According to the problem the wavelength of the light is 5.67 μm.
What is wavelength?Wavelength is the distance between one point on a wave and the next corresponding point of the same phase on the wave. It is usually measured in meters (m) or nanometers (nm). Wavelength is closely related to frequency, as the frequency of a wave is equal to the speed of the wave divided by the wavelength. Wavelengths are used to measure various forms of electromagnetic radiation, such as visible light, radio waves, and x-rays. The wavelength of a wave determines its color, as visible light is made up of a spectrum of different wavelengths.
The angle of diffraction of a given wavelength of light through a grating is given by the equation: θ = (m*λ)/d
where θ is the angle of diffraction, m is the order of diffraction, λ is the wavelength of the light and d is the line spacing of the grating.
In this case, m = 2 (second order principle maximum), d = 300 lines/mm and θ = 18.9 degrees.
Solving for λ gives us: λ = (d*θ)/m
Substituting in the values gives us: λ = (300*18.9)/2 = 5670 nm = 5.67 μm
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. The air bubble formed by explosion inside water perform oscillations with time period, T which depends on pressure (P) density (p) and on energy due to explosion (E). establish the relation between T, P, E and p
The time period of oscillations of an air bubble formed by an explosion inside water depends on the pressure (P), density (p), and energy due to explosion (E). The relationship between these factors can be established using the principles of fluid mechanics and the equations for pressure, density, and energy.
The pressure of a gas inside a bubble is related to its volume and temperature, and can be described by the Ideal Gas Law:
P = (nRT)/V
where n is the number of moles of gas, R is the gas constant, T is the temperature, and V is the volume of the gas.
The density of a fluid is related to its mass and volume, and can be described by the equation:
p = m/V
where m is the mass of the fluid and V is its volume.
The energy due to the explosion can be described by the equation:
E = (1/2)mv^2 + P_0V_0 - P_fV_f
where m is the mass of the gas, v is its velocity, P_0 and V_0 are its initial pressure and volume, and P_f and V_f are its final pressure and volume.
By combining these equations and using the principles of fluid mechanics, it is possible to establish a relationship between the time period of the oscillations (T), pressure (P), energy due to the explosion (E), and density (p). However, the exact relationship will depend on the specific conditions of the explosion and the fluid in which it occurs.
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suppose that the cable from a to b must exert a 8500 n horizontal force on the car to hold it in place. determine the car's weight and also mass in kg
The car's weight and mass in kg if cable from a to b must exert a 8500 n horizontal force on the car to hold it in place is 14722N and 1502.24 kg.
Believe or not, weight is the force that the earth Earth is acting on you. The gravitational acceleration which is multiplied to the mass is deduced by using the equation of Newton's law of solemnity similar that one mass is the mass of the earth Earth.
The free body diagram is shown to the right
Applying the equilibrium equation
[tex]\sum F_s = T- Nsin30\degree = 0\\\\\sumF = Ncos30 - mg = 0\\[/tex]
Setting T = 8500N and solving the equation we get,
N = 17000N
mg = 14722 N
So weight = 14722 N
and mass = 14722/g = 14722/9.8 = 1502.24 kg
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The dependence of the rate constant on temperature is expressed by which equation The Arrhenius equation. The de Broglie equation. The van't Hoff equation. Temperature has no effect on the rate constant.
The dependence of the rate constant on temperature is expressed by the Arrhenius equation.
What is Arrhenius equation?The Arrhenius equation is an equation that is used to describe the relationship between the reaction rate of a chemical reaction and the temperature at which the reaction takes place. The equation states that the reaction rate is equal to the frequency factor (A) multiplied by the product of the Boltzmann constant (k) and the temperature (T) raised to the power of the activation energy (E): rate = A * kT^E.
Therefore, The dependence of the rate constant on temperature is expressed by the Arrhenius equation.
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a 900 kg car driving 20 m/s collides with a 1,100 kg truck going 15 m/s. The vehicle bounce off each other and the car moves back 18 m/s. Whats the velocity of the truck after the collision?
Answer: The velocity of the truck after the collision is 18.2 m/s.
Explanation:
Step 1: Determine the total momentum of the two vehicles before the collision.
Momentum is equal to mass multiplied by velocity (p = mv). The momentum of the car before the collision was 900 kg x 20 m/s = 18,000 kg m/s. The momentum of the truck before the collision was 1,100 kg x 15 m/s = 16,500 kg m/s. The total momentum of the two vehicles before the collision was 34,500 kg m/s.
Step 2: Determine the total momentum of the two vehicles after the collision.
After the collision, the car moves back 18 m/s so the momentum of the car is now 900 kg x (-18 m/s) = -16,200 kg m/s. The momentum of the truck after the collision is not given, so we will use the letter "v" to represent this. The total momentum of the two vehicles after the collision is then -16,200 kg m/s + (1,100 kg x v).
Step 3: Use the law of conservation of momentum to solve for the velocity of the truck.
The law of conservation of momentum states that the total momentum of two objects before a collision is equal to the total momentum of two objects after the collision. This means that 34,500 kg m/s = -16,200 kg m/s + (1,100 kg x v), or 34,500 kg m/s = -16,200 kg m/s + 1,100v. Since we know all of the values on the right side of the equation, we can solve for v by rearranging the equation. We get v = (34,500 kg m/s + 16,200 kg m/s) / 1,100 kg = 18.2 m/s.
Therefore, the Step 1: Determine the total momentum of the two vehicles before the collision.
Momentum is equal to mass multiplied by velocity (p = mv). The momentum of the car before the collision was 900 kg x 20 m/s = 18,000 kg m/s. The momentum of the truck before the collision was 1,100 kg x 15 m/s = 16,500 kg m/s. The total momentum of the two vehicles before the collision was 34,500 kg m/s.
Step 2: Determine the total momentum of the two vehicles after the collision.
After the collision, the car moves back 18 m/s so the momentum of the car is now 900 kg x (-18 m/s) = -16,200 kg m/s. The momentum of the truck after the collision is not given, so we will use the letter "v" to represent this. The total momentum of the two vehicles after the collision is then -16,200 kg m/s + (1,100 kg x v).
Step 3: Use the law of conservation of momentum to solve for the velocity of the truck.
The law of conservation of momentum states that the total momentum of two objects before a collision is equal to the total momentum of two objects after the collision. This means that 34,500 kg m/s = -16,200 kg m/s + (1,100 kg x v), or 34,500 kg m/s = -16,200 kg m/s + 1,100v. Since we know all of the values on the right side of the equation, we can solve for v by rearranging the equation. We get v = (34,500 kg m/s + 16,200 kg m/s) / 1,100 kg = 18.2 m/s.
Therefore, the velocity of the truck after the collision is 18.2 m/s.
under favorable circumstances including reaction time
A car with adequate brakes traveling at 50 mph can be stopped within: 229 feet under ideal conditions, including response time.
How much time does it take to respond?Reaction time is a term used to describe how rapidly an organism responds to a stimuli (RT). The length of time (RT) that passes between the stimulus's presentation and the emergence of the subject's appropriate voluntary response is calculated.
Is IQ a factor in reaction time?Reaction time is a more straightforward measure of the brain's information-processing capacity than intelligence, and it has a mildly positive relationship with IQ. It also has little to do with other, potentially confusing aspects like knowledge, education, or history.
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Based on astronomy from space-based telescopes, select all of the correct statements from the following list.a. Most types of electromagnetic waves are not visible to ground-based telescopes.b. There would be no limit to what an optical telescope in space could see.c. Enough infrared energy gets to Earth's surface that infrared telescopes do not need to be put in orbit.d. The Chandra and Spitzer telescopes are all in space.
Statements which are correct are (a) i.e. Telescopes on the ground cannot see the majority of electromagnetic waves. & (d) The Chandra and Spitzer telescopes are all in space.
(a) Most types of electromagnetic waves, such as X-rays and gamma rays, are not visible to ground-based telescopes due to the absorption of these waves by the Earth's atmosphere. Therefore, to detect these types of electromagnetic waves, space-based telescopes are needed.
(d) This statement is correct. The Chandra X-ray Observatory and the Spitzer Space Telescope are both space-based telescopes designed to observe X-rays and infrared radiation, respectively. Since X-rays and infrared radiation are not visible to ground-based telescopes due to the Earth's atmosphere, space-based telescopes are necessary to detect them.
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In terms of environmental impact, what’s the difference between natural gas, coal, and biomass? Consider both carbon dioxide (CO2) emissions and acid rain
In terms of producing the most energy per unit of carbon dioxide released, natural gas is the best fossil fuel. Because a new crop can be planted after each harvest, biomass is a renewable resource and a low-carbon fuel.
Effects of human activities on the biophysical environment, or environmental challenges, most frequently include. Numerous human activities such as overpopulation, pollution, the burning of fossil fuels, and deforestation have an adverse effect on the physical environment. Climate change, soil erosion, poor air quality, and undrinkable water have all been brought on by changes like these.
There are three main ones that generally have an impact on most of them: the loss of biodiversity, water pollution, ocean acidification, and climate change
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consider two people, one on the earth's surface at the equator and the other at the north pole. which has the larger centripetal acceleration? explain.
Both at the north pole and one at the equator on the surface of the globe. The centripetal acceleration is greater for the person on the equator.
The centripetal acceleration is the acceleration that is directed towards the center of a circular path. On the surface of the Earth, the rotation of the planet around its axis creates a circular path for any object on its surface. This means that any object on the Earth's surface is subject to a centripetal acceleration due to the Earth's rotation.
The magnitude of the centripetal acceleration:-
a = v²/r
here, v tangential velocity of the object
r radius of circular path.
The tangential velocity is velocity of the object in the direction tangent to the circular path.
The tangential velocity of an object on the surface of the Earth is given by:-
v = ω r
here, ω is angular velocity of the Earth's rotation,
r is radius of the Earth.
The angular velocity is rate at which the Earth rotates around its axis.
Since, the radius of the Earth is largest at the equator, an object at the equator has a larger radius than an object at the North Pole. This means that the tangential velocity of an object on the equator is larger than the tangential velocity of an object at the North Pole. Therefore, the centripetal acceleration of an object on the equator is larger than the centripetal acceleration of an object at the North Pole.
In summary, an object on the Earth's surface experiences a centripetal acceleration due to the Earth's rotation, and the magnitude of this acceleration is determined by the tangential velocity and the radius of the circular path. Since the radius is larger at the equator than at the North Pole, an object on the equator has a larger centripetal acceleration than an object at the North Pole.
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when loading a trailer, what percent of the weight should be in the front?
The percentage of weight that should be in front of the trailer is 60%
According to the GMC Trailering Guide, to get the right trailer tongue weight, you should concentrate about 60% of the load evenly on the front half of the trailer. 60% of the trailer load should be in front of the axle and 40% behind it. This ensures the correct weight of the tongs and prevents the trailer from rocking. Weight distribution should be as efficient as possible by placing 60% of the weight in the front of the trailer and 40% in the rear. There are some additional safety precautions to consider.
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The amount of incoming solar radiation available at various places on the earth depends mainly onlongitude.the angle at which the sun's rays strike the earth and the number of daylight hours.the earth's rotation.the absorbance of the earth surface material.
"The amount of incoming solar radiation available at various places on the earth depends mainly on the angle at which the sun's rays strike the earth and the number of daylight hours." Correct option is B.
The amount of incoming solar radiation that is accessible varies on the latitude, day of the year, time of day, and amount of cloud cover. The amount of solar energy received at a particular latitude would not change throughout the year if the earth were not tilted from the perpendicular.
The incoming solar energy received on a daily basis depends primarily on the angle of the Sun above the horizon, the length of time the surface is exposed to the Sun, and atmospheric conditions.
Following the latitude of the Earth we observe that the sun’s rays are more intense at the equator and they become less intense as we approach the poles. At the same time this intensity fluctuates during the seasons due to the Earth's tilted axis (23.5 degree angle), this makes that the hemisphere that is closer to the sun receives more sunlight than the opposing hemisphere.
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Answer: THE ANSWER IS B
a car starts from rest and accelerates at 0.300 m/s2 . what is the speed of the car after it has travelled 25.0 m?
The speed of the car after it has traveled 25.0 m is 3.87 m/s. when the car starts from rest and accelerates at 0.300 m/s2.
At rest, is acceleration zero?A vector quantity is acceleration. We must provide retar-dation—a negative acceleration—to the body in order for it to come to rest. The body's velocity is zero when it comes to a stop. The outcome is that the acceleration is also zero.
Is acceleration zero when at rest?Acceleration is a vector quantity. The body needs to experience retar-dation—a negative acceleration—in order to come to rest. When a body comes to a stop, its velocity is zero. As a result, the acceleration is zero as well.
v^2 = u^2 + 2as
v = final velocity
u = initial velocity = zero [start from rest]
a = acceleration = 0.300 m/s2
s = distance = 25.0 m
v^2 = 0 + 2 [0.300] [25.0]
v^2 = 15
v^2 = 3.87 m/s
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describe the significance of laser irradiance and exposure in vp processes. how do these phenomena affect the solidification, scan pattern, and quality of a vp build? what is the difference between irradiance and exposure?
Laser irradiation of a solution is an effective way to generate crystal nucleinuclei.
Vat polymerisation employs a vat of liquid photopolymer resin in which the model is built layer by layer. Where necessary, an ultraviolet (UV) light is used to cure or harden the resin, while a platform moves the object being made downwards after each new layer is cured.
Because the process employs liquid to form objects, there is no structural support from the material during the build phase, as opposed to powder-based methods, which provide support from the unbound material. In this case, additional support structures are frequently required. Resins are cured using photopolymerisation (Gibson et al., 2010) or UV light, in which the light is directed across the surface of the resin using motorised mirrors (Grenda, 2009). The resin cures or hardens where it comes into contact with light.
The layer thickness lowers the build platform from the top of the resin vat downwards.
The resin is cured layer by layer with a UV light. The platform continues to descend, and new layers are added on top of the previous ones.
Some machines use a blade that moves between layers to provide a smooth resin base on which to build the next layer.
After completion, the resin is drained from the vat and the object is removed.
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Do some research and find out the acceleration due to gravity on the Moon and on each planet
listed in the table. For each site, calculate the gravitational force that would act on a 70-kilogram
person on the surface of each site. Use your calculations to complete the table. Remember to
use the same units in each calculation.
Planet or Moon
Venus
Earth
Earth's Moon
Mars
Jupiter
Acceleration due to gravity
Force experienced by 70 kg
person
The acceleration due to gravity varies according to the size of the planet.
What is the list of the acceleration due to gravity on the moon and on all the planets?Here is a list of the acceleration due to gravity (g) on the Moon and on the planets in our Solar System:
Moon:
g = 1.62 m/s^2
Mercury:
g = 3.7 m/s^2
Venus:
g = 8.87 m/s^2
Earth:
g = 9.8 m/s^2
Mars:
g = 3.71 m/s^2
Jupiter:
g = 24.79 m/s^2
Saturn:
g = 10.44 m/s^2
Uranus:
g = 8.87 m/s^2
Neptune:
g = 11.15 m/s^2
It's worth noting that the acceleration due to gravity is affected by the mass and size of a celestial body. Larger and more massive objects have a stronger gravitational pull and therefore a higher acceleration due to gravity.
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what is the acceleration two point on a velocity time graph which has coordinate (10sec,15ms)and (20sec,35ms)
The acceleration between two point on a velocity time graph which has coordinate (10sec,15ms)and (20sec,35ms) is 2.5m/s².
What is Acceleration?
This is referred to as the rate at which velocity changes with time, in terms of both speed and direction.
Acceleration = change in velocity/change in time
= 35m/s - 15m/s / 20s - 10s
= 25m/s / 10s = 2.5m/s²
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over the course of a single night, does the position of mars in its home constellation change dramatically? group of answer choices yes no
Over the course of a single night, the position of mars in its home constellation do not change dramatically. Correct answer us no.
The position of Mars in its home constellation changes over the course of a single night, but not dramatically. Mars, like all celestial objects, appears to move across the night sky due to the rotation of the Earth.
However, the magnitude of this change in position over the course of a single night is not usually considered to be "dramatic" in astronomical terms. The apparent motion of Mars across the night sky is relatively slow, and its position relative to the surrounding stars may only change by a few degrees over the course of several hours.
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if the mass of the compact car is 1200 kg, what force is needed to make it accelerate to 20 m/s2? show your calculations.
a potter's wheel moves uniformly from rest to an angular speed of 0.19 rev/s in 34.0 s. (a) find its angular acceleration in radians per second per second.
The angular acceleration of the potter's wheel that moves uniformly is [tex]0.0352 radians/s^2[/tex]
Angular acceleration is a measure of how quickly an object's angular velocity changes with respect to time. In this case, the potter's wheel undergoes a uniform acceleration, which means that its angular acceleration remains constant over time.
We can use the following equation to find the angular acceleration of the potter's wheel:
angular acceleration = (final angular speed - initial angular speed) / time
Here, the initial angular speed is zero because the wheel starts from rest. The final angular speed is 0.19 rev/s. The time taken to reach this speed is 34.0 s. We can convert the final angular speed to radians per second using the conversion factor 1 rev/s = 2π radians/s:
final angular speed = 0.19 rev/s * 2π radians/rev = 1.196 radians/s
Put values:
angular acceleration = (1.196 radians/s - 0 radians/s) / 34.0 s = [tex]0.0352 radians/s^2[/tex]
Therefore, the angular acceleration of the potter's wheel is [tex]0.0352 radians/s^2[/tex]
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a curve in a road has a bank angle calculated and posted for 80 km>h. however, the road is covered with ice, so you cautiously plan to drive slower than this limit. what might happen to your car? why?
The automobile may slide or skid sideways off the road if the road is covered with ice and the vehicle is being driven at a slower rate of speed than the indicated bank angle.
A vehicle is kept driving in a circular path without sliding by the centripetal force created by the bank angle of a road bend.
This force is produced by the tire-perpendicular normal force of the road, which acts on the tire.
The bank angle is intended to boost the normal force to counteract the centrifugal force that seeks to pull the automobile away from the curve as speed rises.
On an ice road, however, there is a considerable reduction in the coefficient of friction between the tires and the surface, which also affects the normal force.
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A ball is thrown straight up into the air with initial velocity 62.8 m/s. How fast is the ball moving when it first reaches 56.4 meters?
The final velocity of the ball is equal to 53.06 m/s.
Gravity's acceleration is always constant and downward, but the direction and magnitude of velocity vary. The ball has zero velocity at its highest point in its trajectory, and the magnitude of velocity increases again as the ball falls back towards the earth.
In free fall, an object experiences an acceleration of -9.8 m/s/s. (A downward acceleration is indicated by the - sign.) Whether explicitly stated or not, the acceleration in the kinematic equations for any freely falling object is -9.8 m/s/s.
By using the equation:-
[tex]mgh_{A} + \frac{1}{2} mv^{2} _{A} = mgh_{B} + K.E.[/tex]
We are given:-
[tex]V_{A} =(62.8)m/s[/tex]
g= 9.8 or 10 [tex]m/s^{2}[/tex]
[tex]h_{B} = 56.4[/tex]m.
Putting the values in the equation, we get:-
[tex]\frac{1}{2} mV^{2} _{A} = mg*56.4+\frac{1}{2} mV^{2} _{B} \\=2*(\frac{1}{2} *(62.8)^{2} )= 10*56.4+ \frac{1}{2} V^{2} _{B} \\= (3943.84= 1128+ V^{2} _{B} )\\V_{B} = \sqrt{2815.84} = 53.06 m/s.[/tex]
Hence, the final velocity of the ball is equal to 53.06m/s.
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can someone please help me answer this
Answer:
Rₓ = -10 N
Explanation:
The x-component of a vector R can be found using the following formula:
[tex]\boxed{{R_x = Rcos \ \theta}}[/tex],
where Rₓ is the x-component of R and Θ is the angle between the vector and the positive x-axis.
The 60° angle in the diagram is formed between the vector R and the negative x-axis. The angle we need, Θ, is its supplementary angle. Therefore,
Θ = 180° - 60°
= 120°
Now that we have the value of Θ, we can find its x-component using the formula above:
[tex]{{R_x = Rcos \ \theta}}[/tex]
⇒ [tex]R_x = 20 \times cos(120^{\circ})[/tex]
⇒ [tex]R_x = \bf -10[/tex]
Therefore, the x-component of R is -10 N.
While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself? Neglect air resistance.
While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself? Neglect air resistance.
A. Straight up
B. Slightly backward
C. It is impossible.
D. Slightly forward
If you are running at a constant velocity, you should throw the ball slightly forward with respect to you so that you can catch it yourself. Therefore, option D is correct.
What is constant velocity?An object is said to be moving at constant velocity when it is moving straight and fast. As a result, the item is not accelerating, which is defined as a change in velocity over time in either direction or speed. In other words, there is no net force acting on an item that is travelling at a constant speed.
If there are no forces acting on an object or if those forces are balanced, the object can have a constant velocity. A car is going with constant velocity, for instance, if it is moving at a constant speed in a straight path on a flat road with no external forces acting on it. Likewise, if a thing is going straight line at a constant speed through a vacuum, with no external forces acting upon it, then it is also moving with constant velocity.
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. two blocks are connected by a massless rope as shown below. the mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. the table and the pulley are frictionless. (a) find the acceleration of the system. (b) find the tension in the rope. (c) find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.
A- The acceleration of the system is 1.96 [tex]m/s^{2}[/tex], B- the tension in the rope is 7.84N, C- the velocity at which the hanging mass strikes the ground if it begins at rest and is positioned 1.0 meters off the ground is 1.98 m/s.
A- both the masses will have same magnitude of acceleration a ,
m2g-T = m2a
T=m1a
from both the equations a = m2g/(m1 + m2) = 1.96[tex]m/s^{2}[/tex]. Therefore, The acceleration of the system is 1.96 [tex]m/s^{2}[/tex]
B- T = m1a = 7.84 N is the tension in the rope
C- [tex]V_{f^{2} }[/tex] = [tex]V_{i^{2} }[/tex] + 2as
= 0 + 2am
= 1.98 m/s
Hence , the velocity at which the hanging mass strikes the ground if it begins at rest and is positioned 1.0 meters off the ground is 1.98 m/s.
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The complete question is :
As seen in the illustration below, a massless rope links two blocks. The hanging mass is 1.0 kg, whereas the block on the table weighs 4.0 kg. No friction exists between the table and the pulley. Find the system's acceleration in (a). (b) Determine the rope's tension. (c) determine the speed at which the hanging mass will strike the ground if it begins at rest and is 1.0 m from the ground.
how long will it take the goose to cover a ground distance of 550 km from north to south? (note: even on cloudy nights, many birds can navigate using the earth's magnetic field to fix the north-south direction.)
The goose will take 8.55 hours to travel 550 km at their average speed of 64.3km/h.
The goose average speed is around 40 mph or 64.3 km/h, but in good weather and wind conditions this speed can go up to 70 mph or 112 km/h
Considering the formula:
x = v.t
were x=space, v=speed and t= time. Clearing the equation, to find the hours of flight:
t = x/v = 550km / 64.3km/h = 8.55 hours
Average speed is a useful concept in many fields, including physics, engineering, and transportation. It can be used to calculate travel times, estimate fuel consumption, and analyze the performance of machines and vehicles. However, it is important to note that average speed does not take into account changes in direction or acceleration, which can affect the overall performance of an object.
Average speed is a measure of how quickly an object moves over a certain distance. It is defined as the total distance traveled divided by the time it takes to travel that distance. The units of average speed are usually expressed as distance per unit time, such as kilometers per hour, miles per hour, or meters per second.
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Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 31.3 m/s. He is momentarily blinded by the sunshine. When he recovers, he sees that he is fast overtaking a camper ahead in his lane moving at the slower speed of 13.5 m/s. He hits the brakes as fast as he can (his reaction time is 0.39 s). If he can decelerate at 3.0 m/s2, what is the minimum distance between the driver and the camper when he first sees it so that they do not collide?
The minimum distance between the driver and the camper when he first sees it so that they do not collide is 16.86 m.
What is distance?Distance is the physical length between two points. It is a measure of the space between two objects or a point and a line. Distance can be measured in various ways, such as in meters, kilometers, feet, inches, miles, and even light-years. Distance can also refer to the amount of time or effort required to reach a destination.
The minimum distance between the driver and the camper when he first sees it so that they do not collide is calculated using the following formula:
d = vₒt + (1/2)at²
Where d is the distance, vₒ is the initial velocity, t is the reaction time, and a is the deceleration.
Plugging in the given values, we get:
d = 31.3 m/s × 0.39 s + (1/2) × 3.0 m/s² × (0.39 s)²
d = 16.86 m
Therefore, the minimum distance between the driver and the camper when he first sees it so that they do not collide is 16.86 m.
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a 0.5kg cart is connected horizontally to a spring. there is friction, which damped the amplitude of the motion of the cart with a coefficient of 0.1. if the cart is pulled 2m from the equilibrium position and released. what is the force on the cart by the spring after 2 sec, if the frequency of motion is 1.8hz?
The force on the cart by the spring after 2 sec is -906.4 N.
The motion of the cart connected to a spring can be described by the equation:
x = A × [tex]e^-bt/2m}[/tex] × cos(wt - phi)
where:
x is the displacement, A is the amplitude of the motion b is the damping coefficient m is the mass of the cart w is the angular frequency of the motion phi is the phase angle
To find the force on the cart after 2 seconds, we need to find the values of the amplitude, angular frequency, and phase angle. The amplitude can be found from the initial displacement:
A = 2 m
The angular frequency can be found from the frequency of motion:
w = 2πf = 2π(1.8 Hz) ≈ 11.31 rad/s
The phase angle can be found from the initial conditions. At t = 0, the displacement is 2 m and the velocity is 0, so the phase angle is 0. Therefore: phi = 0
Now we can find the displacement of the cart at t = 2 s:
x = A × [tex]e^-bt/2m}[/tex] × cos(wt - phi)
x = 2 ×[tex]e^{-0.1(2)/2(0.5)}[/tex] × cos(11.31(2) - 0)
x ≈ 1.41 m
The force on the cart by the spring can be found using Hooke's Law, which states that the force is proportional to the displacement:
F = -kx
where k is the spring constant. The spring constant can be found from the angular frequency:
w = √(k/m)
k = m × w²
k = 0.5 kg × (11.31 rad/s)²
k ≈ 642.6 N/m
Therefore, the force on the cart at t = 2 s is:
F = -kx
F = -642.6 N/m × 1.41 m
F ≈ -906.4 N
Note that the negative sign indicates that the force is acting in the opposite direction of the displacement (i.e. restoring force).
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How are atoms in a molecule held together?
through shared electrons
through shared neutrons
through shared protons
through shared energy
The surface of the Sun has a temperature of about 5,800 K. The radius of the Sun is 6.96 108 m. Calculate the total energy radiated by the Sun each second. Assume that the emissivity of the Sun is 0.986. 1 W
The total energy radiated by the Sun each second when emissivity of the Sun is 0.986 is 3.85×10²⁶ W .
The power output of the Sun is given by Stefan's law , i.e
P=σAeT 4
=(5.6696×10⁻⁸ W/m².K⁴ )[4π (6.96×10⁸m)² ]×(0.986)(5800K)⁴
=3.85×10²⁶ W
The total energy radiated by the Sun each second when emissivity of the Sun is 0.986 is 3.85×10²⁶ W .
The radiation radiated from an area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature, according to Stefan Boltzmann's law.
The total energy emitted or radiated by a black body per unit surface area across all wavelengths and per unit time is inversely related to the black body's thermodynamic temperature to the fourth power.
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he fact that the elements of life were produced in stars suggests the following two things: a. (1) these elements have become more common as the universe has grown older. (2) these elements are the most common elements in the universe. b. (1) the universe began in a big bang. (2) the elements of life exist in all stars. c. (1) these elements only exist in very old stars. (2) these elements are extremely rare. d. (1) these elements have become more common as the universe has grown older. (2) these elements should be found in nearly all star systems.
The correct answer is (d): (1) these elements have become more common as the universe has grown older. (2) these elements should be found in nearly all star systems.
The elements that make up life (such as carbon, nitrogen, oxygen, and iron) are created in stars through nuclear fusion and supernova explosions. As the universe has grown older, more and more stars have been born and died, producing these elements and spreading them throughout the universe. Therefore, these elements have become more common over time.
Additionally, because these elements are created in stars, they should be found in nearly all star systems. While the exact abundance of these elements may vary depending on the age and type of star, they should be present in some amount in most stars.
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the tortoise and the hare are running a 1 km race. after running comfortably for 7 s, the hare is so far ahead that he decides to take a nap under a tree, 100 m away from the finish line. if the tortoise is moving constantly at a speed of 0.27 m/s, and the maximum speed of the hare is 15 m/s, how long can the hare afford to nap if he does not want to lose the race?(a) 6.67 s(b) 370 s(c) 3630 s(d) 3690 s
Hare can afford to nap if he does not want to lose the race is of 3690 seconds.
A unit rate is a cost for only one of anything. Unit rate is expressed as a ratio with a denominator of 1. For instance, if you covered 70 yards in 10 seconds, you did so at an average speed of 7 yards per second. Although both of the ratios—70 yards in 10 seconds and 7 yards in one second—are rates, only the latter is a unit rate.
D = 1000m
v = 0.27 m/s
t = 1000/0.27
t = 3703.7 sec
Time taken for 100m for hare.
t = 100/15
t = 6.7 sec
Therefore, total time t is
t = 7 + tn + th
3703.7 = 7 + tn + 6.7
tn = 3703.7 - 13.7
tn = 3690 sec
Therefore, hare afford to nap if he does not want to lose the race is 3690 sec.
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if a voltage of 13 v moves 1.25 c of charge between two points? round the final answer to one decimal place.
If a voltage of 12 V moves 1.25C of charge between two points then charge must be moved with 15 J in joules of energy.
Here, we need to determine how much energy 12 V expends to transport a charge of 1.25 C between two places.
Voltage, which is defined as the effort or energy necessary to transfer a unit charge between two places, is known to drive electrons in a circuit. It can be calculated mathematically by dividing work/energy by charge.
Thus, we have the equation shown below:
[tex]V= \frac{E}{Q}[/tex]
If we rearrange the formula above, we obtain:
[tex]E=(V).(Q) (1)[/tex]
Here,
V→ Voltage
E→ Energy
Q→ Charge
Once the values are entered into equation number 1, we will obtain:
[tex]E = ( 12 V).( 1.25 C ) = 15 J[/tex]
As a result, we discovered that the amount of energy required to move the charge is 15 J in joules.
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Correct question:
How much energy is expended, in joules, if a voltage of12 V moves 1.25C of charge between two points?
