Answer:
The net gravitational force on the mass is
Explanation:
We have by Newton's law of gravity the force of attraction between masses
Applying vales we get
Force of attraction between 135 kg mass and 38 kg mass is
Force of attraction between 435 kg mass and 38 kg mass is
Thus the net force on mass 38.0 kg is
Answer:
B) 1/5 ba^2 T^5
Explanation:
The dissipated energy is given by the work done over the object by the force F=-bv. The work is given by the following formula:
you derivative the function f(x) and replace v by the derivative dx/dt you obtain:
hence, the dissipated energy is 1/5 ba^2 T^5
Answer
given,
intial velocity = 504 mph
wind speed = 219 mph
at an angle of 29◦
from the data
The resultant velocity =
the magnitude of velocity
V = 703.59 m/s
direction
tan θ =
θ = 8.676°
Answer:
La fuerza resultante sobre q₃ es -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N
Explanation:
q₁ = -45 μC = -45 × 10⁻⁶ C
r₁₂ = 30 mm = 30 × 10⁻³ m
q₂ = 25 μC = 25 × 10⁻⁶ C
r₂₃ = 50 mm = 50 × 10⁻³ m
q₃ = 20 μC = 20 × 10⁻⁶ C
k = 9×10⁻⁹ N·m²/C²
Por lo tanto;
r₁₃ = √(50² + 30²) = 10·√(34)
F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴
F₁₂ = -1.125 × 10⁻¹⁴ N
F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j
F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵
Los componentes de F₁₃ son;
-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j
-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i
La fuerza resultante sobre la carga q₃, =
+
∴ = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j
La fuerza resultante sobre q₃ es = -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃,
= √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵
La magnitud de la fuerza resultante sobre q₃, ≈ 1.25 × 10⁻¹⁵ N.