Answer:
+ 5 m/s
Explanation:
change in displacement = ΔX=final position - initial position
ΔX = 0-(-5) =0+5 =+ 5 m
average velocity = ΔX/t
= +5/1
= + 5 m/s
positive sign shows that ball rolls towards right
Answer: -31.36 m/s
Explanation:
This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:
(1)
Where:
is the final velocity of the supply bag
is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)
is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)
is the time
Knowing this, let's solve (1):
(2)
Finally:
Note the negative sign is because the direction of the bag is downwards as well.
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
D=m/V therefore the answer is 120/200 or 0.6
Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns
