Answer:
1.5 Amp is rated for 5 W so it would not be possible
Answer:
Explanation:
Height attained by body = 50 cm
= .5 m
Initial velocity = u
v² = u² - 2gh
0 = u² - 2gh
u² = 2 x 9.8 x .5
u = 3.13 m /s
During the initial period , the muscle stretches by around 10 cm during which force by ground reacts on the body and gives acceleration to achieve velocity of 3.13 m/s from zero .
v² = u² + 2as
3.13² = 0 + 2 a x .10
a = 49 m/s²
reaction by ground R
Net force
R-mg = ma
R= m ( g +a )
= mg + ma
=W + (W/g) x a
W ( 1 + a / g )
= W ( 1 + 49 / 9.8 )
= 6W
Answer:
(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^
Explanation:
Given
W = 0.56N = mg
m = 0.56/g = 0.56/9.8 = 0.057kg
t = 3.00ms = 3.00×10-³s
Impulse is a vector quantity so we would treat it as such
We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that
u = (20.0m/s)ı^−(4.0m/s)ȷ^
ux = 20m/s
uy = – 4.0m/s
F = – (380N)ı^+(110N)ȷ^
Fx = –380N
Fy = 110N
J = impulse = force × time = F×t
So Jx = Fx ×t
Jy = Fy×t
Jx = –380×3×10-³ = -1.14Ns
Jy = 110×3×10-³ = 0.330Ns
Impulse also equals the change in momentum of the body. So
J = m(v–u)
J/m = v – u
V= J/m + u
Vx = Jx/m + ux
Vx = –1.14/0.057 + 20
Vx = -20 + 20 = 0m/s
Vx = 0m/s
Vy= Jy/m + uy
Vy= 0.33/0.057 + (-4.0)
Vy= 5.79 + (-4.0) = 1.79m/s
V = (0m/s)ı^−(1.79m/s)ȷ^
