Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:


Now we find the centripetal acceleration which is given by

where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find

we also have a tangential acceleration, which is given by

where
t = 17.0 s
Substituting values,

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is
.
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '
' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is

Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

Part(c):
If we apply Gauss' law of electrostatics, then

It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>
NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5

Take the square root of both side

<h3>t = 6.42 s</h3>
Therefore, it will take 6.42 s for the ball to get to the ground.
Learn more: brainly.com/question/24903556
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>