
B = 4 * 7
B = 28 N
C = 2 * 7
C = 14 N
∑ y forces = Ay - B = (4*7) - 28 = 0
∑ x forces = Ax - C = (4*7) - 14 = 14 N
so the resultant force = 14 N ( East direction)
Answer:
resultant = 14 N to the right
Explanation:
A→+B→+C→
adding all forces acting on x
and
adding all forces acting on y
A = sqrt(4*7)^2 + (4*7)^2 = 39.6
B = 4 * 7 = 28
C = 2 * 7 = 14
forces acting on x = (4*7) - 28 = 0
forces acting on y = (4*7) - 14 = 14 N
so the resultant = 14 N to the right
an Alpha particle moving in north direction give reasons
Your question has been heard loud and clear.
An alpha particle , can move in any direction randomly. But with a magnetic field , we can deflect the alpha particle in any direction we want.
So , the magnetic field must be placed to the west of the alpha particle , so that the particle gets deflected and moves towards the north direction.
Thank you.
Like AL2006 said reasons for what
What do civic heroes contribute to their society?
the quest for civic virtue
Explanation:
improving schools, the society....to help educators like yourself, the bill of rights institute has written a New classroom-friendly curriculum called Heroes and Villains....
1. A particle is moving along the x-axis. Its position as a function of time is given as x = bt − ct 2 . a) What must be the units of the constants b and c, if x is in meters and t in seconds?
Answer:
Given x = bt-c²
We know that t= time (s)
x= distance (m)
So
bxt= meters
m/s x s= m
And then c= m/s²
And b= m/s
What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface
Answer:
7,790.38 m/s
Explanation:
Given the following :
What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface
Altitude = 203 km
Using the formula :
V = √GM/r
Where G = gravitational constant =6.67×10^-11
Kindly check attached picture for detailed explanation.
You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.
a. What is your average speed?
b. Choose the best explanation from among the following:
1) More time is spent at 15 m/s than at 25 m/s.
2) The average of 15 m/s and 25 m/s is 20 m/s.
3) Less time is spent at 15 m/s than at 25 m/s.
Answer:
A) Average speed = 18.75 m/s
B) More time is spent at 15 m/s than at 25 m/s.
Explanation:
Let the first distance be d1 and the second distance be d2.
We are given;
d1 = 10 km = 10000 m
d2 = 10 km = 10000 m
Speed; v1 = 15 m/s
Speed; v2 = 25 m/s
Now, the formula for distance is; Distance = speed x time
Thus:
d1 = v1 x t1
t1 = d1/v1 = 10000/15 = 666.67 seconds
Also,
d2 = v2 x t2
t2 = d2/v2 = 10000/25 = 400 seconds
Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s
From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;
More time at 15 m/s than at 25 m/s.