Answers
Answer:
i got u
Explanation:
The electromagnetic spectrum and microwaves have played a critical role in many scientific discoveries, including:
1. Cosmic Microwave Background Radiation: In 1964, Penzias and Wilson discovered Cosmic Microwave Background Radiation (CMB) using a microwave antenna. CMB is the remnant radiation from the Big Bang and provides evidence for the theory of the Big Bang.
2. Radio Astronomy: Radio astronomy is the study of celestial objects and phenomena that emit radio waves. Radio telescopes, which are designed to capture radio waves, have allowed scientists to observe and study many celestial objects, such as pulsars, quasars, and black holes.
3. Medical Imaging: Microwaves and other parts of the electromagnetic spectrum have also played a critical role in medical imaging. Magnetic Resonance Imaging (MRI) and X-rays use different parts of the electromagnetic spectrum to generate images of the human body, helping doctors to diagnose and treat medical conditions.
4. Telecommunications: The use of microwaves in telecommunications has revolutionized the way we communicate with each other. Microwaves are used to transmit information through mobile phones, radios, and televisions, and are an essential part of modern communication technology.
5. Microwave Ovens: Microwave ovens are a common household appliance that use microwaves to cook food. The microwaves penetrate the food and cause the water molecules in the food to vibrate, generating heat and cooking the food.
Overall, the electromagnetic spectrum and microwaves have played a significant role in many scientific discoveries, and their use continues to evolve and impact many areas of our lives.
Related Questions
Part A
Your GPS shows that your friend’s house is 10.0 km away (Figure 2). But there is a big hill between your houses and you don’t want to bike there directly. You know your friend’s street is 6.0 km north of your street. How far do you have to ride before turning north to get to your friend’s house?
8 km
Part B
Referring to the diagram in Part A, what is the sine of the angle
θ at the location of the friend's house?
Answers
Answer:
Part A
You have to ride 8.0 km before turning north to get to your friend’s house.
Part B
The sine of the angle θ at the location of the friend's house is 0.8
Explanation:
The remaining part of the question which is an image is attached below
Explanation:
Part A
To determine how far you will ride ride before turning north,
From the diagram, that is the distance of your street.
Let the distance of your street be \(A\)
and the distance of your friend's street be \(B\)
and let the displacement between your friends house and your house be \(C\)
The relation in the diagram shows a right angle triangle.
The sides of the right angle triangle are represented as \(A,B\) and \(C\).
To find \(A\), which is the distance of your street,
From Pythagorean theorem, 'The square of hypotenuse is the sum of squares of the other two sides'
That is,
\(/Hypoyenuse/^{2} = /Adjacent/^{2} + /Opposite/^{2}\)
\(C\) is the hypotenuse, which is the displacement between your friends house and your house,
Hence, \(C = 10.0 km\)
\(B\) is adjacent, which is the distance of your friends street
then, \(B = 6.0 km\)
and \(A\) is the opposite, which is the distance of your house
From Pythagoras theorem, we can then write that,
\(C^{2} = B^{2} + A^{2}\)
Then, \(10.0^{2} = 6.0^{2} + A^{2}\)
\(A^{2} = 100.0 - 36.0\\A^{2} = 64.0\\A = \sqrt{64.0}\)
\(A = 8.0km\)
Hence, you have to ride 8.0 km before turning north to get to your friend’s house.
Part B
To find the sine of the angle θ at the location of the friend's house,
In the diagram, the sine of the angle θ is given by
\(Sin\theta = \frac{Opposite}{Hypotenuse}\)
Hence, \(Sin\theta = \frac{A}{C}\)
Then,
\(Sin\theta = \frac{8.0}{10}\)
\(Sin\theta = 0.8\)
Hence, the sine of the angle θ at the location of the friend's house is 0.8
A. The amount of distance you have to ride before turning North to get to your friend’s house is 8 kilometers.
B. The sine of the angle (θ) at the location of your friend's house is 0.8.
Let your friend's house be a.Let your friend's street be b.Let the distance between your house and your friend be c.Given the following data:
Distance c = 10 kmDistance a = 6 kmA. To determine the amount of distance you have to ride before turning North to get to your friend’s house, we would apply Pythagorean's theorem:
Mathematically, Pythagorean's theorem is given by the formula:
\(c^2 = a^2 + b^2\\\\10^2 =6^2+b^2\\\\100=36+b^2\\\\b^2 =100-36\\\\b^2 =64\\\\b=\sqrt{64}\)
b = 8 kilometers
B. To find the sine of the angle (θ) at the location of the friend's house:
Mathematically, the sine of an angle is given by the formula:
\(Sin\theta = \frac{opposite}{hypotenuse}\)
Substituting the given parameters into the formula, we have;
\(Sin\theta = \frac{8}{10} \\\\Sin\theta = 0.8\)
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When carbon and oxygen atoms combine, energy is
a. taken in by the reaction
b. given off by the reaction
c. not involved
Answers
When carbon and oxygen atoms combine, energy is given off by the reaction option (b).
A carbon-oxygen link is a polar covalent connection formed by carbon and oxygen atoms. Several inorganic compounds, such as carbon oxides and oxohalides, carbonates and metal carbonyls, and organic compounds, such as alcohols, ethers, carbonyl compounds, and oxalates, include carbon-oxygen bonds.
Each atom shares three electrons in order to establish a triple bond between carbon and oxygen. Six valence electrons are needed for bonding and are referred to as bonding electrons, whereas four valence electrons are non-bonding electrons and are present as two electron pairs in the carbon and oxygen atoms.
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Water flows steadily from an open tank as shown in the figure. (Figure 1)The elevation of point 1 is 10.0 m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80×10−2 m2 ; at point 3, where the water is discharged, it is 1.60×10−2 m2 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. What is the gauge pressure p_gauge at point 2?
Answers
The gauge pressure at point 2 is 98100 Pa or 9.81 x\(10^4\) Pa, which is equivalent to 6.97 x\(10^4\) Pa when rounded to two significant figures.
Step 1: Identification of the given data:
- Elevation at point 1 (h1) = 10.0 m
- Elevation at points 2 and 3 (h2 = h3) = 2.00 m
- Cross-sectional area at point 2 (A2) = 0.0480 \(m^2\)
- Cross-sectional area at point 3 (A3) = 0.0160 \(m^2\)
Step 2: Determination of the discharge rate:
As mentioned earlier, the discharge rate (Q) is given by Q = A2 * v2, and since the velocity at point 2 (v2) is negligible, the discharge rate will be 0.
Therefore, the discharge rate is 0 cubic meters per second.
Step 3: Determination of the gauge pressure at point 2:
To find the gauge pressure at point 2, we'll use Bernoulli's equation:
P1 + (1/2)ρ\(v1^2\) + ρgh1 = P2 + (1/2)ρ\(v2^2\) + ρgh2
Since the velocity at point 2 (v2) is negligible, the term (1/2)ρ\(v2^2\) can be ignored.
The equation simplifies to:
Patm + ρgh1 = P2 + ρgh2
We want to find the gauge pressure at point 2, so we'll subtract the atmospheric pressure (Patm) from P2:
\(P_g_a_u_g_e\) = P2 - Patm
Now let's substitute the given values into the equation:
\(P_g_a_u_g_e\) = (Patm + ρgh1) - Patm
\(P_g_a_u_g_e\) = ρgh1
Plugging in the values:
\(P_g_a_u_g_e\) = (1000 kg/m^3) * (9.81 \(m/s^2\)) * (10.0 m)
\(P_g_a_u_g_e\) = 98100 Pa
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Mass is a physical property that describes the amount of matter in a substance. Mass is commonly measured in grams or kilograms using a triple-beam balance or an electronic balance. Weight is a physical property of matter that
A. is measured in the metric system as pounds.
B. is always equivalent to an object's mass.
C. considers the gravitational force acting on an object.
D. can be measured using a graduated cylinder. Reset Submit
Answers
Answer:
C
Explanation:
it's the measure of the force of gravity pulling the object towards the centre of the Earth and this value differs at various locations away from the equator and highest at the poles
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula
KE = one half.mv2, where m is the mass and v is the speed (velocity). Assume that the speed of the soda bottle falling from a height of
0.8 m will be 4 m/s, and use this speed for each calculation.
Record your calculations in Table A of your Student Guide.
When the mass of the bottle is 0.125 kg, the KE is
✔ 1
kg m2/s2.
When the mass of the bottle is 0.250 kg, the KE is
✔ 2
kg m2/s2.
When the mass of the bottle is 0.375 kg, the KE is
✔ 3
kg m2/s2.
When the mass of the bottle is 0.500 kg, the KE is
✔ 4
kg m2/s2.
Answer included in Question
Answers
Answer:
kinetic energy is given as KE = (0.5) m v²given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg KE = (0.5) m v² = (0.5) (0.125) (4)²
Explanation:
Answer:
1. 0.5 2. 2 3. 3.75 4. 5
Explanation:
The speed of light is 3×10^8 meters per second, which means that light can travel 300 million meters in just one second. How far can light travel in one minute?
Answers
Answer:
(1.8 × 10^9) meters in one minute
Explanation:
To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.
The speed of light is 3 × 10^8 meters per second.
There are 60 seconds in a minute.
Therefore, the distance light can travel in one minute is:
Distance = Speed × Time
Distance = (3 × 10^8 meters per second) × (60 seconds)
Calculating this, we get:
Distance = 3 × 10^8 meters/second × 60 seconds
Distance = 18 × 10^8 meters
Distance = 1.8 × 10^9 meters
So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.
Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answers
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
When both distance and direction are indicated, it becomes a
Answers
Distance travelled by an object in a specific direction is called Velocity.
Velocity is a vector quantity.
Velocity= Time/Displacement
.
prepare a report on why a vehicle needs to be maintained/serviced after a certain period of time. How is servicing different in a petrol/diesel and electric vehicle?
Answers
Vehicles need to be serviced for several reasons such as preventing costly repairs and improving fuel economy.
Why should cars be maintained and / or serviced ?First, regular maintenance can help to prevent costly repairs down the road. Second, maintenance can help to improve fuel economy and emissions. Third, maintenance can help to keep your vehicle safe and reliable.
The servicing requirements for petrol/diesel and electric vehicles differ in a number of ways. Petrol/diesel vehicles require oil changes more frequently than electric vehicles. This is because petrol/diesel engines use oil to lubricate the moving parts, while electric motors do not. Petrol/diesel vehicles also require tune-ups more frequently than electric vehicles.
This is because petrol/diesel engines have more moving parts that need to be synchronized, while electric motors have fewer moving parts.
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Consider the arrangement of two fixed point charges, equal in magnitude, shown in the figure. Which of the following statements arc correct for the initial motion of a third charge if it is released from rest in the vicinity of the two charges shown? A positive charge at point d will accelerate up. A positive charge at point b will accelerate up. A negative charge at point c will accelerate up. A positive charge at point a will accelerate toward the lower-left. A negative charge at point c will accelerate toward the lower-right.
Answers
The option (B and C) are true others are false.
What is acceleration ?
Acceleration is the rate at which velocity changes, as well as how quickly velocity changes over time. instantaneous acceleration is an acceleration at a single moment instantaneous acceleration divided by time deceleration
What is motion ?
The movement of atomic or molecular building blocks in a certain direction is referred to as molecular motion. Temperature and heat both have an impact on molecular mobility. This is so because the measurement of temperature, which reflects molecular motion and average kinetic energy, is used to determine molecular motion.
A is false; a positive charge at point d will be attracted to the nearby negative more than it is repelled by the more distance positive.
B is true
C is true
D is false; the dipole field at point a is directed up (it curves from the lower positive charge to the upper negative one, but is directly upward here) so the charge will initially accelerate up
E is false; similar to question D, the field is up at point c so the negative charge will accelerate initially straight down.
Therefore, the option (B and C) are true others are false.
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why the speed of light decreases as it passes from air into another substance?
Answers
Answer:
If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.
Estimat the number and wattage of lamps. which would be required to illuminate a workshop space 60x1.5 meteres by means of lamps mounted 5 metres above the working Plane The average illumination required is about 100 wt. coefficient of utilisation = 0.4 luminous efficiency 16 lumens per watt. Assume a space-height ratio of unity and a cundle Power depreciation of 20%
Answers
The number and wattage of lamps required to illuminate the workshop would be approximately 8 lamps and 70 watts respectively.
Wattage calculationTo estimate the number and wattage of lamps required to illuminate a workshop space of 60x1.5 meters, we can follow these steps:
Calculate the area of the workshop:
Area = length x widthArea = 60m x 1.5mArea = 90 square metersDetermine the total lumens required:
Lumens = area x average illuminationLumens = 90 sq m x 100 luxLumens = 9000 lumensAdjust for the coefficient of utilization and luminous efficiency:
Effective lumens = lumens / (coefficient of utilization x luminous efficiency)Effective lumens = 9000 / (0.4 x 16)Effective lumens = 1406.25 lumensAdjust for space-height ratio and candle power depreciation:
Effective lumens per lamp = effective lumens x space-height ratio x (1 - depreciation)Effective lumens per lamp = 1406.25 x 1 x (1 - 0.2)Effective lumens per lamp = 1125 lumensDetermine the number of lamps required:
Number of lamps = total lumens required / effective lumens per lampNumber of lamps = 9000 / 1125Number of lamps = 8 lamps (rounded up)Determine the wattage of each lamp:
Wattage per lamp = effective lumens per lamp / luminous efficiencyWattage per lamp = 1125 / 16Wattage per lamp = 70.3 watts (rounded up)Therefore, approximately 8 lamps with a wattage of 70 watts each would be required to illuminate the workshop space.
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What is the displacement of the particle in the time interval 7 seconds to 8 seconds?OA. O metersОВ.1.5 metersOC. 3 metersOD. 7 meters
Answers
Given,
A velocity-time graph.
The area under the curve of a velocity-time graph of an object gives us the displacement of the object.
From the graph, we can see that the area under the curve from 7 seconds to 8 seconds is a triangle.
The height of the triangle is h=6 m/s.
And the base of the triangle is b=1 s.
The area of a triangle is given by,
\(A=\frac{1}{2}bh\)On substituting the known values,
\(\begin{gathered} A=0.5\times6\times1 \\ =3\text{ m} \end{gathered}\)Therefore the displacement of the particle in the time interval 7 s to 8 s is 3 meters.
Thus, the correct answer is option C.
Increasing the potential difference across the two plates of a capacitor will _________________.Group of answer choicesincrease the chargedecrease the chargeincrease the capacitancedecrease the capacitance
Answers
We know,
Q=CV
where C is the capacitance Q is the charge and V is the potential difference.
So with increase in potential difference the charge increases while teh capacitance decreases.
So the answers are:
increase the charge
decrease the capacitance
Which is not a difference between chemical reactions and nuclear reactions?
A.The subatomic particles involved in the reaction is different.
B.The amount of energy absorbed during the reaction is different.
C.The masses of the reactants and products differ.
D.The amount of energy released during the reaction is different.
Answers
Answer:
B.The amount of energy absorbed during the reaction is different
Explanation:
I just took the test
3.14=0.81(3.89) + 1/2 a (3.89)^2
Answers
The given equation can be solved as,
\(\begin{gathered} 3.14=(0.81)(3.89)+\frac{1}{2}a(3.89)^2 \\ a(7.566)=3.14-3.15 \\ a=\frac{-0.01}{7.566} \\ \approx-0.0013 \end{gathered}\)Therefore, the value of a is -0.0013.
What is an example of a combined approach in psychology?
Answers
Answer:
The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality
Explanation:
Answer:
The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality
2. For electric circuit shown in Figure find currents in each resistor.
Answers
The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.
What is the current flowing in each resistor?The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;
The total voltage in loop 1 is calculated as;
2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0
6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0
The current flowing in loop 2 is calculated as;
I = V/R
I₂ = ( 6 V - 4 V ) / (3 + 4)
I₂ = 0.286 A
The value of the current flowing in loop 1 is calculated as;
6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0
6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0
6 - 3I₁ - 3₁ + 0.858 = 0
-6I₁ = -6.858
I₁ = 6.858 / 6
I₁ = 1.14 A
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Describe how water can be both physical and chemical weathering.
Answers
Explanation:
Both because water can fall in holes and then freeze. However, water can also be chemically react with other elements and substances to wear something away.
Most of the bending of light in the eye is done at the air-cornea interface. By what angle (θcornea) is the beam of light shown in Figure deviated as it passes from air to the cornea if the incident angle is θi = 23.6°? The refractive index of air is nair = 1.00, the refractive index of the cornea is ncornea = 1.38. Explain why θcornea< θi?
Answers
The angle (θcornea) when light passes from air to cornea is 16.86°
What is Snell's law?It states that the ratio of sine of angle of incidence and angle of refraction is equal to the refractive index of second medium to the first medium.
sini/sinr =n₂ / n₁
Most of the bending of light in the eye is done at the air-cornea interface. The beam of light deviated as it passes from air to the cornea if the incident angle is θi = 23.6°.
Given the angle of incidence i = 23.6°, refractive index of air n₁ =1, refractive index of cornea n₂ = 1.38, then the angle of refraction at cornea is
sinr = sini x (n₁/n₂)
Plug the values, we get
sinr = sin23.6 x (1/1.38)
sinr = 16.86°
The angle of refraction is less than angle of incidence due to refraction.
Thus, the angle (θcornea) when light passes from air to cornea is 16.86°
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Which of the following accurately details the relationship between wavelength of illumination, lens quality, magnification, and the limit of resolution for light microscopy?A) As shorter wavelength illumination and better quality lenses are used, magnification will increase and the limit of resolution will decrease.B) As shorter wavelength illumination and better quality lenses are used, magnification will decrease and the limit of resolution will decrease.C) As shorter wavelength illumination and better quality lenses are used, magnification will increase and the limit of resolution will increase.D) Magnification and limit of resolution are independent of wavelength of illumination and lens quality for light microscopy.
Answers
C) As shorter wavelength illumination and better quality lenses are used, magnification will increase and the limit of resolution will increase.
What is wavelength illumination?Wavelength illumination is a way of measuring the intensity of a light source in terms of the wavelength of the light being emitted.
Light microscopes use light of different wavelengths and high-quality lenses to magnify and enhance images. Shorter wavelength illumination, such as UV light, can provide greater resolution because it has a higher frequency than longer wavelengths, such as visible light. Additionally, higher quality lenses provide greater magnification and allow for a larger field of view. Thus, as shorter wavelength illumination and better quality lenses are used, magnification will increase and the limit of resolution will also increase.
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Question 6 of 10
Which statement describes one way that nuclear fusion differs from nuclear
fission?
O A. Nuclear fusion occurs spontaneously on Earth.
B. Nuclear fusion is used to power submarines.
O C. Sustained processes of nuclear fusion occur only in stars.
O D. Nuclear fusion is used in power plants to produce electricity.
Answers
Answer:
did you get the answer??
Sustained processes of nuclear fusion occur only in stars. This statement describes one way that nuclear fusion differs from nuclear fission. Hence option C is correct.
When two lighter nuclei are fused together to form heavier nuclei which is called as fusion reaction. It generally involves hydrogen nuclei, In Fusion process two hydrogen nuclei are fused together and helium nucleus is formed, In this process and enormous amount of energy is released. Fusion reaction occurs only in the star like sun. In sun, hydrogen is a fuel which converts into helium that is why we can see that sun is burning.
It is almost impossible to get fusion reaction on the earth, that is why uranium is used in nuclear power plant. In nuclear power plant uranium is converted into two lighter nuclei which is called as fission.
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The sound from a clarinet at a distance of 5 m from a sound level meter is found to be 52 dB. If
the frequency is 1000 Hz, find (a) the sound loudness level in phons, (b) the sound intensity in
watts/meter2, and (c) the power of the source in watts.
Answers
There are 90 phones of volume, 10-7 W/m2 of sound intensity, and 0.0314 watts of source power.
Which frequency is the simplest?A straightforward frequency analysis compares the values of the fields you provide and generates a report listing each value for those fields along with the frequency at which each value occurs.
How often does sound occur?The rate at which a sound power wave repeats itself, also known as frequency or pitch, is measured in cycles per second. Bullfrog calls and cricket chirps have lower frequencies than drum beats and whistles, respectively.
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Brainliest if correct
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Answer:
B is the correct answer of the question
imagine imagining an imagination.
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Answer:
We’re imagining imagining imagining an imagination...
In two paragraph of lines each, describe how you
would simplify, for your
grade 10 learners the
following topics
1.1 Electricity and magnetism.
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The question above wants to assess your ability to simplify complex matters. In that case, I can't answer this question for you, but I'll show you how to answer it.
First, you should research electricity and magnetism. This research will make you understand the subject and all its elements. You can do this research in textbooks and digital platforms aimed at children and teenagers since these media usually present subjects in a more simplified way.
After doing this research, you will be able to write a text about electricity and magnetism as follows:
Present the main focus of this subject.Present the elements that make up this subject.Present how this subject applies in the real world.Present the importance of this subject in human life and society.Remember that you must use simple language, without the use of technical terms, but with common terms.
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The wave in the liquid travels towards the surface at an angle. Fig 9.2 shows the centres of the compressions of the sound wave in liquid. Some compressions shown have reached the liquid-air boundary. The parts of these compressions in the air are not shown on Fig 9.2 These waves are also reflected at the boundary. Draw on the diagram the reflected wavefronts.
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The image of the reflected sound wavefronts traveling through a liquid is found in the attachment.
What are reflected sound wavefronts?Reflected sound wavefronts refer to the waves of sound that are bounced back or reflected off of a surface such as a wall, ceiling, or floor.
When a sound wave travels through the air and encounters a surface, some of the sound energy is absorbed by the surface, while some of it is reflected back into the air.
These reflected sound waves can interfere with the original sound waves, leading to complex patterns of sound intensity and phase.
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You see a boat sitting at the end of a dock. Ten minutes later you see the same boat in a cove to the right of the dock. You did not see the boat move.
However, you know that the boat moved because its
relative to the dock changed.
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The boat must have moved, despite not being seen to move, because its relative position to the dock has changed. This phenomenon is known as relative motion .
Everything is always in motion, but the way we perceive it depends on our frame of reference.
In this scenario, the dock was the frame of reference for the initial position of the boat. When the boat moved to the cove, its position relative to the dock changed, and the dock was no longer an appropriate frame of reference. The boat's motion is now relative to the cove instead.
It is important to note that relative motion depends on the chosen frame of reference. If we were to choose the boat as the frame of reference, then it would be the dock that appears to move, not the boat. This is because motion is always relative to a chosen frame of reference.
In conclusion, the boat must have moved because its position relative to the dock changed. The concept of relative motion reminds us that motion is always relative to a chosen frame of reference, and that the way we perceive motion depends on our chosen frame of reference.
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Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?
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Answer:
d= 23.25 m
Explanation:
Assuming no other external forces acting on the disk, total mechanical energy must be conserved.Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:\(E_{o} = K_{transo} + K_{roto} (1)\)
When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:\(E_{f} = m*g*h (2)\)
Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:\(h = d* sin 37 (3)\)
Replacing (3) in (2):\(E_{f} = m*g* d * sin 37 (4)\)
Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:\(v = \omega * R (5)\)
For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².The rotational kinetic energy of a rotating rigid body can be written as follows:\(K_{rot} = \frac{1}{2}* I * \omega^{2} (6)\)
Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:\(E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)\)
Since (4) and (7) must be equal each other, we can solve for d as follows:\(d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)\)
Replacing by the values, we finally get:\(d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.\)
A spacecraft travels directly from Earth to the Moon. Where is the vector sum of the gravitational forces exerted on the spacecraft by Earth and by the Moon zero?
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Gravitational force, F is directly proportional to the mass of the object.
The mass of the satellite is the same for both earth and the moon.
So, Gravitational force of earth > gravitational force of the moon
Also, the distance is inversely proportional to the gravitational force.
Thus, the vector sum of gravitational force due to earth and due to the moon on the satellite will be zero, when the satellite is more than halfway to the moon.