The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
The correct option is
a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.
Explanation:
The Krebs citric acid cycle happens within the mitochondrial matrix and generates a pool of energy (ATP, NADH, and FADH2) from the oxidization of pyruvate, the tip product of metabolism. Pyruvate is transported into the mitochondria and loses dioxide to make acetyl-CoA, a 2-carbon molecule.
Answer:
1.8 × 10⁻⁸ Hm
Explanation:
Given that:
The refractive index of the film = 19
The wavelength of the light = 136.8 μ m
The thickness can be calculated by using the formula shown below as:
Where, n is the refractive index of the film
is the wavelength
So, thickness is:
Thickness = 1.8 μ m
Since,
1 μ m = 10⁻⁸ Hm
So,
Thickness = 1.8 × 10⁻⁸ Hm
The answer is b
300,000 km
