Answer:
To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.
Explanation:
Given ,
F = 130N
M = 24kg
A = ?
F = m× a
then ,
130N = 24kg ×a
a = 130/24 = 5 m/s.
Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Answer: D. Energy released from atoms.
200N is the answer (at least thats what I think)
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
