First, you find the velocity at each component. The general equation is:
a = (v2 - v1)/t
a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s
a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s
To find the final speed, find the resultant velocity by taking the hypotenuse.
v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
A. 22 Million...........................................................
Answer:
t ’= , v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’=
t ’=
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’=
t ’= 547.19 s
Answer:
0.06 Nm
Explanation:
mass of object, m = 3 kg
radius of gyration, k = 0.2 m
angular acceleration, α = 0.5 rad/s^2
Moment of inertia of the object
I = 3 x 0.2 x 0.2 = 0.12 kg m^2
The relaton between the torque and teh moment off inertia is
τ = I α
Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia
τ = 0.12 x 0.5 = 0.06 Nm
Protons is 16 and Neutrons also 16