This problem is looking for the minimum value of μs that is
necessary to achieve the record time. To solve this problem:
Assuming the front wheels are off the ground for the entire
¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².
For a constant acceleration, distance = 402.3
m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2
µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18
Answer:
A. 181.24 N
Explanation:
The magnitude of hte electrostatic force between two charged objects is given by the equation

where
k is the Coulomb's constant
q1, q2 are the magnitudes of the two charges
r is the separation between the charges
In this problem, we have:
is the magnitude of the 1st charge
is the magnitude of the 2nd charge
r = 2.5 cm = 0.025 m is the separation between the charges
Therefore, the magnitude of the electric force is:

So, the closest answer is
A) 181.24 N
Answer:
v = 5.9 x 10⁷ m/s
Explanation:
The kinetic energy of the electron in terms of potential difference is given as:
--------------- equation (1)
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
V = Potential Difference = 9.9 KV = 9900 Volts
The kinetic energy in general is given as:
--------- equation (2)
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of electron = ?
Therefore, comparing equation (1) and equation (2), we get:

<u>v = 5.9 x 10⁷ m/s</u>
Answer:
Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell
Answer:
E.
Explanation:
In a galvanic cell, electrons flow from the anothe to the cathode.
I hope you got the answer