Answer:
a) 0.750
Explanation:
When the unpolarized light passes through the first polarizer, it becomes polarized along the axis of transmission of the polarizer itself.
Then, the light passes through the second polarizer, whose axis of transmission is inclined by an angle [tex]\theta[/tex] with respect to the direction of polarization of the light.
Calling [tex]I_0[/tex] the initial intensity of the light, the intensity of light passing through the second filter is
[tex]I=I_0 cos^2 \theta[/tex]
where
[tex]\theta=30^{\circ}[/tex]
Solving the formula for [tex]\frac{I}{I_0}[/tex], which is the fraction of the incident intensity transmitted through the second polarizer, we find
[tex]\frac{I}{I_0}=cos^2 \theta = cos^2 30^{\circ}=0.750[/tex]
When unpolarized light passes through two polarizers whose transmission axes are at an angle of 30.0 degrees with respect to each other, the fraction of the incident intensity transmitted through the polarizers is 0.75.
Explanation:When unpolarized light passes through two polarizers whose transmission axes are at an angle of 30.0 degrees with respect to each other, the fraction of the incident intensity transmitted through the polarizers can be calculated using Malus' Law.
Malus' Law states that the intensity of the transmitted light is equal to the initial intensity multiplied by the square of the cosine of the angle between the transmission axes of the polarizers.
In this case, the angle between the transmission axes is 30.0 degrees, so the fraction of the incident intensity transmitted through the polarizers is (cos(30.0))² = 0.75.
A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a car is initially traveling at 20 m/s (45 mi/h), (a) How much time does it take the car to stop? (b) What is its stopping distance? (c) What is the deacceleration? (d) How big is the net force to be applied to stop this car? (e) Calculate the work done by this force (Work = Force * distance). (f) During the stopping process, what happens to the car's kinetic energy?
(a) 4.0 s
The acceleration of the car is given by
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time interval
For this car, we have
v = 0 (the final speed is zero since the car comes to a stop)
u = 20 m/s is the initial velocity
[tex]a=-5.0 m/s^2[/tex] is the deceleration of the car
Solving the equation for t, we find the time needed to stop the car:
[tex]t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s[/tex]
(b) 40 m
The stopping distance of the car can be calculated by using the equation
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 20 m/s is the initial velocity
a = -5.0 m/s^2 is the acceleration of the car
d is the stopping distance
Solving the equation for d, we find
[tex]d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m[/tex]
(c) [tex]-5.0 m/s^2[/tex]
The deceleration is given by the problem, and its value is [tex]-5.0 m/s^2[/tex].
(d) 5000 N
The net force applied on the car is given by
[tex]F=ma[/tex]
where
m is the mass of the car
a is the magnitude of the acceleration
For this car, we have
m = 1000 kg is the mass
[tex]a=5.0 m/s^2[/tex] is the magnitude of the acceleration
Solving the formula, we find
[tex]F=(1000 kg)(5.0 m/s^2)=5000 N[/tex]
(e) [tex]2.0\cdot 10^5 J[/tex]
The work done by the force applied by the car is
[tex]W=Fd[/tex]
where
F is the force applied
d is the total distance covered
Here we have
F = 5000 N
d = 40 m (stopping distance)
So, the work done is
[tex]W=(5000 N)(40 m)=2.0\cdot 10^5 J[/tex]
(f) The kinetic energy is converted into thermal energy
Explanation:
when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).
A car traveling along the highway needs a certain amount of force exerted on it to stop it in a
certain distance. More stopping force is required when the car has
A) more mass.
B) more momentum.
C) less stopping distance.
D) all of these
E) none of these
A car traveling along the highway needs a certain amount of force exerted on it to stop it in a certain distance.
More stopping force is required when the car has more mass, or more momentum, or less stopping distance. (D)
Answer:
More stopping force is required when the car has D) all of these.
Explanation:
Let's explain some equations and concepts in order to answer the question:
The second law of Newton states that a force exerted on an object is directly proportional to the mass of the object and the acceleration of the object. Mathematically, we can write the following equation :[tex][/tex]
[tex]F=m.a[/tex] (I)
Where ''F'' is the force
Where ''m'' is the mass
And where ''a'' is the acceleration.
Now, we can define the momentum as :
[tex]p=m.v[/tex] (II)
The momentum ''p'' is a vector magnitude.
''m'' is the mass of the object
And ''v'' is the velocity vector.
Finally, let's explain the following motion equation :
[tex]Vf=Vi-a.t[/tex] (III)
Vf is final speed
Vi is initial speed
''a'' is the acceleration of the object.
Notice that we write a ''-'' in the ''a.t'' term because we assume that the object is stopping. Therefore, its acceleration is negative. ''t'' is the time in which the object will stop.
Let's proceed analyzing each option :
A) more massIf the car has more mass, therefore by looking at the equation (I), the stopping force will be greater.
B) more momentumBy looking the equation (II), if the car has more momentum therefore it has more mass or more speed (or both).
If it has more mass the stopping force required must be greater.
Otherwise, if it has more speed, by looking at the equation (III) and assuming that Vf = 0 (because we need the car to stop) ⇒
[tex]0=Vi-a.t[/tex]
[tex]Vi=a.t[/tex]
If [tex]Vi[/tex] is greater and assuming that time must be the same, therefore the acceleration will be greater. So, if acceleration increases, the stopping force increases (looking at equation (I) ).
Finally, C) less stopping distanceIf the car has less stopping distance, therefore the magnitude of the acceleration vector must be greater (in order to stop the car faster). By looking the equation (I) we conclude that the stopping force will be greater.
The correct option is D) all of these
Why is it important to have a control setup in an experiment?
Explanation:
A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.
The reflective quality of a surface is known as its
The albedo is an amount that expresses the percentage of radiation a surface reflects with respect to the incident radiation.
In other words:
This amount allows us to know the level of radiation that reflects a surface compared to the total radiation it receives.
According to this, light surfaces such as snow covered ground or white sand will have a higher albedo than dark surfaces such as carbon covered ground. It is also important to note, the albedo will be higher on glossy surfaces than on matte surfaces.
It should be noted that the albedo of the Earth is on average about [tex]37\%[/tex], which means that part of the radiation received by the Sun is absorbed and another part reflected back to space.
A Carnot Engine operates between a hot reservoir temperature of 215 degrees C and a cold reservoir temperature of 20 degrees C. If the engine draws 1000 J from the hot reservoir per cycle, how much work will it do per cycle?
a, 1000 J
b, 100 J
c, 400 J
d, 600 J
e, 900 J
Answer:
c. 400 J
Explanation:
The efficiency of a Carnot Engine is given by:
[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]
where in this case we have
[tex]T_C = 20^{\circ} +273 =293 K[/tex] is the temperature of the cold reservoir
[tex]T_H = 215^{\circ} +273 =488 K[/tex] is the temperature of the hot reservoir
Substituting into the equation,
[tex]\eta = 1 - \frac{293 K}{488 K}=0.40[/tex]
But the efficiency can also be written as
[tex]\eta = \frac{W_{out}}{Q_{in}}[/tex]
where
[tex]W_{out}[/tex] is the useful work in output
[tex]Q_{in}[/tex] is the heat absorbed by the hot reservoir
Here,
[tex]Q_{in} = 1000 J[/tex]
So solving the formula for [tex]W_{out}[/tex] we find
[tex]W_{out} = \eta Q_{in} = (0.40)(1000 J)=400 J[/tex]
The Carnot Engine absorbs 1000 Joules from the hot reservoir and has an efficiency of 0.4. The work done by the engine per cycle is the product of the absorbed heat and the efficiency, which equates to 400 Joules.
Explanation:The efficiency of a Carnot Engine is determined by the temperatures of the hot and cold reservoirs. Specifically, efficiency (η) = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Note that these temperatures must be in Kelvin for the formula to work properly.
In this case, we need to convert the temperatures from degrees Celsius to Kelvin: Th = 215°C + 273.15 = 488.15 K and Tc = 20°C + 273.15 = 293.15 K. We then substitute these values into the formula to get the efficiency: η = 1 - (293.15 K /488.15 K) ≈ 0.4
The work done by the engine (W) is the product of the heat (Q) absorbed from the hot reservoir and the efficiency: W = η x Q. Substituting the given heat of 1000 Joules and the calculated efficiency, we get W = 0.4 x 1000 Joules = 400 Joules. Therefore, the amount of work done per cycle by the engine is 400 Joules (Option c).
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Watt which is a unit of power is equal to
Answer:
A joule divided by a second
Explanation:
i googled it m8 wasnt that hard to find
Answer: a joule divided by a second
Explanation:
Determine the wavelength of light having energy of 2.25 × 10−19 j.
Answer:
[tex]8.84\cdot 10^{-7} m[/tex]
Explanation:
The energy a single photon of an electromagnetic wave is given by
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
In this problem, we have
[tex]E=2.25\cdot 10^{-19} J[/tex] is the energy of the photon
So we can re-arrange the equation to find the wavelength:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.25\cdot 10^{-19} J}=8.84\cdot 10^{-7} m[/tex]
By what mechanism is a person injured when he or she falls from a significant height? Select one: a. Kinetic energy is converted to potential energy; the potential energy is then converted into the work of bringing the body to a stop. b. As the person falls, the amount of kinetic energy is converted into work; work is then converted to kinetic energy upon impact. c. Potential energy is created as the person is falling; the potential energy is then converted into kinetic energy upon impact. d. Potential energy is converted to kinetic energy; the kinetic energy is then converted into the work of bringing the body to a stop.
Answer:
d. Potential energy is converted to kinetic energy; the kinetic energy is then converted into the work of bringing the body to a stop.
Explanation:
- At the beginning of the falls, when the person is still at a certain height h, the person has gravitational potential energy:
U = mgh
where m is the mass of the person, g the acceleration due to gravity, h the height above the ground.
- As the person falls down, h decreases, so the potential energy decreases; according to the law of conservation of energy, potential energy is converted into kinetic energy, since the speed of the person increases:
[tex]K=\frac{1}{2}mv^2[/tex]
where v is the speed.
- Just before hitting the ground, all the potential energy has been converted into kinetic energy
- When the person hits the ground, he/she comes to a stop: so work is done by the ground on the person, because the ground applied a force required to stop the person, and the kinetic energy "lost" by the person is equal to the work done by the ground to bring the body to a stop.
1. Provide one example of an area with high ecosystem diversity and one example of an area with low ecosystem diversity.
Explanation:
An example of a high ecosystem diversity is a rainforest or seashore.
An example of a low ecosystem diversity is a desert or farmland.
The rainforests are considered as an example of high ecosystem diversity and on the other hand, typical deserts and farmlands are considered as low ecosystem diversity.
What is an ecosystem?A geographical area where one can observe plants, animals, and other living organisms to reside, forming a part of a complete landscape, is known as an ecosystem.
An example of a high ecosystem diversity is a rainforest or seashore. As one in the Amazon basin in South America, rainforests provide varieties of biodiversity like coral reefs, species of snakes, fishes, monkeys, and many others.
An example of a low ecosystem diversity is a desert or farmland. These parts of land generally have only two to three varieties of grasses, some dandelion flowers, and few inhabitants.
Thus, we can conclude that the rainforests are considered as an example of high ecosystem diversity and on the other hand typical deserts and farmlands are considered as low ecosystem diversity.
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