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Mrrafil [7]
2 years ago
4

A cold soda initially at 2ºC gains 18 kJ of heat in a room at 20ºC during a 15-min period. What is the average rate of heat tran

sfer during this process?
Physics
1 answer:
KengaRu [80]2 years ago
5 0

Answer:

  q= 20 W                

Explanation:

Given that

Initial temperature ,T₁ = 2 ºC

Heat gains ,Q = 18 kJ

The final temperature ,T₂ = 20 ºC

time ,t= 15 min

We know that

1 min = 60 s

t= 15 x 60 = 900 s

The average rate of heat transfer is give as

q=\dfrac{Q}{t}

q=\dfrac{18}{900}\ kW

q=0.02 kW

q= 20 W

Therefore the rate of heat transfer will be 20 W.

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A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The mater
Rom4ik [11]
The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr²  x 0.02
= 0.04πr²

Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh

Total cost:
C = 0.04πr² + 0.02πrh

We know that the volume of the can is:
V = πr²h
h = 500/πr²

Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r

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0 = 0.08πr - 10/r²
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4 0
2 years ago
A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
svetlana [45]

Answer:

7 deg

Explanation:

m = mass of the rod = 1.4 kg

W = weight of the rod = mg = (1.4) (9.8) = 13.72 N

k_{L} = spring constant for left spring = 59 Nm^{-1}

k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

L = length of the rod = 0.75 m

\theta = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

3 0
2 years ago
Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9
Semmy [17]
<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west. Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up? Barbara is going due south at 5.9 m/s, so she's at (0,-5.9) Neil is going due west at 1.4 m/s, so he's at (-1.4,0) Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this. So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0). The magnitude of Neil's velocity as seen by Barbara is sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s The angle of his vector relative to due west will be atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>
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The answer would be D. Eating nutrient-dense foods
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2 years ago
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