Hola!
93 millones de millas equivalen a
1,50 x 10¹¹ metros. Para calcular este valor, necesitamos saber la equivalencia entre millas y metros (1 milla=1609,34m), y aplicar el siguiente factor de conversión. Para expresar el resultado en potencias de 10, se debe correr la coma a la izquierda hasta obtener un valor de una sola unidad y multiplicar este valor por 10 elevado a la cantidad de espacios que la coma se tuvo que correr a la izquierda:
Saludos!
Answer:
Option (c) Acceleration is 1/2x the original acceleration
Explanation:
To know the the correct answer to the question, we shall determine the acceleration of the box in each case.
Case 1:
Mass (m) = m
Force (F) = 20 N
Acceleration 1 (a₁) =?
F = ma₁
20 = m × a₁
Divide both side by m
a₁ = 20 / m
Case 2:
Mass (m) = 2m
Force (F) = 20 N
Acceleration 2 (a₂) =?
F = ma₂
20 = 2m × a₂
Divide both side by 2m
a₂ = 20 / 2m
a₂ = 10 / m
Finally, we shall determine the acceleration of the box after the mass was doubled. This can be obtained as illustrated below:
Acceleration 1 (a₁) = 20 / m
Acceleration 2 (a₂) = 10 / m
a₂ : a₁ = 10 / m : 20 / m
a₂ / a₁ = 10 / m ÷ 20 / m
a₂ / a₁ = (10 / m) × (m / 20)
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
Thus, the acceleration of the box after the mass was doubled is ½ times the original acceleration.
Answer:
Explanation:
mass of car, m = 1000 kg
initial velocity, u = 20 m/s
final velocity, v = 0 m/s
distance, s = 120 m
Let a be the acceleration of motion
use third equation of motion
v² = u² + 2 as
0 = 20 x 20 + 2 x a x 120
a = - 1.67 m/s²
Let F be the force
Force, F mass x acceleration
F = - 1000 x 1.67
F = - 1666.67 N
The direction of force is towards south and the magnitude of force is 1666.67 N.
Answer:
6 x 10⁻¹⁵ J
Explanation:
d = distance between the plates = 1.5 cm = 0.015 m
E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m
q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C
Force on the electron due to electric field is given as
F = q E
F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)
F = 4 x 10⁻¹³ N
KE₀ = initial kinetic energy of electron at negative plate = 0 J
KE = final kinetic energy of electron at positive plate = ?
Using work-change in kinetic energy
F d = KE - KE₀
(4 x 10⁻¹³) (0.015) = KE - 0
KE = 6 x 10⁻¹⁵ J