B
Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is 
where a = 1.2 km north
and b = 1.6 km east
magnitude = 2 km
Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
Answer:
4
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
= Mass of Earth
= Mass of Moon
r = Distance between Earth and Moon
Old gravitational force
New gravitational force
Dividing the equations
The ratio is 
The new force would be 4 times the old force
A and c...............................
Answer:
a = 0.55 m / s²
Explanation:
The centripetal acceleration is given by the relation
a = v² / r
angular and linear velocities are related
v = w r
we substitute
a = w² r
In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system
w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s
let's calculate
a = 0.105² 50.0
a = 0.55 m / s²
