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Lorico [155]
3 years ago
6

A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly

to 100°C just before it starts to boil. (a) What amount of heat is absorbed by the water in reaching 100°C? (b) The water then boils until all of it has evaporated, turning to water vapor at 100°C. How much heat does the water absorb in this process?
Physics
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer:

100°heat

Explanation:

since when i calculate this and that, the answer is 100° heat.

sorry if it is inconvenient

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LuckyWell [14K]
You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.
4 0
3 years ago
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a sta
yuradex [85]

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

7 0
3 years ago
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha
jasenka [17]

Explanation:

It is given that,

Mass of an object, m=4\times 10^5\ kg

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The formula for the time period of spring is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

Where

k is the force constant

k=\dfrac{4\pi ^2 m}{T^2}

k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}

k=2.74\times 10^6\ N/m

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

U=\dfrac{1}{2}kx^2

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Hence, this is the required solution.

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olasank [31]
Hope this helped! i’m not sure if it’s right. it honestly bored so i’m answering questions.

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2 years ago
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