The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
Explanation:
Expression for times period of a satellite can be given as follows
Time period T = 1.8 x 60 x 60
= 6480
T² = 
where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.
6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM
GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²
= 3.96 X 10¹⁴
Expression for acceleration due to gravity
g = GM / R² where R is radius of satellite
20 = 3.96 X 10¹⁴ / R²
R² = 3.96 X 10¹⁴ / 20
= 1.98 x 10¹³ m
R= 4.45 x 10⁶ m
Answer:
La rapidez media es 25 m/s en ambos casos.
Explanation:
Podemos definir como rapidez media al cociente entre la distancia total recorrida y el tiempo que se tardó en recorrer dicha distancia.
Así tenemos:
Rapidez media = Distancia/tiempo.
Entonces si el guepardo recorre 100m en 4 segundos, su rapidez media es:
Rapidez media = 100m/4s = 25 m/s
En el caso de que el guepardo recorre 50 metros en 2 segundos, su rapidez media será:
rapidez media = 50m/2s = 25m/s
Es el mismo resultado, pues recorrió la mitad de distancia en la mitad de tiempo.
Answer:
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