Answer:
45 W/m^2
Explanation:
Intensity of light, Io = 90 W/m^2
According to the law of Malus
The average value of Cos^θ is half
So, I = Io/2
I = 90 /2
I = 45 W/m^2
Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².
Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.
Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².
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0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts
The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.
Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.
The given data in the problem is;
V is the voltage = ? Volt (V)
Electric current (I)= 15.8 amps (A)
P is the power =10.0 kilowatts =10⁴ watt
The formula for the power is given as;
The voltage of the electricity will be 63.29 V.
Hence, option B is correct.
To learn more about the electric power, refer to the link;
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Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.
10,000 W / 15.8 A = V
632.9, or 633.
Answer: i think you should place it on the red line
Explanation:
hope this helps
and need brainliest
Answer:13.6 cm
Explanation:
Given
v(image distance)=-8.5 m
height of object=6 mm
height of image =37.5 cm
and magnification of concave mirror is given by
u=13.6 cm
so object is at a distance of 13.6 cm from mirror.
for focal length
f=-13.4 cm
thus radius of curvature of mirror is R=2f=26.8 cm
The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.
To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:
1/f = 1/do + 1/di
Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.
With the given information, we have:
do = ?
di = 8.50 m
Using the magnification formula:
magnification = -di/do
By substituting the values we know, we can solve for do:
37.5 cm / 6.00 mm = -8.50 m / do
Solving for do, we find that do ≈ - 0.85 m.
Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.
To find the radius of curvature for the concave mirror, we use the mirror formula:
1/f = 1/do + 1/di
With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:
1/f = 1/-0.85 + 1/8.50
1/f ≈ -1.1765
Solving for f, we find that the focal length is approximately 0.85 m.
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Answer:
The no. of electrons is
Solution:
According to the question:
The rate at which the charge is delivered is given by:
Now,
No. of electrons, n can be calculated from the following relation:
Q = ne
where
e = electronic charge =
Thus
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.
Answer:
1, 2, 4, 5 are correct
Explanation:
1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).
2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.
3) This is not true because the positively charged particles in the nucleus are called protons.
4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.
5) This is true because opposite charges attract each other. And electron is a negative charge.
6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J
The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.
The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.
The given data in the problem is;
E is the electrical energy
P is the power output = 400 W
t is the time period = 5 minutes
The power output is given as;
Hence the electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.
To learn more about the power output refer to the link;
Answer:
The answer is C. 120,000 J.
Explanation: