Answer:
a = 1.962 m/s²
Explanation:
The frictional force applied on the car can be given by the following formula:
where,
f = frictional force = ma (according to Newton's Second Law)
μ = coefficient of friction = 0.2
R = Normal Reaction = Weight of car = mg
Therefore, equation becomes:
where,
a = maximum acceleration = ?
μ = coefficient of friction = 0.2
g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>a = 1.962 m/s²</u>
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
I would say the decision not to have children or to have only one child is becoming more common.
Answer:
a.)
b.)
Explanation:
<u>Electric Field and Potential Difference</u>
There are several conditions that must be met for a spark to be created into an air gap. Once the physical conditions are fixed, a minimum electric field is necessary for the spark to be initiated. Let s be the separation between the electrodes and V their potential difference. The electric field is
a.)
Solving for V
The separation is
Thus the potential difference is
b.) If the separation was greater, the applied voltage needs to be greater if the electric field has to be constant. One possible measure to keep electrodes as close as possible is to build them as sharp edges. It gives the spark an easier path to travel to.
If the separation is 0.05 inches =0.00127 m
If the mass of the original sample of Galium-68 is 10 mg, then . . .
== After one half-life, (1/2)(10 mg) = 5 mg remain
== After the second half-life, (1/2)(5 mg) = 2.5 mg remain
== After the third half-life, (1/2)(2.5 mg) = 1.25 mg remain
It doesn't matter how long the half-life is.