Answer:
The actual distance is [tex]d_a = 0.3233\ m[/tex]
Explanation:
From the question we are told that
The distance of the can is d = 43 cm = 0.43 m
Generally the actual distance is mathematically represented as
[tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]
Where [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is
[tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]
So
[tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]
[tex]d_a = 0.3233\ m[/tex]
The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.
79.70% Mg-24 (23.9872 amu), 10.13% Mg-25 (24.9886 amu), and 10.17% Mg-26 (25.9846 amu).
A. How many neutrons are in Mg-25?
B. What is the average atomic mass of magnesium in these rocks?
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
Answer:
A. number of neutrons of Magnesium Mg = 13
B. The average mass of Mg = 22.29 amu
C. the magnesium composition on Mars is not the same as that on Earth.
Explanation:
Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.
The atomic number of Magnesium Mg = 12
The atomic number of an element is the number of protons present in the atomic nucleus of the element
i.e Atomic number = number of protons = 12
The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.
Mass number = number of protons + number of neutrons
Given that the mass number of Mg = 25
Then;
25 = 12 + number of neutrons
25 - 12 = number of neutrons
13 = number of neutrons
number of neutrons of Magnesium Mg = 13
B. What is the average atomic mass of magnesium in these rocks?
The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.
Therefore the average atomic mass of magnesium can be calculated as:
= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]
= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]
= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]
The average mass of Mg = 22.29 amu
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.
There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.
A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?
Answer:
0.03143 Gy
Explanation:
Mass of the human body = 70 kg
Mass of potassium in the human body = 140 g
chemical atomic mass of potassium = 39.1
From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms
Thus,
140g of potassium will contain;
(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms
We are told that the natural abundance of one of the 40K isotopes is 0.012%.
Thus;
Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms
Formula for activity of K-40 is given as;
Activity = (0.693 × number of K-40 atoms)/half life
Activity = (0.693 × 7.2276 × 10^(19))/1300000000
Activity = 3.85 × 10^(10)
We are told that each decay deposits 1.0 MeV of energy into the body.
Thus;
Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV
Now, 1 MeV = 1.602 × 10^(-13) joules
Thus;
Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J
1 Gy = 1 J/kg
Thus;
Yearly dose = 2.25/70 = 0.03143 Gy
A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Answer:
b
Explanation:
An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?
Answer:
1.6×10²⁰
Explanation:
An ampere is a Coulomb per second.
1 A = 1 C / s
The amount of charge after 5 seconds is:
5.0 A × 5 s = 25 C
The number of electrons is:
25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons
The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.
Answer:
The radius of the earth is [tex]r = 6365.4 \ km[/tex]
Explanation:
From the question we are told that
The distance at Alexandria is [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]
The angle of the sun is [tex]\theta = 7.2 ^o[/tex]
So we want to first obtain the circumference of the earth
So let assume that the earth is circular ([tex]360 ^o[/tex])
Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many [tex](7.2 ^o)[/tex] are in [tex]360^o[/tex]
i.e [tex]N = \frac{360}{7.2}[/tex]
=> [tex]N = 50[/tex]
With this value we can evaluate the circumference as
[tex]c = 50 * 800[/tex]
[tex]c = 40000 \ km[/tex]
Generally circumference is mathematically represented as
[tex]c = 2\pi r[/tex]
[tex]40000 = 2 * 3.142 * r[/tex]
=> [tex]r = 6365.4 \ km[/tex]
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Answer:
6000 counts per secondExplanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample
A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
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Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction
Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction
Answer:
Explanation:
Using the Continuity equation
v X A = v' xA'
so if A is 1/2of A' then A velocity must be 2 times the A'
after-contraction v = 2 x 5.0m/s = 10m/s
Using the Bernoulli equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
, the "h" terms cancel
3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²
p₂ = 342500pa
How are electricity and magnets connected
Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.
PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above
At what minimum angle will you get total internal reflection of light traveling in diamond and reflected from ethanol? °
Answer:
34°
Using the relation
θᶜ = sin^-1(n₂/n₁),
where n1= the refractive index of light is propagating from a medium
And n2 = refractive index of medium into which light is entering
So we know that
refractive index of diamond at 589nm = 2.41= n₁
refractive index of ethanol at 589nm and 20°C = 1.36= n₂
Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°
Explanation:
Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?1. Increasing the wavelength of the laser.2. Increasing the distance to the screen.3. Increasing the frequency of the laser.4. Increasing the number of lines per length.
Answer:
Increase in frequency of the laser
Explanation:
Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin
You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?
Answer:
Explanation:
The moment of inertia of the disk I = 1/2 m R² where R is radius of the disc and m is its mass .
putting the values
I = .5 x 95 x .38²
= 6.86 kg m²
n = 87 rpm = 87 / 60 rps
n = 1.45 rps
angular velocity ω = 2π n , n is frequency of rotation .
= 2 x 3.14 x 1.45
= 9.106 radian /s
frictional force = 16 x .2
= 3.2 N
torque created by frictional force = 3.2 x .38
= 1.216 N.m
angular acceleration = torque / moment of inertia
= - 3.2 / 6.86
α = - 0.4665 rad /s²
b ) ω² = ω₀² + 2 α θ , where α is angular acceleration
0 = 9.106² - 2 x .4665 θ
θ = 88.87 radian
no of turns = 88.87 / 2π
= 14.15 turns
light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb
Answer:
121ohmsExplanation:
Formula used for calculating power P = current * voltage
P = IV
From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;
P = IV
P =(V/R)V
P = V²/R
Given parameters
Power rating of the bulb P = 100 Watts
Source voltage V = 110V
Required
Resistance of the bulb R
Substituting the given parameters into the formula for calculating power to get Resistance R;
P = V²/R
100 = 110²/R
R = 110²/100
R = 110 * 110/100
R = 12100/100
R = 121 ohms
Hence, the resistance of this bulb is 121 ohms
A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?
Answer:
386.13 N
Explanation:
The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).
Therefore, KE of the ball
[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]
Now, workdone in moving the glove
W= Fd
where F = Force applied, d = displacement of the glove= 0.23 cm.
88.81 = F×0.23
F= 88.81/0.23 = 386.13 N
1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?
Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.
Explanation:
Answer:
Light amplification by stimulated emission of radiation
Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood
Answer:
metal conducts heat better than wood.
Explanation:
Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.
If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)
Answer:
The uncertainty in momentum is 5.25x 10^25Jsm
Explanation:
We know that
h bar = h/2π
So
1.05x 10^34=h/2pπ
h=1.05x 10^ 34(2π)=6.597x 10^-34Js
dp=(6.597x10^-34/4pπ)/(1x10^-10)
=5.25x10^-25 Jsm
A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is
Answer:
At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage
Explanation;
Because at series connection the battery and resistor have equal voltage
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards
Answer:
His angular velocity will increase.
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = [tex]I[/tex]'ω'
where
[tex]I[/tex]' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = [tex]mr'^{2}[/tex]
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].
From
[tex]I[/tex]'ω' = [tex]I[/tex]ω
since [tex]I[/tex] is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)
Answer:
t = 23.6 min
Explanation:
First we need to find the mass of air in the room:
m = ρV
where,
m = mass of air in the room = ?
ρ = density of air at room temperature = 1.2041 kg/m³
V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³
Therefore,
m = (1.2041 kg/m³)(3247.2 m³)
m = 3909.95 kg
Now, we find the amount of energy consumed to heat the room:
E = m C ΔT
where,
E = Energy consumed = ?
C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C
ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C
Therefore,
E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)
E = 29324.62 KJ
Now, the time period can be calculated as:
P = E/t
t = E/P
where,
t = Time needed = ?
P = Power of heater = 20.7 KW
Therefore,
t = 29324.62 KJ/20.7 KW
t = (1416.65 s)(1 min/60 s)
t = 23.6 min
A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
1.34 mm
Explanation:
A double slit experiment is conducted with a light which has a wavelength of 620 nm
The fringes are separated 2.3 mm apart
The light is changed to a wavelength length of 360 nm
Let x represent the fringe spacing as a result of the change in wavelength
Therefore,the fringe spacing can be calculated as follows
2.3mm/x= 620nm/360nm
Multiply both sides
x × 620= 2.3×360
620x= 828
x= 828/620
x= 1.34 mm
The temperature of the hot spots caused by the impact of transferred matter onto the surface of a pulsar can be 108 K. What is the peak wavelength in the blackbody spectrum of such a spot, and in what range of the electromagnetic spectrum does it occur
Given that,
Temperature = 10⁸ K
We need to calculate the peak wavelength in the blackbody spectrum
Using formula of peak wavelength
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]
Where, T= temperature
Put the value into the formula
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]
[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]
[tex]peak\ wavelength = 290\ nm[/tex]
This range of wavelength is ultraviolet.
Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
