Answer:
true blood season and answers the question of whether you are not the intended recipient you are not the intended recipient you are not the intended recipient
Answer:
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
c) True. Information is missing to perform the calculation
Explanation:
Let's consider solving this exercise before seeing the final statements.
We use Newton's second law Rotational
τ = I α
T r = I α
T gR = I α
Alf = T R / I (1)
T = α I / R
Now let's use Newton's second law in the mass that descends
W- T = m a
a = (m g -T) / m
The two accelerations need related
a = R α
α = a / R
a = (m g - α I / R) / m
R α = g - α I /m R
α (R + I / mR) = g
α = g / R (1 + I / mR²)
We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant
Let's review the claims
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
b) False. Missing data for calculation
c) True. Information is missing to perform the calculation
d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
