Answer:
a = αR
Then We apply the Newton's second law of motion:
∑ F = ma
F - mg sinθ - f = ma.
F - mg sinθ - μmg cosθ = ma
Using given data we have:
F - 3217 = 470a ........................... (1)
Apply the Newton's law for rotation:
∑τ = I α
FR - (mg sinθ)R = 
then: F - mg sinθ = 3ma / 2.
F - 2774 = 705a .......................... (2)
solving 1 and 2 we get:
F = 4103 N
a = 1.885 m/s²
b) In this case the time is given by:
t = 
it comes from: d = 
starting from rest vo = 0
t = 
= 3.09 s
Answer: 7.5 ml
Explanation: Density= mass/ volume.
So, volume = mass / density
= 15g / 2 g/ml
= 7.5 ml
friction and the density of the air.
I just shot my shot at this little 52 ms far
Answer:
1.6s
Explanation:
Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,
To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.
Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh
Where h = 5.2sin25
h = 2.2 m
P.E = 1.2 × 9.8 × 2.2
P.E = 25.84 j
At the bottom, K.E = P.E
1/2mv^2 = 25.84
Substitutes mass into the formula
1.2 × V^2 = 51.69
V^2 = 51.69/1.2
V^2 = 43.07
V = 6.56 m/s
Using the third equation of motion
V^2 = U^2 + 2as
Since the object started from rest,
U = 0
6.56^2 = 2 × a × 5.2
43.07 = 10.4a
a = 43.07/10.4
a = 4.14 m/s^2
Using the first equation of motion,
V = U + at
Where U = 0
6.56 = 4.14t
t = 6.56/4.14
t = 1.58s
Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s
