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-BARSIC- [3]
3 years ago
11

How to solve these two questions? ​

Physics
1 answer:
hammer [34]3 years ago
3 0

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

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Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

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Which of the following statements are true? Positively charged objects attract other positively charged objects. Negatively char
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A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
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Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

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