Answer:
28.6 m/s
Explanation:
the verified expert clearly isnt an expert no shade tho
How fast must a 3000 kg elephant move to have the same kinetic energy as a 65 spinter running at 10ms?
Q. : How fast must a 3000 kg elephant move to have the same kinetic energy as a 65.0 kg sprinter running at 10m/s?
Answer:
The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
Explanation:
Kinetic Energy: This is the energy of a body in motion. It can be expressed mathematically as
Ek = 1/2mv²
Since the kinetic energy the elephant = the kinetic energy of the sprinter.
1/2m₁v₁² = 1/2m₂v₂².......................... Equation 1
Making v₁ the subject of the equation,
v₁ = √(m₂v₂²/m₁)................................. Equation 2
Where v₁ = speed of the elephant, m₁ = mass of the elephant, v₂ = speed of the sprinter, m₂ = mass of the sprinter.
Given: m₁ = 3000 kg, m₂ 65 kg, v₂ = 10 m/s
Substituting these values into equation 2,
v₁ =√(65×10²/3000)
v₁ = [tex]\sqrt{(6500)/3000[/tex]
v₁ =[tex]\sqrt{2.167}[/tex]
v = 1.47 m/s
Therefore The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
A car travels 60 km in the first 2 hours and 68 km in the next 2 hours. What’s the cars average speed?
Answer:
The car's average speed is 32 kilometers per hour
Explanation:
1. Let's review the information given to us to answer the question correctly:
First two hours = 60 kilometers
Next two hours = 68 kilometers
2. What is the car's average speed?
Total distance traveled by the car = 60 + 68
Total distance traveled by the car = 128
Total time of travel = 2 + 2 hours
Total time of travel = 4 hours
Average speed = Total distance/Total time
Replacing with the real values, we have:
Average speed = 128/4
Average speed = 32 kilometers per hour
2 Physics Questions WILL GIVE BRAINLIEST!!
A book is being pushed along a table at a constant speed. The coefficient of static friction is 0.75, and the coefficient of kinetic friction is 0.30. The book has a mass of 0.50 kg. What is the force of friction?
0.15 N
1.5 N
3.7 N
0.38 N
Suppose the mass of each of two spherical objects is doubled, and the distance between their centers is three times as large. How is the magnitude of the force of gravity between them affected?
It is 4/9 as large
It is 3/2 as large
It is 4/3 as large
It is 2/3 as large
It is 2/9 as large
Answer:
a) The force of friction is 1.5 N.
b) The new force is 4/9 times as large as the original force.
Explanation:
It is given that the book is moving on the table.
∴ The friction acting on the block should be kinetic.
∴ Coeffecient of kinetic friction , k = 0.3
The force of friction = kN , where 'N' is the normal force acting on the surface where friction is acting.
In this case , N = mg , where 'm' is the mass of the book and 'g' is the acceleration due to gravity.
∴ Force of friction , f = kmg = 0.3×0.5×10 = 1.5N
b) We know that ,
Gravitational force , F ∝ m , where 'm' is the mass of the body.
F ∝ [tex]x^{2}[/tex] where 'x' is the distance between the two bodies.
Since mass of both the bodies is doubled , the force will be 4 times as large compared to the original force F from the first proportionality equation.
However , as the distance is 3 times as large , the force will be 9 times as small as compared to the original force F
∴ The new force is 4/9 times as large as the old force
You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 13.2 cm. When the cylinder is rotating at 1.63 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
Answer:
centripetal acceleration[tex](a_{c})=13.8m/s^{2}[/tex]
Explanation:
convert 1.63rev/sec to rpm by multiplying by 60
= 1.63*60=97.8rpm
Convert this to rad/sec
1rpm =π/30 rad/sec
97.8rpm = 97.8 * (π/30 rad/sec)
=10.25rad/sec
linear velocity= angular velocity *radius
radius =13.2cm=13.2/100=0.132m
v=rω
v= 0.132*10.25
v=1.35m/s
centripetal acceleration = [tex]\frac{v^{2} }{r}[/tex]
[tex]a_{c}=\frac{1.35^{2} }{0.132}[/tex]
[tex]a_{c}=13.8m/s^{2}[/tex]
Answer:
[tex]a_{c}[/tex] = 13.8 m/s².
Explanation:
The acceleration centripetal [tex]a_{c}[/tex] is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex] (1)
where v: is the tangential speed and r: is the container radius
The tangential speed is equal to:
[tex] v = \omega \cdot r [/tex] (2)
where ω: is the angular velocity
Since 1 revolution is equal to 2π rad, the velocity (equation 2) is:
[tex] v = 1.63 \frac{rev}{s} \cdot \frac{2\pi rad}{1rev} \cdot 0.132m = 1.35m/s [/tex]
Now, by entering the velocity value calculated into equation (1) we can find the acceleration centripetal:
[tex] a_{c} = \frac{(1.35m/s)^{2}}{0.132m} = 13.8m/s^{2} [/tex]
I hope it helps you!
You have just been appointed as director of your company's corporate training division. The CEO of your company has been displeased with your company's prior training programs, so you are tasked with rehauling the entire training division. You convene a meeting of all training division managers to decide on the types of training that the division will implement. One of your managers is a firm supporter of e-training programs for employees in your company's international offices. He touts the benefits of e-programs by stressing that ________.
Answer:
He touts the benefits of e-programs by stressing that:
E-training are highly adjustable and flexible as well as employees can complete their training according to their ease.
Explanation:
The concept of E-training is becoming more and more popular now-a-days as it has the following advantages:
You can train the people all over the world with the help of internet.You can complete the training by remaining in your comfort zone.You can get the feedback of the people in a structured manner in less time.You can evaluate the performance in the real time.Due to these reasons, the manager wants the company to do the e-training programs.
Describe how the water cycle is driven by the energy of the sun
Explanation:
Sun is the is the cause of entire water cycle because it is responsible for for its major two components that are
1. Condensation
2. Evaporation
The Sun's heat cause the evaporation( water converting into vapor) of water from the earth. This water ends up in atmosphere as water vapor. It cools and rise becoming cloud, and this eventually condenses to water droplets in form of rain or dwe.
What is Icm, the moment of inertia of the disk around its center of mass? You should know this formula well.
[tex]I_{cm} =mr^{2}[/tex] is the moment of inertia of rigid body along its centre of mass.
The moment of inertia of a body is the rotational analog of mass in linear motion. If we consider a rigid body of mass m, assuming its mass concentrated at its center of mass (cm) which is at a distance of r from the axis of rotation of the body, then the moment of inertia of the rotating body is given by:
[tex]I_{cm} =mr^{2}[/tex]
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The moment of inertia (Icm) of a disk about an axis through its center is 0.5 * mass * radius^2. It plays a significant role in the study of rotational motion.
Explanation:The moment of inertia of a disk about an axis passing through its center of mass, perpendicular to the plane of the disk (also known as Icm) is calculated using the formula:
Icm = 0.5 * mass * radius^2
where the mass is how much matter is in the disk, and the radius is the distance from the center of the disk to its edge. This formula is important as it helps determine the rotational motion of an object.
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The most likely place to find abundant manganese nodules is on the: a. abyssal plain far from a continent. b. continental rise. continental shelf. c. crest of a mid-ocean ridge. d. All of the above locations contain manganese nodules.
Answer:
a. abyssal plain far from a continent
Explanation:
The sediments of the abyssal plains contain of the abundant mineral resources and notably polymetallic nodules and re also called the manganese nodules and these are on the shape of potatoes and distributed at the depths of about 4000 meters. And they lie at the international line of the pacific oceans stretching in an area of 118°–157°, and from 9°–16°N, that is 3 million m square is currently under the noddle exploration and discovery.What BEST completes the following sentence.The dust and gas that escapes from a comet creates a/an _____________________________.
Answer:
Explanation:
The dust and gases that escape from a comet creates a coma.
Coma can also be defined as an unclear envelope around the comet, formed when the comet passes near the sun. Sun temperature melts the comet ice and thus give comet a fuzzy appearance when viewed telescope and distinguishes it from a star.
This technique can be useful to find the size of the comet of different size
A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK = 0.58, find the angle of the incline, in degrees?
Answer:
30.11°
Explanation:
[tex]\mu[/tex] = Coefficient of friction = 0.58
g = Acceleration due to gravity = 9.81 m/s²
[tex]\theta[/tex] = Angle of incline
As the forces of the system are conserved we have
[tex]mgsin\theta-\mu mgcos\theta=0\\\Rightarrow mgsin\theta=\mu mgcos\theta\\\Rightarrow sin\theta=\mu cos\theta\\\Rightarrow \mu=\dfrac{sin\theta}{cos\theta}\\\Rightarrow \mu=tan\theta\\\Rightarrow \theta=tan^{-1}\mu\\\Rightarrow \theta=tan^{-1}0.58\\\Rightarrow \theta=30.11^{\circ}[/tex]
The angle of incline is 30.11°
A particle P is projected from the origin O so that it moves along the x-axis. At time t after projection, the velocity of the particle v is given by v = 3t^2 -24t +45. ) Find an expression for the displacement of P in the first 3 seconds of its motion using calculus(integration).
Answer:
x = t³ − 12t² + 45t
Explanation:
Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 24t + 45) dt
x = t³ − 12t² + 45t + C
The particle is originally at the origin, so at t = 0, x = 0.
0 = (0)³ − 12(0)² + 45(0) + C
0 = C
x = t³ − 12t² + 45t
At 35.0°C and 3.00 atm pressure, a gas has a volume of 1.40 L. What pressure does the gas have at 0.00°C and a volume of 0.950 L? Which equation should you use? A balloon containing 0.500 mol Ar at 0.00°C and 65.0 kPa pressure is expanded by adding more argon. How many moles of argon are added to bring the sample to a final volume of 60.0 L at 30.0°C and 45.0 kPa? What is the original volume of the gas? L
Answer :
The final pressure of gas will be, 3.92 atm
The original volume of gas is, 17.46 L
The number of moles of argon gas added is, 0.57 mol.
Explanation :
Part 1 :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 3.00 atm
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = 1.40 L
[tex]V_2[/tex] = final volume of gas = 0.950 L
[tex]T_1[/tex] = initial temperature of gas = [tex]35.0^oC=273+35.0=308K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]0.00^oC=273+0.00=273K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{3.00atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}[/tex]
[tex]P_2=3.92atm[/tex]
Therefore, the final pressure of gas will be, 3.92 atm
Part 2 :
First we have to calculate the original volume of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 65.0 kPa
V = volume of gas = 3.06 L
T = temperature of gas = [tex]0.00^oC=273+0.00=273K[/tex]
n = number of moles of gas = 0.500 mol
R = gas constant = 8.314 kPa.L/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex](65.0kPa)\times V=(0.500mol)\times (8.314kPa.L/mol.K)\times (273K)[/tex]
[tex]V=17.46L[/tex]
Now we have to calculate the final moles of sample of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 45.0 kPa
V = volume of gas = 60.0 L
T = temperature of gas = [tex]30.0^oC=273+30.0=303K[/tex]
n = number of moles of gas = ?
R = gas constant = 8.314 kPa.L/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex](45.0kPa)\times (60.0L)=n\times (8.314kPa.L/mol.K)\times (303K)[/tex]
[tex]n=1.07mol[/tex]
Now we have to calculate the number of moles of argon gas added.
Moles of argon gas added = Final moles of gas - Initial moles of gas
Moles of argon gas added = 1.07 mol - 0.500 mol
Moles of argon gas added = 0.57 mol
Thus, the number of moles of argon gas added is, 0.57 mol.
Which friction-reducing technologies are used in the Variable Compression Turbo Engine?
Friction-reducing technologies used in the Variable Compression Turbo Engine are Diamond-like coating on valve lifters, micro finishing on crankshaft and camshaft and mirror bore coating on cylinder wall.
Explanation:
Variable compression is a technology to adjust the compression of an internal combustion engine while the engine is in operation. At this time friction may occur that need to be reduced. To reduce this friction some technologies are used like
Diamond-like coating on valve liftersMicro finishing on crankshaft and camshaftMirror bore coating on cylinder wallA hydrogen free diamond like carbon coating is applied to an engine valve lifter to reduce mechanical loss. Micro finishing on crankshaft and camshaft achieves improvement in geometric parameters such as roundness. Mirror bore coating on cylinder wall raises energy efficiency by reducing the friction inside the engine.
Friction-reducing technologies in Variable Compression Turbo engines include gasoline direct injection, variable valve timing, and multi-valve configurations, all enhanced by advanced computer controls. These technologies help to optimize engine efficiency by minimizing frictional losses.
Explanation:The Variable Compression Turbo (VCT) Engine employs several friction-reducing technologies to enhance performance and efficiency. One key technology used in such engines is gasoline direct injection, which enables precise control over the fuel-air mixture and improves combustion efficiency. Another technology is variable valve timing, which optimizes the opening and closing of the engine's valves to match the engine's speed and load, reducing mechanical friction and improving efficiency.
In addition to these, the VCT engines may use multi-valve configurations that increase the engine's ability to breathe by allowing for more intake and exhaust flow, further reducing frictional losses. Enhanced computer controls also play a critical role in adjusting the compression ratio and monitoring engine performance to minimize friction. The use of advanced materials and engineering solutions contribute to reducing heat transfer into the environment, although it cannot be eliminated entirely due to the second law of thermodynamics.
a parallel RC circuit has a capacitive reactance of 962 ohms and a resistance of 1,200 ohms what is the impedance of this circuit
A. 751 ohm
B. 653 ohms
C. 968 ohm
D, 1,254 ohm
Answer:
A. 751 ohm
Explanation:
Impedance: This is the total opposition to the flow of current in an a.c circuit by any or all of the three circuit elements ( R, L, C). The unit of impedance is Ohms (Ω). The impedance in a parallel circuit is gives a s
Z = RXₐ/√(Xₐ² + R²)............................... Equation 1
Where Z = The impedance of the a.c circuit, Xₐ = capacitive reactance, R = resistance.
Given: Xₐ = 962 Ω, R = 1200 Ω
Substituting these values into equation 1,
Z = 962×1200/√(962² + 1200²)
Z = 1154400/√(925444 + 1440000)
Z = 1154400/√(925444+1440000
Z = 1154400/1538
Z = 750.59 Ω
Z≈ 751 Ω
Therefore the impedance of the circuit = 751 Ω
The right option is A. 751 ohm
Why is it unwise to stir a pot of soup with a metal spoon?
Answer:
To prevent the soup from cooling soon and prevent our hands from high temperature of the spoon by the contact of soup.
Explanation:
It is unwise to stir a pot of soup with a metal spoon because the metal has a much lower specific heat capacity than that of water i.e. it requires lesser heat energy to raise its temperature.Soup being a blend of edible fats and vegetables forms a layer of fat and starch on the surface preventing much of the heat from escaping to the environment and thus delaying the process of loosing the heat to the ambiance.When a spoon is put into soup it becomes an accessible good conductor path for the heat to escape easily as a result it also increases the temperature of spoon significantly due to its low specific heat capacity.What is stellar parallax?a.It describes the fact that stars are actually moving relative to one another, even though to our eyes the stars appear fixed in the constellations. b.It is the change in the set of constellations that we see at different times of year in the evening sky. c.It is the slight back-and-forth shifting of star positions that occurs as we view the stars from different positions in Earth's orbit of the Sun. d.It is the daily rise and set of the stars.
Answer:
Option C
C. It is the slight back-and-forth shifting of star positions that occurs as we view the stars from different positions in Earth's orbit of the Sun.
Explanation:
The angle due to the change in position of a nearby object against the background stars it is known as parallax.
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth(as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km) is defined as 1 astronomical unit.
[tex]d(pc) = \frac{1}{p('')}[/tex] (1)
Equation (1) represents the distance in a unit known as parsec (pc).
Key terms:
Parsec: Parallax of arc second
Stellar parallax is the small apparent shift in the position of a star resulting from the change in the Earth's position as it orbits the sun. This effect is similar to the apparent motion of nearby objects against a distant background when you're moving.
Explanation:Stellar parallax is the slight back-and-forth shifting of star positions that occurs as we view the stars from different positions in Earth's orbit of the Sun. To put it in simple terms, as Earth revolves around the Sun, we see nearby stars from slightly different angles at different times of year. This apparent shift in position of the star is what we refer to as the stellar parallax effect.
Think of this as similar to the parallax effect you get when you're travelling in a car. Objects closer to you seem to move faster than the objects farther away. In the case of stellar parallax, the stars closer to us appear to move against the background of distant stars.
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A string is 9.00 m long with a mass of 7.00 g. The string is held taut with a tension of 200.00 N applied to the string. A pulse is sent down the string. How long (in ms) does it take the pulse to travel the 9.00 m of the string?
Answer:
It takes the pulse 25.1 ms to travel the 9.00 m of the string
Explanation:
Force (F) = Mass( m) × Acceleration.( a)
F = ma............... equation 1
Making a the subject of equation 1 above,
a = F/m .............. equation 2
Where F = T = 200 N, m = 7.00 g = 0.007 kg,
Substituting these values into equation 2,
a = 200/0.007 = 28571.43 m/s².
Using one of the equation of motion,
S = ut + 1/2at²................... equation 2
Where S = distance, u = initial velocity, t = time, a = acceleration.
u = 0.
Therefore, S = 1/2at²................. equation 3
Making t the subject of equation 3,
t = √ (2S/a).................. equation 4
Where S = 9.00 m , a = 28571.43 m/s²
Substituting this values into equation 4,
t = √{(2×9)/28571.43}
t = √(18/28571.43
t = √0.00063
t = 0.0251 s
t = 25. 1 ms
Therefore it takes the pulse 25.1 ms to travel the 9.00 m of the string.
Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.6 m/s how much work does Sam do to stop it?
Answer:
Sam will do 1152 J of work to stop the boat
Explanation:
Work: This is defined as the product of force and distance, the S.I unit of work is Joules. At any point in science, during calculation Energy and worked can be interchange because they have the same unit.
E = W = 1/2mv²................ Equation 1
Where E = energy, W = work, m = mass, v = velocity.
Given: m = 900 kg, v = 1.6 m/s
Substituting these values into equation 1
W = 1/2(900)(1.6)²
W = 450×2.56
W = 1152 J.
Therefore Sam will do 1152 J of work to stop the boat
You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving when it is halfway down the building? (assume no friction)
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 0.5 x 44 = 22 m
Substituting
v² = u² + 2as
v² = 0² + 2 x 9.81 x 22
v² = 431.64
v = 20.78 m/s
Velocity at 22 m = 20.78 m/s
The balloon is moving when it is halfway down the building at 20.78 m/s.
When an alkali metal is involved in a chemical reaction with a nonmetal, what charge will its atoms most likely form? 1- 2- 2+ 1+
Answer:
+1 ion
Explanation:
Alkali metals are metals that are found in Group I of the periodic table. Their electronic configuration is such that their valence shell in grounds state is always holding only one electron which they always lose when reacting with non-metals. A loss in an electron makes the atom electrically imbalanced and hence becoming a +1 ion.
1. A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope attached to a strong tree limb. The child and tube is drawn back until it makes a 17.4-degree angle with the vertical. The child is released and allowed to swing to and from. Assuming negligible friction, determine the child's speed at his lowest point in the trajectory.
2. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of 39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a horizontal distance of 25.1 cm.
Answer
1) given,
mass of child = 21.3 Kg
length of the rope = L = 7.28 m
angle made with vertical = 17.4°
speed of the child at the bottom most point
using conservation energy
[tex]m g h = \dfrac{1}{2}mv^2[/tex]
[tex] v = \sqrt{2gh}[/tex]
H = L cos θ
H = 7.28 x cos 17.4°
H = 6.94
where
L = H + h
h = L - H
h = 7.28 - 6.94 = 0.34 m
now,
[tex] v = \sqrt{2\times 9.8 \times 0.34}[/tex]
v = 2.58 m/s
2) given,
mass of baseball = 163 g = 0.163 kg
initial speed = 39.8 m/s
final speed = 0
horizontal distance = 25.1 cm = 0.251 m
Force = ?
using equation of motion
v = u + at
0 = 39.8 + at
at = -39.8 m/s
using equation
[tex]s = ut + \dfrac{1}{2}at^2[/tex]
[tex]0.251= 39.8 t - 0.5 \times 39.8 t [/tex]
t = 0.0126 s
using formula of impulse
I = F x t
I = m(v - u)
F x 0.0126 = 0.163 x (-39.8)
F = 514.87 N
Final answer:
To determine the child's speed at his lowest point in the trajectory, we can use the conservation of mechanical energy by equating the gravitational potential energy to the sum of the kinetic and elastic potential energy. By using the equations for kinetic energy and tension, we can calculate the child's speed and the force of tension in the rope at the lowest point.
Explanation:
To determine the child's speed at his lowest point in the trajectory, we can use the conservation of mechanical energy. At the highest point, the gravitational potential energy is equal to the sum of the kinetic energy and the elastic potential energy of the rope. At the lowest point, the gravitational potential energy is zero, so the kinetic energy is equal to the elastic potential energy. Using this information, we can calculate the child's speed at the lowest point using the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the child, and v is the child's velocity.
Since we are given the mass of the child and the length of the rope, we can also calculate the force of tension in the rope at the lowest point using the equation:
T = mg + (mv^2)/L
where T is the tension, m is the mass of the child, g is the acceleration due to gravity, v is the child's velocity, and L is the length of the rope.
By using these equations, we can determine the child's speed at his lowest point in the trajectory.
A degausser creates a magnetic field that erases data from magnetic storage media?
Explanation:
Degausser is a device used by automatically manipulating the alignment of magnetic domains on the medium to avoid data stored on computer and laptop hard disks, floppy disks, and magnetic cassette. There was a mistake. Therefore, a degausser is used to remove all audio, video and data signals from magnetic store media completely.
Which telescope has a better (smaller) angular resolution: a 2-m telescope observing visible light (wavelength 5.0×10-7 m) or a 10-m radio telescope observing radio waves (wavelength 2.1×10-2 m)?
The 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Explanation:In order to determine which telescope has a better (smaller) angular resolution, we need to calculate the angular resolution for each telescope. The formula for angular resolution is given by θ ≈ 1.22λ/D, where λ is the wavelength of light and D is the diameter of the telescope's aperture.
For the 2-m telescope observing visible light with a wavelength of 5.0×10-7 m, the angular resolution is approximately 1.22×(5.0×10-7)/(2) = 3.05×10-7 radians.
For the 10-m radio telescope observing radio waves with a wavelength of 2.1×10-2 m, the angular resolution is approximately 1.22×(2.1×10-2)/(10) = 2.54×10-3 radians.
Therefore, the 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Final answer:
The 2-m optical telescope observing visible light has a better (smaller) angular resolution compared to the 10-m radio telescope observing radio waves.
Explanation:
To determine which telescope has a better (smaller) angular resolution, we can use the formula for resolving power (θ), which is given by: θ ≈ λ / D, where λ is the wavelength and D is the diameter of the telescope's aperture. For the 2-m telescope observing visible light (λ = 5.0×10⁻⁷ m), the angular resolution would be θ ≈ 5.0×10⁻⁷ m / 2 m = 2.5×10⁻⁷ radians. For the 10-m radio telescope observing radio waves (λ = 2.1×10⁻² m), the angular resolution would be θ ≈ 2.1×10⁻² m / 10 m = 2.1×10⁻³ radians. Thus, the 2-m optical telescope has a smaller (better) angular resolution than the 10-m radio telescope.
What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Incomplete question.The complete question is
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Answer:
Total gauge pressure= 3724Pa or 3.724kPa
Explanation:
Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.
Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.
Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa
If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Answer:
Venus is definitely the answer
Explanation:
If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Venus is known as the earth twin because it is similar in shape to the earth. Although, it is not the closest to the sun, it traps the suns heat in its atmosphere making it the hottest planet in the solar system. the temperature is as high as 465 degrees Celsius. Venus is known to be the second brightest body in the solar system, the moon being the first. A day in venus could be longer than a year on the earth. It revolves in the opposite direction around the sun compared to other planets. It has an atmospheric pressure which ninety times grater than what we have here on earth.
While troubleshooting a network problem, a technician realized it could be resolved by opening a port on a firewall. The technician opened the port and verified the system was now working. However, an attacker accessed this port and launched a successful attack. What could have prevented this problem?A. Patch management processes
B. Vulnerability management processes
C. Configuration management processes
D. Change management processes
An amplifier has an input power of 5 mW. The output power is 100 mW. What is the decibel gain to the nearest decibel?
Answer:
The decibel gain to the nearest decibel = 13 db
Explanation:
Gain of an amplifier: This is defined as the measure of the ability of an amplifier to increase the power of a signal from the input to the output port by adding energy converted from some power supply to the signal.
It is represented mathematically as.
db = 10×log (p₂/p₁)............................. Equation 1
Where dp = decibel gain of the amplifier, p₁ = input power of the amplifier, p₂ = output power of the amplifier.
Given: p₁ = 5 mW, p₂ = 100 mW.
Substituting these values into equation 1
db = 10×log(100/5)
db = 10×log(20)
dp = 10 ×1.301
dp = 13 db gain
Therefore the decibel gain to the nearest decibel = 13 db
Based on the research of Albert Einstein, what change would most likely result in stopping the emission of electrons from this metal? a) an increase in the intensity of the light b) a decrease in the intensity of the light c) the use of light that has a higher frequency d) the use of light that has a lower frequency.
Answer:
option (d)
Explanation:
In the experiment of photo electric emission, it starts only when the incident energy has some minimum frequency so that the electrons just emits from the cathode surface. Such frequency is called threshold frequency.
So, if we use light of frequency lower than the threshold frequency, no photo electric emission takes place.
Thus, option (d) is correct.
Answer:
the use of light that has a lower frequency
Explanation:
what's your current physics grade
Answer: grade 8 (uk grading system)
When you shoulder your shotgun, what part of your body should fit snugly against the stock?
Answer:
Cheek
Explanation:
Typically the scope is a bead on the gun's edge. Our eye must be in alignment with the muzzle, so that the proper placement of our head on the stock is crucial. The stock will fit snugly to our cheek as we put the pistol to our face with our head on other side just above gun's center line.
The correct answer is cheek.