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nevsk [136]
3 years ago
14

PLS HELP FIND THE AVREGE KENETIC ENERGY!!

Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer: if you gogle it it will tell you the awenser

lanation:

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Which use combustion to release thermal energy?
julia-pushkina [17]
Heat of combustion.<span> The calorific value is the total energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon or other organic molecule reacting with oxygen to form carbon dioxide and water and release heat.</span>
8 0
3 years ago
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You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o
Mama L [17]

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

8 0
3 years ago
A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl
raketka [301]
The wavelength of the note is \lambda = 39.1 cm = 0.391 m. Since the speed of the wave is the speed of sound, c=344 m/s, the frequency of the note is
f= \frac{c}{\lambda}=879.8 Hz

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }
where \mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
L= \frac{1}{2f}  \sqrt{ \frac{T}{\mu} } =0.31 m
8 0
3 years ago
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
Rina8888 [55]
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r 
 v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min). 
 (32.68 rev/s)(60 s/min)=<span> <span>1960.74 rev/min
</span></span>
5 0
3 years ago
In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or
Vesnalui [34]

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

4 0
2 years ago
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