Answer:Electromagnetic Energy Example One
activity: cellphones
type of electromagnetic: radio waves
description: we all use our phones to make phone calls and to send a text!
Electromagnetic Energy Example two
activity: microwave
type of electromagnetic: microwave radiation
description: The microwave radiation is absorbed by water molecules in the food which converts to heat intern heats the food do to high levels of radiation being emitted into the food!
Explanation:
i hope this helps you sorry if it doesn't
The radius of the sphere in meters is ,r =
Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,
r/10 = tan [arctan(1/2)] = tan (1/2)
Since, tan (θ/2) = 1-cos(θ) / sin(θ)
we find that,
r/10 = ![\sqrt{5} -2](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D%20-2)
Hence, r = ![10\sqrt{5} -20](https://tex.z-dn.net/?f=10%5Csqrt%7B5%7D%20-20)
So, the radius of the sphere in meters is ,r =
Learn more about radius (r) of the sphere here;
brainly.com/question/14100787
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1,000 W = 1 kW
100 W = 0.1 kW
(0.1 kW) x (6 h) = 0.6 kWh <=== energy
(0.6 kWh) x (£0.1359/kWh) = £0.0815 <=== cost of it
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = ![\sqrt{\frac{4 * 0.00677 }{\pi } }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B4%20%2A%200.00677%20%7D%7B%5Cpi%20%7D%20%7D)
d = 9.28 cm