Answer:
Following are the responses to these question:
Explanation:
Since the
is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance
for capacities
for

When 
then
therefore,
remains at the same so, the maximum current remains the in same ckt.
"<em>The different types of radiation are defined by the the amount of </em><em>energy</em><em> found in the photons. Radio </em><em>waves</em><em> have photons with low energies, microwave photons have a little </em><em>more energy</em><em> than radio </em><em>waves</em><em>, infrared photons have still </em><em>more</em><em>, then visible, ultraviolet, X-rays, and, the </em><em>most</em><em> energetic of all, gamma-rays.</em>"
Answer:
4.32
Explanation:
The centripetal acceleration of any object is given as
A(cr) = v²/r, where
A(c) = the centripetal acceleration
v = the linear acceleration
r = the given radius, 1.9 m
Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that
9.8 = v²/1.9
v² = 1.9 * 9.8
v² = 18.62
v = √18.62
v = 4.32 m/s
This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s
Answer:
F = 39.36 N
Explanation:
given,
initial speed, u = 38 m/s
final speed, v = 0 m/s
mass of ball = 0.145 Kg
time, t = 0.14 s
Force = ?
using impulse formula
J = change in momentum
J = F x t
m(v - u) = F x t
0.145 x (0 - (-38)) = F x 0.14
F x 0.14 = 5.51
F = 39.36 N
force exerted by the ball is equal to 39.36 N.
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.