Given what we know, we can confirm that if the wavelength of light were increased, its frequency and therefore energy would decrease.
<h3>Why do frequency and energy decrease?</h3>
The frequency of light waves is measured through the wavelengths. Shorter waves result in higher frequencies, and since these waves are increasing in length, the frequency will decrease. Since the frequency decreases, the waves lose kinetic energy, and thus the energy decreases.
Therefore, we can confirm that if the wavelength of light were increased, its frequency and therefore energy would decrease.
To learn more about wavelengths visit:
brainly.com/question/7143261?referrer=searchResults
Answer:
The current flowing through the lightbulb was too high. Increasing resistance would decrease the current
Explanation:
Resistance is defined as the opposition to the flow of current across a resistor in an electric circuit. This means that resistors in a circuit opposes the motion of a current. Increase in the value if a resistor will decrease the amount of current that will pass through the resistor.
The higher the value of the resistance, the lower the current and vice versa.
Based on the conclusion above, we could infer that the resistance of the bulb was tested because 'the current flowing through the lightbulb was too high. Increasing resistance would decrease the current.'
Answer:
0.546 ![\hat k](https://tex.z-dn.net/?f=%5Chat%20k)
Explanation:
From the given information:
The force on a given current-carrying conductor is:
![F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})](https://tex.z-dn.net/?f=F%20%3D%20I%20%28%20%5CL%20%20%5Climits%20%5E%20%7B%5Cto%20%7D%20%5Ctimes%20B%20%5E%7B%5Cto%7D%29%5C%5C%20%5C%5C%20dF%20%3D%20I%28dL%5Climits%20%5E%20%7B%5Cto%20%7D%20%5Ctimes%20B%20%5E%7B%5Cto%7D%29)
where the length usually in negative (x) direction can be computed as
![\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i](https://tex.z-dn.net/?f=%5CL%20%5E%20%7B%5Cto%20%7D%20%20%3D%20-x%5Chat%20i%20%5C%5CdL%5Climits%20%5E%20%7B%5Cto%20%7D-%20dx%5Chat%20i)
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
![\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})](https://tex.z-dn.net/?f=%5Cint%20dF%20%3D%20%5Cint%20%5E3_1%20I%20%28%20dL%5E%7B%5Cto%7D%20%5Ctimes%20B%20%5E%7B%5Cto%7D%29)
![F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)](https://tex.z-dn.net/?f=F%20%3D%20I%20%5Cint%5E3_1%20%28%20-dx%20%5Chat%20i%20%29%20%5Ctimes%20%28%204.0%20%5Chat%20i%20%2B%209.0%20%5C%20x%5E2%20%5Chat%20j%29)
![F = I \int^3_1 - 9.0x^2 \ dx \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%5Cint%5E3_1%20%20-%209.0x%5E2%20%5C%20dx%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT ![\hat k](https://tex.z-dn.net/?f=%5Chat%20k)
F = 0.546 ![\hat k](https://tex.z-dn.net/?f=%5Chat%20k)
A person walks up a flight of stairs √
Wind lifts a balloon into the air √
A weightlifter holds a barbell straight overhead √
Gravity is used in most real world examples because if not of gravity everyone would be floating all around ^-^
Hope this helps ;p
True .......................................