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Fiesta28 [93]
3 years ago
7

Energy can be transferred from one _______ to another and one __________ to another. *

Physics
2 answers:
Dmitrij [34]3 years ago
8 0

Answer:

Source, form

Explanation:

yaroslaw [1]3 years ago
7 0

Answer:

Explanation:

Energy can be transferred from one _<u>source</u>_ to another and one __<u>form</u>____ to another.

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You have finger tips but you dont have toe tips but you can tip toe explain pls
Arada [10]

Answer:

ummmmmmmm

Explanation:

this might be some type of trick question bro lol

or someone tryna play with you if the teacher gave you this question bro she deserves to be fired

4 0
3 years ago
The density of iron is 7.8 g/cm3, what is the mass of 5cm3 of it?
marissa [1.9K]
<h2>Answer:   39.0 g</h2>

Explanation:

Mass = volume × density

⇒ mass of iron = (5 cm³ ) (7.8 g/cm³)

                         = 39.0 g

4 0
3 years ago
ASAP!! PLZZ HELP ME
kozerog [31]
The brush passes current from a power source into the commutator.  Therefore they both must carry electric current, B
4 0
3 years ago
Read 2 more answers
A parallel plate air capacitor of value C, is charged to a potential V, between
k0ka [10]

Answer:

(i) C=3Co

(ii) V=Vo/3

(iii) Decrease

Explanation:

(i) capacitance increase K times.

(ii) As charge remain constant so by using Q=CV

you will get potential decrease by K time.

(iii) Decrease due to induced electric field inside

dielectric material..

E=Eo-Eind

Explanation:

heyaaa hope it helps ☺️✌️

4 0
3 years ago
A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor
Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

Hence, 1.16 cm were cut from the end of the second pipe

6 0
3 years ago
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