There is no <span>radioactive decay</span>
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 =
/ W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
Answer:
d
a balanced force acted on it and propelled it to 4,000 km/hr
Explanation:
For the neutrons star which is moving in outer space at 4,000 km/hr, it could only be possible as a result of the balanced force which had already acted on it. <em>This is based on newton's law of motion which states that 'To every action, there is equal and opposite reaction'. </em>
Answer:
Explanation:
Area A of the coil = .1 x .1 = .01 m²
no of turns n = 5
magnetic field B = .5 t²
Flux Φ perpendicular to plane passing through it.= nBA sin30
rate of change of flux
dΦ/dt = nAdBsin30 / dt
= nA d/dt (.5t²x .5 )
= nA x 2 x .25 x t
At t = 4s
dΦ/dt = nA x 2
= 5x .01 x 2
= .1
current = induced emf / resistance
= .1 / 4
= .025 A
= 25 mA.