Which of these statements best describes a double-replacement reaction?

Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.

miss Akunina [59]

Answer:

Δt = $\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 + Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

$\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$ = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0

3 years ago

Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter

stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

$m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

$m_c$ = mass of copper = 227 g

$m_w$ = mass of water = 844 g

$m_a$ = mass of aluminum = 155 g

$C_c$ = specific heat capacity of calorimeter = 385 J/kg.°C

$C_w$ = specific heat capacity of water = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum = 890 J/kg.°C

$\Delta T_c$ = change in temperature of copper = 283°C - T

$\Delta T_w$ = change in temperature of water = T - 14.6°C

$\Delta T_a$ = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

$(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}$

8 0

2 years ago

A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c

natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

$mg = \rho V_{displaced} g$

$V_{displaced} = \frac{m}{\rho}$

$V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3$

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

$V\rho_w = m_{ice}$

$V = \frac{0.200}{1000} = 2\times 10^{-4} m^3$

So here extra volume that rise in the level will be given as

$\Dleta V = V - V_{displaced}$

$\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}$

$(\pi (0.039^2) h = 0.41 \times 10^{-4}$

$h = 0.86 cm$

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0

3 years ago

Vectors vs Scalar

IrinaK [193]

Answer:

I wish I knew

Explanation:

Ummm

7 0

2 years ago

Does a wall or door have more inertia?

suter [353]

Answer:

wall

Explanation:

4 0

3 years ago

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