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KATRIN_1 [288]
3 years ago
9

The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary t

o increase the object's temperature from 6°C to 12°C? 4Q 2Q Q 3Q
Physics
2 answers:
Tresset [83]3 years ago
5 0

Answer:

The quantity of heat required to increase the temperature of the object from  6°C to 12°C is 3Q

Explanation:

Heat capacity is the quantity of heat required to increase the temperature of an object.

Q = mcΔθ

where;

Q is the quantity of heat

m is the mass of the object

c is specific heat capacity of the object

Δθ is change in temperature = T₂ - T₁

For the first sentence of this question;

Q = mc(6-4)

Q = mc(2)

Q = 2mc

For the second sentence of this question;

Let Q₂ be the quantity of heat required to increase the temperature of the object from  6°C to 12°C

Q₂ = mcΔθ

Q₂ = mc(12-6)

Q₂ = mc(6)

Q₂ = 6mc

Q₂ = 3(2mc)

Recall, Q = 2mc

Thus, Q₂ = 3Q

bixtya [17]3 years ago
4 0

Answer:

The answer is 3Q

Explanation:

The metal temperature increases in a linear way, we could get a difference between final and initial temperature

DT=FinalTemperature-InitialTemperature

We get a temperature difference of 2 degrees per each heat addition.  If we add the same heat 3 times more, it will increase to 12 degrees

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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

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What type of motion does this graph represent?
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An airplane pilot flies due west at a speed of 216 km/hr with respect to the air. After flying for a half an hour, the pilot fin
Yuki888 [10]

Answer:

speed wind  Vw = 54.04 km / h   θ = 87.9º

Explanation:

We have a speed vector composition exercise

In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south

Let's add the vectors in each coordinate axis

   

X axis (East-West)

      -Xvion - Xw = -119

      Xw = -Xavion + 119

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Y Axis (North-South)

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Let's use the Pythagorean theorem and trigonometry to compose the answer

 Vw = √ (Vwx² + Vwy²)

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  Vw = 54.04 km / h

  tan θ = Vwy / Vwx

  tan θ = 54/2 = 27

  θ = Tan⁻¹ 1 27

  θ = 87.9º

The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º  west from the south

5 0
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