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UkoKoshka [18]
2 years ago
13

What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

electric field of magnitude 700 N/C ?
Physics
1 answer:
tekilochka [14]2 years ago
8 0

Answer:

charge will be equal to 2.03\times 10^{-5}C  

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity g=9.8m/sec^2

Electric field E = 700 N/C

Electric force will be equal to F=qE, here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C

So charge will be equal to 2.03\times 10^{-5}C

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A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
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a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

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