Acceleration........................................
Answer:
200 mL
Explanation:
Given that,
Initial volume, V₁ = 300 mL
Initial pressure, P₁ = 0.5 kPa
Final pressure, P₂ = 0.75 kPa
We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

V₂ is the final volume

So, the final volume of the sample is 200 mL.
Answer:
Explanation:
ignore air resistance
Let t be the time of fall for the dropped stone.
½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²
4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)
4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716
0 = 21.56t - 71.148
t = 71.148/21.56 = 3.3 s
h = ½(9.8)3.3² = 53.361 = 53 m
or
h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m
Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b