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3 years ago

PLEASE HELP! FOR BRAINLIEST!!!

Physics

2 answers:

Stells [14]3 years ago

7 0

Answer:

I think it's C I'm so sorry if I'm wrong.

Explanation:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

<h2>A or Alex > Chai > Jimmy > Sanni</h2>

Explanation:

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Here's why this is the answer,

__________________________________________________________

Alex's Range = 3.20 - 0.90 = 2.30

Sanni's Range = 1.35 - 1.30 = 0.05

Chai's Range = 2.10 - 1.05 = 1.05

Jimmy's Range = 2.20 - 1.55 = 0.65

__________________________________________________________

Greatest to Least

Alex > Chai > Jimmy > Sanni

Here's why this is correct.

Alex = 2.30

Chai = 1.05

Jimmy = 0.65

Sanni = 0.05

I do not see the answer I have on this question so I think it is A.

__________________________________________________________

Hope this helps! <3

__________________________________________________________

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C) Use a battery with more voltage.

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2 years ago

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Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in

Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

$t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C$

Therefore, the final temperature of the tea is 7.39⁰C.

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3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.