A force is a pull or a push true or false?

Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down

VashaNatasha [74]

U can always just do the classic roller coaster going up an incline and create some sort of story from that.

8 0

3 years ago

A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam

lyudmila [28]

<span>P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>

7 0

3 years ago

(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of

NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

$F = \dfrac{kq_1q_2}{r^2}$

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

$F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}$

Force due to Q₃ :

$F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}$

8.3. The net force on the particle at Q₂ is the vector

$\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}$

Its magnitude is

$\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}$

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

$\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}$

where we subtract 180° because $\vec F$ terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.