Answer:
The net force acting on the object is doubled while the mass of the object is held constant. What will be the new acceleration? An object has an acceleration of 12.0 m/s^2. The net force acting on the object is halved (decreased to one half its original value) while the mass of the object is held constant.
Answer:
Explanation:
a )
initial velocity u = 45 m/s
acceleration a = - 5 m/s²
final velocity v = 0
v = u - at
0 = 45 - 5 t
t = 9 s
b )
s = ut - 1/2 at²
= 45 x 9 - .5 x5x 9²
405 - 202.5
202.5 m
2 )
a )
s = ut + 1/2 a t²
u = 0
s = 1/2 at²
= .5 x 9.54 x 6.5²
= 201.5 m
b )
v = u + at
= 0 + 9.54 x 6.5
= 62.01 m / s
3
a )
acceleration = (v - u) / t
= (34 - 42) / 2.4
= - 3.33 m /s²
b )
v² = u² - 2 a s
34² = 42² - 2 x 3.33² s
s = 27.41 m
c )
Average velocity
Total displacement / time
= 27.41 / 2.4
= 11.42 m /s
4 )
a )
v = u + at
v = 0 + 3 x 4
= 12 m /s
b )
s = ut + 1/2 a t²
= o + .5 x 3 x 4²
= 24 m
The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
