The angle of the wedge is 30°.
Answer:
5.88 ft/s
Explanation:
a) The block will slide down due to it's weight.
initial velocity u= 0
final velocity, v
acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²
Sliding displacement, s = 3ft
Use third equation of motion:
substitute the values and solve for v
b) Use conservation of momentum:
Initial momentum of the system = 0
final momentum = (15) ( 9.8)+ (25)(v')
v' = 5.88 ft/s
Incandescent light is a glowing white light produced by heat. An incandescent light bulb works by heating a filament in the bulb. Fluorescent light is a bright light produced by electricity flowing through a tube filled with ionized gas. Fluorescent light bulbs are more energy-efficient than incandescent bulbs
Answer:
Static Friction - acts on objects when they are resting on a surface
Sliding Friction - friction that acts on objects when they are sliding over a surface
Rolling Friction - friction that acts on objects when they are rolling over a surface
Fluid Friction - friction that acts on objects that are moving through a fluid
Explanation:
Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book
, car parked on a hill.
Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.
Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.
Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon, sucking water through a straw, submarine moving through water.
Answer:
hope this helps you're welcome
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂
, b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 / - 1 /)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r