A sound wave has a frequency of 425Hz. What is the period of this wave? a) 0.00235 b) 0.807 c) 425 d) 850
1 answer:
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Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>
Explanation:
a) Acceleration due to gravity
Here we need to find acceleration due to gravity of Sun,
G = 6.67259 x 10⁻¹¹ N m²/kg²
Mass of sun, M = 1.989 × 10³⁰ kg
Radius of sun, r = 6.957 x 10⁸ m
Substituting,
Acceleration due to gravity on the surface of the Sun = 274.21 m/s²
b) Acceleration due to gravity in earth = 9.81 m/s²
Ratio of gravity = 274.21/9.81 = 27.95
Weight = mg
Factor of increase in weight = 27.95