Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer:
Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
Explanation:
Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
Answer: Katherine Johnson's knowledge of (mathematics) was instrumental in the return of the Apollo astronauts from the Moon to Earth.
Answer:
induced emf = 28.65 mV
Explanation:
given data
diameter = 7.3 cm
magnetic field = 0.61
time period = 0.13 s
to find out
magnitude of the induced emf
solution
we know radius is diameter / 2
radius = 7.3 / 2
radius = 3.65 m
so induced emf is dπ/dt = Adb/dt
induced emf = A × ΔB / Δt
induced emf = πr² × ΔB / Δt
induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13
induced emf = 0.0286538 V
so induced emf = 28.65 mV
Answer:
space , small amount of gravity is found in the space ,infact we can say that there is no gravity in the space
