Answer:
27,000 m
450 m/s
Explanation:
Assuming the initial velocity is 0 m/s:
v₀ = 0 m/s
a = 15 m/s²
t = 60 s
A) Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²
Δy = 27,000 m
B) Find: v_avg
v_avg = Δy / t
v_avg = 27,000 m / 60 s
v_avg = 450 m/s
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
The image distance is 30 cm
image height = - 5 cm
Explanation:
The formula for calculating the image distance is expressed as
1/f = 1/u + 1/v
where
f is the focal length
u is the object distance
v is the image distance
From the information given,
u = 30
f = 15
By substituting these values into the formula,
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
Taking the reciprocal of both sides,
v = 30
The image distance is 30 cm
magnification = image height/object height = - v/u
Given that object height = 5 cm, then
image height/5 = - 30/30 = - 1
image height = - 5 * 1
image height = - 5 cm
