Answer:
Magnitude of C=
Angle=arctan(Ay+By/Ax+Bx)
Explanation:
Let
A = Axi + Ayj
B = Bxi + Byj
where i and j are unit vectors in the direction of x and y axes respectively.
C=A+B
C = (Ax + Bx)i + (Ay + By)j
The magnitude of vector C can be calculated by adding the square of magnitude of its componets and then taking sqaure root.
Magnitude of C=
And the angle will be
Angle=arctan(Ay+By/Ax+Bx)
The volume will be <u>increased</u>
<h2>Explanation</h2>Look at the formula
If the mass increase the density will be increased because of their direct relationship. But in the case of constant mass. The volume increase the density will decrease because there is an inverse relationship exists between them. In the inverse relationship, the two objects perform differently. for example, if you are going to inflate the balloon, its volume will increase but density decrease
Answer:
1. The stone will strike the ground 49.46 m from the base of the cliff
2. A) Approximately 0.542 seconds
B) Approximately 3.69 m/s
3. A) The time the ball spends in the air is approximately 4.0775 s
B) The horizontal range is approximately 141.25 m.
Explanation:
1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)
∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds
The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m
The stone will strike the ground 49.46 m from the base of the cliff
2. A) The time the ball spends in the air = t = √(h/(1/2 × g)
∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds
B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s
The initial horizontal velocity ≈ 3.69 m/s
3. A) The time the ball spends in the air is given by the following equation;
t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s
t ≈ 4.0775 s
B) The horizontal range, R, of the ball is given by the equation for the range of a projectile as follows;
Substituting the known values, gives;
The horizontal range ≈ 141.25 m.