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2 years ago

A rock is kicked horizontally off a dock into the bay 3

Physics

1 answer:

Maslowich2 years ago

6 0

Answer:

The true statement is;

(b) The rock takes about the same time to hit the water in each case

Explanation:

The given information are;

The direction in which the rock is kicked = Horizontally

The number of different initial horizontal with which the rock as kicked = different horizontal velocity

The time taken for the rock to hit the water is given by the following equation for projectile motion;

y = y₀ + v₀× sin(θ₀)×t - 1/2×g×t²

Where;

θ = The angle the rock makes with the horizontal when it was kicked = 0

y = The height through which the rock drops

y₀ = The initial height of the rock

t = The time taken for the ball to hit the water

g = The acceleration due to gravity ≈ 9.81 m/s²

For the different speeds, v₁, v₂, and v₃, we have;

y = y₀ + v₁× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₂× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₃× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

Therefore, the time it takes for the ball to hit the water is given as follows;

$t = \sqrt{\dfrac{(y_0 - y) \times 2}{g} }$

Which is the same for each case

However, the speed, v, with which the rock hits the water is given by the following equation;

$v = \sqrt{v_x^2 + v_y^2}$

$v_y$ = The vertical velocity = v₀ × sin(θ₀) - g×t = v₀ × sin(0) - g×t = -g×t

$v_y$ = -g×t

Therefore, $v_y$ , is the same for all three times

While vₓ = v₁, v₂, and v₃, for each of the three times.

Therefore, $v = \sqrt{v_x^2 + v_y^2}$ the speed with which the rock hits the water is different for all three times.

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A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat

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Answer:

Approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ (assuming that the boiling point of water in this experiment is $100\; \rm ^\circ C\!$ .)

Explanation:

Latent heat of condensation/evaporation of water: $2260\; \rm J \cdot g^{-1}$ .

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to $\rm J \cdot g^{-1}$ .

Specific heat of water: $4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}$ .

Specific heat of copper: $0.39\; \rm J \cdot g^{-1}\cdot K^{-1}$ .

The temperature of this calorimeter and the $250\; \rm g$ of water that it initially contains increased from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ . Calculate the amount of energy that would be absorbed:

$\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}$ .

$\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}$ .

Hence, it would take an extra $585\; \rm J + 31500\; \rm J = 32085\; \rm J$ of energy to increase the temperature of the calorimeter and the $250\; \rm g$ of water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

Assume that it would take $x$ grams of steam at $100\; \rm ^\circ C$ ensure that the equilibrium temperature of the system is $50\; \rm ^\circ C$ .

In other words, $x\; \rm g$ of steam at $100\; \rm ^\circ C$ would need to release $32085\; \rm J$ as it condenses (releases latent heat) and cools down to $50\; \rm ^\circ C$ .

Latent heat of condensation from $x\; \rm g$ of steam: $2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J$ .

Energy released when that $x\; {\rm g}$ of water from the steam cools down from $100\; \rm ^\circ C$ to $50\; \rm ^\circ C$ :

$\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}$ .

These two parts of energy should add up to $32085\; \rm J$ . That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

$(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J$ .

Solve for $x$ :

$x \approx 13$ .

Hence, it would take approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ for the equilibrium temperature of the system to be $50\; \rm ^\circ C$ .

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