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3 years ago

5 Ohm 3 Ohm 2 Ohm R=?

Physics

1 answer:

Tema [17]3 years ago

5 0

Answer:

6.2 ohm

Explanation:

Let R1 = 5 ohm

R2= 3 ohm

R3= 2 ohm

Since R3 and R2 are parallel then net resistance R' is given by

1/R' = 1/R2 + 1/R3

1/R' = 1/3 + 1/2

1/R' = 5/6

then

R' = 6/5 = 1.2 ohm

Now R1 and R' are in series, so

R = R1 + R'

R = 5 + 1.2

R = 6.2 ohm

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At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m

Finger [1]

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$

Applying these equations to the given variables yields

$E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}$

Divide the first line to the second line:

$\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m$

Using this distance in either of the equations give the magnitude of the charge.

$12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C$

8 0

3 years ago

In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

Fynjy0 [20]

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

8 0

3 years ago

You decide it is time to clean your pool since summer is quickly approaching. Your pool maintenance guide specifies that the chl

kotykmax [81]

Answer:

Concentration of Cl2: 1,048 ppm

Explanation:

The unit of measurement molarity (M) represents the same magnitud as mole/L (number of moles of the substance in one liter of solution). The concentration in ppm represents the same as mg/L (number of miligrammes of the substance in one liter of solution). In order to convert from M to ppm we have to consider, then, mass and volume. We can see that in both cases the volume is expressed in liters, so we don't have to do anything to change it. The problem comes when you see that you have to convert moles, from the molatiry, to miligrammes, to get the result in mg/L, meaning ppm.

First of all, notice that the substance is Cl2, so we need to find a relationship between the number of moles and its mass. If you look in the periodic table you'll see that the atomic mass for Cl is 35,4 g/mole (grammes of Cl in one mole). So, there is a way to relate the moles to the mass of the substance and it is represented on the equation below:

$mass=mole*atomic mass$

We want to find the mass (m) and we know the amount of moles of Cl2 in the solution (moles=2,96x10^-5), so, if we use the values known on the equation above we get that:

$m=2,96x10^-5 moles*35,4 g/mole\\ m=1,048x10^-3 g$

Remember that these grammes are found in one liter of solution. So, this means we have 1,048x10^-3 g/L. Previously we said that ppm=mg/L, so all that's left to do it to convert grammes to miligrammes:

1 g = 1000 mg

If we multiply both sides by 1,048x10^-3:

1 * 1,048x10^-3 g = 1000 * 1,048x10^-3 mg

1,048x10^-3 g = 1,048 mg

Knowing that this amount of mass was found in one liter, we get that the amount of substance in the solution is:

1,048 mg/L

Knowing that <em>mg/L=ppm</em>, then the concentration of Cl2 in ppm is:

4 0

3 years ago

If a calorie is equivalent to 4.184 joules how many joules are contained in 250 kg calories slice of pizza

Andreyy89

In order to answer this, we will set up a simple ratio as such:

1 calorie = 4.184 joules

1 kilocalorie = 1000 calories

1 kilocalorie = 4,184 joules

250 kilocalories = x joules

Cross multiplying the second and third equations, we get:

x joules = 4,184 * 250

250 kilocalories are equivalent to 1,046 kJ

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3 years ago

Choose the best tool for each scenario.

julsineya [31]

Answer:

The first one is graph and the second one is electronic balance, third computer then data table and the last is graduated cylinder.

Explanation:

cuz i got it right

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3 years ago

5 Ohm 3 Ohm 2 Ohm R=?​

5 Ohm 3 Ohm 2 Ohm R=?