The diesel engine has no ignition system. (C)
It depends on the compression of the fuel/air mixture to ignite the mixture.
This may be a big part of the reason that Diesel engines always
sound to me as if they are falling apart.
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e

N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e

= 1250W
d) Final peak=
P= Ik/e
= 
P = 2.5 × 10⁷W
Answer:
the answer to this question is 2,4,3,1
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:

where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:

where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:

- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:

- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

- Replacing by the givens in (5), we can solve for μsmín, as follows:

Answer:
12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.
Explanation:
In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:




12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.