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2 years ago

7

With a power supply able to produce 125 a of current, how long must the nails stay in the electrolysis chamber to receive 1.33×1

07 c of charge?

1 answer:

KengaRu [80]2 years ago

3 0

<span>With a power supply able to produce 125 an of current, then the time on how long must the nails stay in the electrolysis chamber to receive 1.33×107 c of charge is 10 hours, then a minute of 29, and lastly, 20 more seconds. </span>

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Two students, Jenny and Cho, are investigating motion.

IgorLugansk [536]

Answer:

1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

pls mark brainliest

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10 months ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

2 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

2 years ago

A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k

levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

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2 years ago

Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e

My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

$PV =NkT$

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

$N = 1$

$k = 1.38*10^{-23}J/K$

$T = 27\°C = 27\°C + 273 = 300K$

$P = 1atm = 101325Pa$

Rearrange the equation to find V we have,

$V = \frac{NkT}{P}$

$V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}$

$V = 4.0858*10^{-26}m^3$

We know that length of a cube is given by

$V = L^3$

Therefore the Length would be given as,

$L = V^{1/3}$

$L = (4.0858*10^{-26})^{1/3}$

$L = 3.445*10^{-9}m$

Therefore each length of the cube is 3.44nm

7 0

2 years ago