Answer:
pf = 198.8 kg*m/s
θ = 46.8º N of E.
Explanation:
- Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
- If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

- We can do exactly the same for the initial momentum along the y-axis:

- The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

- We can repeat the process for the y-axis, as follows:

- Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

- In the same way, we can find the component of the final momentum along the y-axis, as follows:

- With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

- The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

- Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
- We can find this angle applying the definition of tangent of an angle, as follows:

⇒ θ = tg⁻¹ (1.06) = 46.8º N of E
Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) 
The equation for velocity v and a constant acceleration a is:
(ii) 
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) 
(iv) 
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) 
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

1. 
Explanation:
We have:
voltage in the primary coil
voltage in the secondary coil
The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.
2. 5.7 V
We can solve the problem by using the transformer equation:

where:
Np = 400 is the number of turns in the primary coil
Ns = 19 is the number of turns in the secondary coil
Vp = 120 V is the voltage in the primary coil
Vs = ? is the voltage in the secondary coil
Re-arranging the formula and substituting the numbers, we find:

Answer:
2,500 feet (760 meters)
Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>
Answer:
When air resistance equals the weight of an object, the object has reached free fall.
Explanation:
- When an object has only force acting on it as gravity then, it experiences free fall.
- During free fall all the forces except gravity is balanced by one another.
- In the question, object's weight is balanced by air resistance so it is in the state of free fall.
- At the null point of free fall, object experiences weightlessness i.e. it feels like object is not attracted by any force.