Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m
Answer:
The tension in the two ropes are;
T1 = 23.37N T2 = 35.47N
Explanation:
Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity
W = 4.2×10 = 42N
The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.
Using triangular law of force and sine rule to get the tension we have;
If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;
T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}
T1/sin33.8° = T2/sin57.6° = 42/sin88.6°
From the equality;
T1/sin33.8° = 42/sin88.6°
T1 = sin33.8°×42/sin88.6°
T1 = 23.37N
To get T2,
T2/sin57.6°= 42/sin88.6°
T2 = sin57.6°×42/sin88.6°
T2 = 35.47N
Note: Check attachment for diagram.
Answer:
correct answer is D
Explanation:
answer is perihelion and aphelion
It would crash into the sun