Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
Answer:
Explanation:
I think this answer would be 1m to the left.
Answer:
there are no pictures attached.
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.
Hi there!
We can begin by solving for the linear acceleration as we are given sufficient values to do so.
We can use the following equation:
vf = vi + at
Plug in given values:
4 = 9.7 + 4.4a
Solve for a:
a = -1.295 m/s²
We can use the following equation to convert from linear to angular acceleration:
a = αr
a/r = α
Thus:
-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.
Now, we can find the angular displacement using the following:
θ = ωit + 1/2αt²
We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:
v = ωr
v/r = ω
9.7/0.61 = 15.9 rad/sec
Plug into the equation:
θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad