Addition of 2vector gives you 1large vector quantity
Answer:
(b) To get m3 to slide, m1 must be increased, never decreased.
Explanation:
Lab experiments require attentiveness. If there is one thing missed or not taken seriously whole experiment could go wrong. In this case to slide m3 there should be more weight at m1. If the weight of m1 is lesser than m3 then the object will not slide. It will remain at the point where there is more weight. To slide an object there must be less frictional surface and more weight placed at the desired end point.
Answer:
7 Newton
Explanation:
Dado
Longitud de la cuerda = 50 m
El cable se dobla en 0,058 m.
Masa de roedor = 350 gramos = 0,35 kg
T = m * a + m × v2 / r
Sustituyendo los valores dados obtenemos
T = 0,35 (10 + 10)
T = 0,35 * 20
T = 7 Newton
Answer:
(B) The total internal energy of the helium is 4888.6 Joules
(C) The total work done by the helium is 2959.25 Joules
(D) The final volume of the helium is 0.066 cubic meter
Explanation:
(B) ∆U = P(V2 - V1)
From ideal gas equation, PV = nRT
T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3
P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal
∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules
(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal
Assuming a closed system
(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules
(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter
Hello!
momentum is p=mv so if the velocity is decreased by a factor of 2, which is 1/2 the value, then the momentum of that object will decrease by the same factorial.
Hope this helps, any questions please ask. Thank you!
