You have to add 1.7 and 2.2 together!
1.7+2.2=3.9μF
Q = mcΔT
<span>q = 55.8g x 0.450J/gC x 23.5C </span>
<span>q = 590. J ................ to three significant digits
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Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
It is possible for on object to be going at 100 miles per hour, but still have a velocity. This is because the object going at 100 miles per hour has speed, which is a scalar quantity, which is defined by only magnitude, but the velocity of the object can be 0, since velocity is a vector quantity which is defined by both magnitude and direction.
Since this object only has magnitude and no direction (which is not given), then the velocity can be 0
The correct answer is
A. Chain reaction
This is how the chain reaction works in a process of nuclear fission:
1) Slow-moving neutrons reach heavy, unstable, radioactive nuclei (such as Uranium-235)
2) The neutrons interact with the nuclei of Uranium, which break apart producing smaller nuclei and additional neutrons
3) The neutrons produced in the reaction are slowed down by a moderator (e.g. water) and they interact with new nuclei of Uranium, causing new fission reactions)
4) The new neutrons interact with new nuclei of Uranium, and so on...