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3 years ago

Phase change requires...

1 answer:

amm18123 years ago

3 0

<span>Evaporation involves a liquid becoming a gas and sublimation is the change of a solid directly to a gas.Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing.</span>

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William tell must split the apple atop his son's head from a distance of 24 m. when he aims directly at the apple, the arrow is

mezya [45]

Let x = the angle of elevation for shooting the arrow.

Assume
g = 9.8 m/s²
No wind resistance
The vertical launch velocity is 25.1 sin(x) m/s
The horizontal velocity is 25.1 cos(x) m/s

The time of flight is
24/[25.1 cos(x)] s = 0.9562 sec(x) s

Therefore
0.5*[0.9562 sec(x)]*(9.8) = 25.1 sin(x)
4.6854 = 25.1* sin(x)cos(x)
sin(2x) = 0.3733
2x = sin⁻¹ 0.3733 = 21.92 deg
x = 10.96 deg

Answer: 11 degrees (nearest integer)

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4 years ago

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Clara is shopping in a grocery store with her friend. Each has their own shopping cart in which they want to place their items.

BARSIC [14]

she must apply more force to her cart

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3 years ago

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What is motion? give 3 examples

Tomtit [17]

Answer:

Motion definition: the action or process of moving or being moved.

Examples:

1. moving a hand

2. riding a bicycle

3. running

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3 years ago

How are color and size related

DiKsa [7]

They’re both describing factors of something

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4 years ago

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

2 years ago