Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
(1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
(2)
Qin is calculate by using the charge density:
(3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
Next, you use the results of (3), (2) and (1):
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
hence, the electric field is 1580594.95 N/C
Answer: Point Z only
Explanation: Just took the quiz
Answer:
Explanation:
The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:
F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that
f = μ
where μ is the coefficient of friction, and is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:
so
so the weight of the sled is
1.0 × 10³ with the correct number of sig dig there. Now to find f:
f = (.025)(1.0 × 10³) so
f = 25 to the correct number of sig fig. Now on to our "real" equation:
F - f = ma and
230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.
230 - 25 will round to the tens place giving us 210. Then
210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:
a = 2.0 m/sec²
Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + 
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
Answer:
Force = mass × acceleration
Acceleration:
From first Newton's equation of motion:
Change = v - u:
