What happens when you mix water and baking soda

How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$

While the area of the second square is

$A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C$

So, the total area (and the total charge) is

$Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C$

3 0

3 years ago

A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the sk

malfutka [58]

Answer:

The mass of the cargo is $M = 188.43 \ kg$

Explanation:

From the question we are told that

The radius of the spherical balloon is $r = 7.40 \ m$

The mass of the balloon is $m = 990\ kg$

The volume of the spherical balloon is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

substituting values

$V = \frac{4}{3} * 3.142 *(7.40)^3$

$V = 1697.6 \ m^3$

The total mass the balloon can lift is mathematically represented as

$m = V (\rho_h - \rho_a)$

where $\rho_h$ is the density of helium with a value of

$\rho_h = 0.179 \ kg /m^3$

and $\rho_a$ is the density of air with a value of

$\rho_ a = 1.29 \ kg / m^3$

substituting values

$m = 1697.6 ( 1.29 - 0.179)$

$m = 1886.0 \ kg$

Now the mass of the cargo is mathematically evaluated as

$M = 1886.0 - 1697.6$

$M = 188.43 \ kg$

5 0

3 years ago

A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the

uysha [10]

Answer:

3.14 × 10⁻⁴ m³ /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as

d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.

Speed in the first segment of the pipe is

v1 = 4 m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow rate = Q = A1 v1 = A2 v2 = A3 v3.

Q = A1v1

=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s

3 0

2 years ago

Which is an example of an unintentional injury? A student decides to cut herself. Two people have a domestic dispute that leads

slavikrds [6]

I think its the last one, a student slips on the ice in front of school and sprains his ankle. An example of a natural fiber could be cotton B.

7 0

3 years ago

a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0

2 years ago