Physics, 14.04.2020 19:36, 20jacksone

Gravity pulls gases and dust particles in space toward one another, squeezing them into a smaller and smaller space. The temperature at the center rises high enough for fusion of hydrogen to begin. What stage in a stars life cycle is this describing?
A. The birth of a star.
B. The death of a star.
C. Black Hole.
D. Stellar Nebula.

Answers: 2

Show answers

Answers

answered: Guest

Fitness is the ability to perform only very light levels of physical activity.

answered: Guest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

answered: Guest

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

answered: Guest

The answer is 4f======since electrostatic force is given by $f_{elec} = \dfrac{kq_1 q_2}{r^2}$ we let f be the electrostatic force when r = 1 m $f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2$ if we let $f_{0.5}$ denote the electrostatic force when the distance is decreased so that r = 0.5m, then $\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}$ we established earlier that f = kq₁q₂. therefore $f_{0.5} = 4 f$ the force will now be 4f.