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
#SPJ4
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
Answer:
The average net force on the truck is 375 Newtons.
Explanation:
Using Newton's 3rd equation of motion, we have :
×a×s
where, v = final velocity = 25 m/s
u = initial velocity = 20 m/s
a = acceleration
s = distance traveled = 300 m
Using these values in the above equation, we get acceleration = 0.375 m/
Using Newton's second law, we have:
F=m×a
where m = mass = 1000 kg
a= acceleration = 0.375 m/
Putting values we have F=375 N
Sound waves travel faster through <em>solids</em> than they do through gases or liquids. <em>(C) </em>They don't travel through vacuum at all.
Example:
Speed of sound in normal air . . . around 340 m/s
Speed of sound in water . . . around 1,480 m/s
Speed of sound in iron . . . around 5,120 m/s
Answer:
0.5 m/s north
Explanation:
Take east to be +x, west to be -x, north to be +y, and south to be -y.
His displacement in the x direction is:
x = 20 m − 20 m = 0 m
His displacement in the y direction is:
y = 10 m
His total displacement is therefore 10 m north.
His velocity is equal to displacement divided by time.
v = 10 m north / 20 s
v = 0.5 m/s north
