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Nadusha1986 [10]
3 years ago
5

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is​

Physics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

5

Explanation:

54.34+45.66=100.00 (you have to use the .00 because when you add to numbers you keep the number of decimals)

so you get 100.00

all numbers that are not 0 are sig figs so 1 is a sig fig

If a number ends with a 0 after a decimal place that 0 is a sig fig

all numbers between two sig figs are sig figs so that would make all of the numbers sig figs

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Answer:

240 N

Explanation:

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Naya [18.7K]

Explanation:

Particle moving in a circular path with a constant speed.

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Serga [27]

Answer:

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Explanation:

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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

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4 0
3 years ago
parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet
Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J /m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

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- Where the Electric field strength E between capacitor plates is given by:

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Hence,

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Where, ε = 8.854 * 10^-12

                               V^2 = 2*u*d^2 / ε

                               V = d*sqrt ( 2*u / ε )

Plug in values:

                               V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

                               V = 576 V

4 0
3 years ago
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