How do find the force of buoyancy
1 answer:
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Answer:
Explanation:
here we know that car starts from rest and continue to accelerate till it will reach to maximum speed of 20 m/s
so we will have
so car will accelerate till t = 10 then it will move with uniform speed
so the distance moved by the car till it accelerates is given as
now it will move with uniform speed for next 36 s
so we have
so total distance moved by the car is given as
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