Answer:
0.694 m
Explanation:
Case 1 : When only mass of 2.82 kg is hanged from spring
m = mass hanged from the spring = 2.82 kg
x = stretch caused in the spring = 0.331 m
k = spring constant
Using equilibrium of force in vertical direction
Spring force = weight of the mass
k x = m g
k (0.331) = (2.82) (9.8)
k = 83.5 N/m
Case 2 : When both masses are hanged from spring
m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg
x = stretch caused in the spring = ?
k = spring constant = 83.5 N/m
Using equilibrium of force in vertical direction
Spring force = weight of the mass
k x = m g
(83.5) x = (5.91) (9.8)
x = 0.694 m
Answer:
The answer will be the first one.
Explanation:
Divide
The working equation to be used for this is written below:
E = kQ/d²
where
E is the electric field
k is a constant equal to 8.99 x 10⁹ N m²/C²
Q is the charge
d is the distance
E = (8.99 x 10⁹ N m²/C²)(17×10⁻⁹ C)/(0.05 m)²
E = 61,132 N/C
Answer:
circuito paralelo
Explanation:
Siempre el circuito en paralelo dara una resistencia menor. Recuerda que las resistencias se suman en el circuito en serie, an cambio en el circuito en paralelo, la corriente se bifurca de manera de circular con mayor intensidad por las ramas que tengan menos resistencia, y tal situacion llevara siempre a producir una menor resistencia equivalente.
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.
L must stay the same before and after the professor brings the dumbbells closer to himself.
His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2
His final moment of inertia is 2.2kg•m^2.
Calculate the initial angular velocity:
L = 3.0π
Final angular velocity:
L = 2.2w
Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:
3.0π = 2.2w
w = 1.4π rad/s
The rotational energy is given by:
KE = 0.5Iw^2
Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J
Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J
There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
