Physics, 30.07.2021 14:00 maganns14

Which type of reaction does this diagram? Nuclear fusion because nuclei combine to form a heavier nucleus nuclear fission because an atom splitting into large fragments of comparable mass
nuclear fusion because a large amount of energy is being released
nuclear fission because the resulting products aren't radioactive

Which type of reaction does this diagram? Nuclear fusion because nuclei combine to form a heavier n

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Answer from: Quest

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

Answer from: Quest

Initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is 69.48 m/s downward

Answer from: Quest

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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