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3 years ago

How many significant figures should be written in the sum 4.65g +9.032g+580.0078+540.439g?

Physics

1 answer:

Ne4ueva [31]3 years ago

7 0

Answer: 1.13 X 10^3 g or 1130 g

Explanation: When you add up everything, you will get 1134.1288 g = 1.11341218 X 10^3. But while adding significant figures, we always look for the least significant figures (here 4.65 has least sig. figures i.e. 3) and take that as a reference to how many significant figures we should have in the answer. (Pretty hard to explain!) Hope you got it!!

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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3 years ago

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Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number o

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Answer:

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1(x, t) = Asin(kx - ωt)

y1(x, t) = 7sin(3x - 2.5t)

Wave 2 has a wave function:

y2(x, t) = Asin(kx - ωt + ϕ)

y2(x, t) = 7sin(3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y(x, t) = 7sin(3x - 2.5t) + 7sin(3x - 2.5t + π/12)

y(x, t) = 7[sin(3x - 2.5t) + sin(3x - 2.5t + π/12)]

Using trigonometric function:

sin(a) + sin(b) = 2cos[(a - b)/2]sin[(a + b)/2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y(x, t) = (2*7)[cos(π/24)sin(3x - 2.5t + π/24)]

Therefore, when x = 0.53cm and t = 2s,

y(x, t) = (2*7)[cos(π/24)sin{(3*0.53) - (2.5*2)+ π/24}]

y(x, t) = 14 * 0.9914 * 0.0713

y(x, t) = 0.99 m

The height of the resultant wave is 0.99cm

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3 years ago

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Answer:

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the 3 basic concepts of government that the English brought with them to North America are listed in the answer

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