What mass of water can be heated from 25.0° C to 50.0° C by the addition of 2825 J of heat energy if the specific heat capacity
1 answer:
Answer:
Mass, m = 27g
Explanation:
<u>Given the following data;</u>
Initial Temperature, T1 = 25°C
Final temperature, T2 = 50°C
Quantity of heat = 2825J
Specific heat capacity of water = 4.184
Heat capacity is given by the formula;
Where,
- Q represents the heat capacity.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
Making mass, m the subject of formula, we have;
Change in temperature, dt = T2 - T1
Change in temperature, dt = 50-25 = 25°C
Substituting into the equation, we have;
Mass, m = 27g
Therefore, the mass of water that can be added is 27 grams.
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Hooke's Law
k = spring constant
x = stretch
F = force
Input the value
Answer:
a) 24.33 m of distance.
b) 34.55° east of the south.
Explanation:
The question is incomplete. The whole exercise is the following:
<em>"Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 28.0 m in a direction 60.0° west of north. Jane walks 12.0 m in a direction 30.0° south of west. They then stop and turn to face each other. </em>
<em>(a) What is the distance between them? </em>
<em>(b) In what direction should Ricardo walk to go directly toward Jane?</em><em>"</em>
Now that we know what we need to do in this question, let's head for every part of the problem.
<u>a) Distance between Ricardo and Jane</u>
In this case, we need to analyze the given data:
Ricardo (which we will call R) is 28 m from the starting point at 60° west of north, and Jane (J) is 12 m at 30° south of west. So the distance between them, will be the point where they both stop and face each other. This point can be seen in the image attached (See picture).
Let's call the distance between them as "D", to get the distance of D, according to the picture will be:
D = J - R (1)
However, as they are facing in different angles and directions, we cannot do the difference of their values distance just like that. In order to do that, we need to calculate the components in the "x" and "y" axis of each vector. In that way, we can get the components of x and y of the Distance D, and then, the whole distance between them will be:
D = √Dx² + Dy² (2)
So, let's get the components of x and y of R and J.
For Ricardo (R):
Rx = R sin60° = 28 sin60° = -24.25 m
Ry = R cos60° = 28 cos60° = 14 m
The sign "-" it's because R it's on the second quadrant, therefore in x, we'll have to add the negative.
For Jane (J):
Jx = J cos30° = 12 cos30° = -10.39 m
Jy = J sin30° = 12 sin30° = -6 m
Again, the negative is added because J is on the third quadrant.
Now that we have the components, let's calculate vector D using expression (1):
Dx = -10.39 - (-24.25) = 13.86 m
Dy = -6 - 14 = -20 m
Now, using expression (2) we can finally know the distance between Jane And Ricardo:
D = √(-20)² + (13.86)²
<h2>D = 24.33 m</h2>This is the distance between Jane and Ricardo.
b) Direction of Ricardo walking to Jane
In this case, we already have the components of x and y of the distance between them, so, to know the direction:
Tanα = Dy/Dx
α = tan⁻¹ (Dy/Dx)
Replacing the values we have:
α = tan⁻¹ (-20/13.86)
α = 55.45°
Which should south of east or:
β = 90 - 55.45
<h2>β = 34.55°</h2>Ricardo should walk 34.55° east of south
Hope this helps
