Answer:
a. Microwaves—3 and infrared waves—1
Explanation:
Microwaves and infrared waves are both part of the electromagnetic spectrum, but they have different frequency and wavelength.
In particular:
- Microwaves are long-wavelength electromagnetic waves, with wavelength between 1 mm and 1 m. Their wavelength is longer than visible light
- Infrared waves are also long-wavelength electromagnetic waves, but their wavelength is shorter than microwaves: between 700 nm and 1 mm. Their wavelength is also longer than visible light.
The two types of waves are also used for different purposes. In particular:
- Infrared waves are emitted by any hot object, and their intensity depends on the temperature of the object. Therefore, they are used in astronomy to show the heat released by astronomical objects (option 1)
- Microwaves are used to study the Cosmic Microwave Background (CMB). This is electromagnetic radiation that permeates the whole universe, and its wavelength depends inversely on the local temperature. Therefore, areas with longer wavelength have lower temperature, and viceversa. Therefore, microwaves are used to measure temperature differences in space (option 3).
The complex, highly technical formula for capacitors is
<em>Q = C V</em>
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
<em>Charge = 72 Coulombs</em>
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average <u><em>lightning bolt</em></u>, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around <u><em>15 coulombs</em></u> of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
Resistance = (voltage) / (current)
Resistance = (12v) / (0.33 A)
Resistance = (12/0.33) ohms
<em>Resistance = 36.4 ohms</em>
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
