What is flens, the focal length of the lens? if the lens is converging flens is positive. it the lens is diverging, flens is neg
1 answer:
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Answer:
Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)
1) Todas las fuerzas están en la misma dirección.
Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:
F = 48N + 60N + 30N = 138N
Y por la segunda ley de Newton sabemos que:
F = m*a
fuerza igual a masa por aceleración.
Entonces la aceleración está dada por:
a = F/m = 138N/12kg = 11.5 m/s^2
2) Segundo caso, suponemos que F1 es opuesta a F2 y F3
En este caso, la fuerza neta será:
F = F2 + F3 - F1 = 60N + 30N - 48N = 42N
En este caso, la aceleración será:
a = 42N/12kg = 3.5 m/s^2
Answer:
For situation (a)
net charge E = E₊₂ + E₋₅ + E₋₃
E = K(q/d²)
where K = 8.99e9
d = 5.7cm = 5.7e-2m
Therefore,
E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)
E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)
E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)
thus
E = E₊₂ + E₋₅ + E₋₃
= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)
= 7.88e6(x) + 7.88e6(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) net magnitude = 1.115e6 @ 45° above +x axis
for situation (b)
net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆
E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)
E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)
E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)
E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)
thus,
E = E₊₄ + E₊₁ + E₋₁ + E₊₆
= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)
= 7.88e5(x) + 7.88e5(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) and (b) the net magnitude = 1.242e6 @ 45° above +x axis
Explanation:
I attached a sample image, i hope that corresponds to your question
Answer:
Explanation:
The angular momentum of an object in rotation is given by
where
I is the moment of inertia
is the angular speed
In this problem, initially we have
is the moment of inertia of the wheel
is the initial angular speed
So the initial angular momentum is
Later, a counterclockwise torque of
is applied
So the angular acceleration of the wheel is:
in the direction opposite to the initial rotation.
As a result, the final angular velocity of the wheel will be:
where
t = 4.0 is the time interval
Solving,
which means that now the wheel is rotating in the counterclockwise direction.
Therefore, the new angular momentum of the wheel is:
So, the change in angular momentum is: