Answer:
517.5Ns
Explanation:
F=(MV - MU)/t
where MV - MU is the change in momentum,
therefore, MV - MU = Ft
= 345 X 1.
= 517.5Ns
Answer:
562.5J
Explanation:
The following were obtained from the question:
F = 45N
d = 12.5m
w =?
The work done can be achieved by using
w = F x d
w = 45 x 12.5
w = 562.5J
Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N
Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s
