(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².
(b) The speed of the electron after the given time is 4.78 x 10⁵ m/s.
<h3>
Acceleration of the electron</h3>
The acceleration of the electron is calculated as follows;
F = qE
ma = qE
a = qE/m
a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)
a = 5.62 x 10¹³ m/s²
<h3>Speed of the electron</h3>
v = at
v = 5.62 x 10¹³ m/s² x 8.50 x 10⁻⁹ s
v = 4.78 x 10⁵ m/s
Learn more about speed here: brainly.com/question/4931057
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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).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
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I think its 13...........
Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
Answer:
B -- Cobalt and iron
Explanation:
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