a) create an expression
for the ball's initial horizontal velocity, V0x, in terms of the variables
given in the problem statement.
v0x = vf * cos(Θf)
<span>
b) calculate the ball's initial vertical velocity, V0y, in
m/s</span>
v0x = 32.4m/s * cos(-25.5º)
= 29.2 m/s <span>
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93
m/s
the vertical velocity when the ball was caught.
(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78
m²/s²
v0y = 17.37 m/s
c) calculate the magnitude of the ball's initial velocity,
v0, in m/s</span>
v0 = sqrt (v0y^2 +
v0x^2)
v0 = sqrt (17.37^2 + 29.2^2)
m/s
v0 = 33.98 m/s
<span>
d) find the angle, theta0, in degrees above the horizontal at
which which the ball left the bat.</span>
tan Θ = v0y/v0x
<span>Θ = arctan(17.37/29.2) =
30.75º above horizontal</span>
Answer:
deep in the earth's crust
Explanation:
Sedimentary rocks become metamorphic in the rock cycle when they are subjected to heat and pressure from burial. The high temperatures are produced when the Earth's tectonic plates move around, producing heat. And when they collide, they build mountains and metamorphose.
Igneous rocks form as magma cools below ground or lava cools on the surface. Sedimentary rocks are made from the eroded particles of other rocks or from mineral deposits left when water evaporates. Metamorphic rocks form when any existing rock undergoes intense and prolonged exposure to heat and pressure.
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6