The passing of heat through a material while the material itself stays in place is conduction.
Answer: C
Explanation:
Energy stays the same, but there’s more kinetic energy the less it moves and mor kinetic energy the more it moves
<span>To answer this question, the equation that we will be using is:
y = A cos bx + c
where A = amplitude, b = 2 pi/Period, Period = 12 hrs, c = midline,
x = t and y = f(t)
A = 1/2 (Xmax - Xmin)
12 - 2 / 2 = 10/2 = 5
b = 2 pi / 12 = pi/6
c = 1/2 (Xmax + Xmin)
12+2/2 = 7
answer: f(t) = 5 cos pi/6 t + 7 </span>
Answer:
x =4.5 10⁴ m
Explanation:
To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law
m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg
q = -200 nc (1C / 10 9 nC) = -200 10-9 C
Let's calculate the acceleration
F = ma
F = q E
a = qE / m
a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶
a = 1 10² m / s²
Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)
Vf² = Vo² -2 a x
0 = Vo² - 2 a x
x = Vo² / 2a
x = 3000²/ 2100
x =4.5 10⁴ m
This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial
Second part
In this case Newton's second law is applied on the y axis
F -W = 0
F = w = mg
E q = mg
E = mg / q
E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹
E = 9.8 10⁵ C
The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E
Answer:
(a) 20 m
(b) 6 m/s²
(c) Between t=0 and t=2, the body moves to the left.
Between t=2 and t=4, the body moves to the right.
Explanation:
v = 3t² − 6t
x(0) = 4
(a) Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 6t) dt
x = t³ − 3t² + C
Use initial condition to find value of C.
4 = 0³ − 3(0)² + C
4 = C
x = t³ − 3t² + 4
Find position at t = 4.
x = 4³ − 3(4)² + 4
x = 20
(b) Acceleration is the derivative of velocity.
a = dv/dt
a = 6t − 6
Find acceleration at t = 2.
a = 6(2) − 6
a = 6
(c) v = 3t² − 6t
v = 3t (t − 2)
The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.
When 0 < t < 2, v < 0.
When t > 2, v > 0.
