Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Answer:
Angular acceleration = 23.68 rad / s²
Explanation:
Given that,
acceleration = 9m/s²
Therefore acceleration of string is 9m/s²
since string is constant in length
cylinder of radius 38.0 cm = 0.38m
Angular acceleration = a / r
Angular acceleration = 9 / 0.38
= 23.68 rad / s²
Angular acceleration = 23.68 rad / s²
Temperature can change the state from solid to liquid causing it to melting, liquid to gas causing vaporization or a solid to a gas causing sublimation. Pressure alone cannot change the state of matter.
Yo sup??
Average velocity=total distance covered/total time taken
total distance covered=4 + 8=12 miles
total time taken=6 hours
Therefore
average velocity=12/6
=2 miles/hour
Hope this helps
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) =
=
= 392 J
(C) average frictional force = 
- change in KE (ΔKE) = initial KE - final KE
- ΔKE =
-
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] =
=
= 212.33 N