Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ +
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
2. Newton's laws of motion
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as

here we know
m = 3.5 kg
k = 270 N/m
now we have


Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as



Part b)
Maximum speed of SHM at its mean position is given as


Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
C Camera. I think this because you can make timelapses with cameras which makes it easy to see.