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2 years ago

If a man with

1 answer:

6 0

Lots of factors play a role. Firstly if he is presbyopic he won't be able to read without specs at normal reading distances. Many years ago a chap by the name of Donders published a table which shows a linear relationship between age and ability to focus. The ready made reader market relies on his findings to suggest the power you need for reading at different ages. Secondly if he is far sighted (needing correction at distance to see clearly) it will also influence his ability to see close. For example +3.00D distance Rx at an age of 40 years will be very different than the same prescription at age 20 years due to his ability to accommodate. My suggestion is to have comprehensive eye exam to find out what you need at the specific working distance (computer or laptop or book reading all have different reading distances).

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Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge

gulaghasi [49]

Answer:

a) $E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) zero

Explanation:

a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:

$E_T=E_1+E_2$

E1: electric field of charge 1

E2: electric field of charge 2

It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:

$E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\$

r13: distance between charges 1 and 3

r12: charge between charges 2 and 3

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Thus, you first calculate the distance r13 and r23, and also the angles:

$r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°$

Next, you replace the values of all parameters in order to calculate E1 and E2:

$E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}$

finally, you obtain for ET:

$E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.

4 0

3 years ago

The earth's hydrosphere includes which of the following?

Oduvanchick [21]

The answer is all of these

5 0

3 years ago

Read 2 more answers

A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700

Alekssandra [29.7K]

A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700 boards per day. recently the production facility has been producing 950 boards per day. The design capacity utilization is (950/100) *100 = 95 %

3 0

3 years ago

What unit can be used to measure length or distance

strojnjashka [21]

Answer:

The most common units that we use to measure length in the metric system are the millimeter, centimeter, meter, and kilometer.

Explanation:

7 0

3 years ago

How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

irinina [24]

Answer:

Explanation:

Work = Force times displacement. Therefore,

W = 3150(75.5) so

W = 238000 N*m

6 0

2 years ago