Answer:
130 km at 35.38 degrees north of east
Explanation:
Suppose the HQ is at the origin (x = 0, y = 0)
So the coordinates of the helicopter after the 1st flight is
After the 2nd flight its coordinate would be:
So in order to fly back to its HQ it must fly a distance and direction of
north of east
Answer:
Glucose and Oxygen
Explanation:
Cellular respiration is the process whereby cells derives energy by the use of glucose and oxygen.
Organisms that use cellular respiration to produce their energy are known as heterotophs. They derive the glucose from food materials obtained from plant sources. They use the oxygen from the environment to liberate energy from the glucose obtained from feeding on plant materials.
Cellular respiration can be simply expressed as shown below:
GLUCOSE + OXYGEN → CO₂ + H₂O + ATP
The reactants are glucose and oxygen.
The products are CO₂, water and ATP
The current intensity is defined as the amount of charge flowing through a certain point of a wire divided by the time interval:
where Q is the charge and
is the time. Re-arranging the formula, we have
for the compressor in our problem, the intensity of current is I=66.1 A, while the time is
, so the amount of charge that crosses a certain point of the circuit during this time is
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
Answer: q = 2.781e-9C = 2.781nC
E=200C
Explanation:
E = Qd/(2πEor^3)
Where
E=Electric field intensity
Q=Charge
d=distance between the dipole=0.008m
Eo=permitivitty
400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)
Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008
q = 2.781e-9C = 2.781nC
b)
Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:
E = kq*2sin θ/r^2
= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2
= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2
=200 C
