Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
1 answer:
Answer:
av=0.333m/s, U=3.3466J
b.
Explanation:
a. let be the mass of block A, and
be the mass of block B. The initial velocity of A,
-The initial momentum =Final momentum since there's no external net forces.
Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):
-Applying the conservation of momentum. The blocks have the same velocity after collision:
#Total Mechanical energy before and after the elastic collision is equal:
Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s
b. Taking the end collision:
From a above,
We plug these values in the equation:
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Answer:
a).β=0.53 T
a).β=0.40 T
Explanation:
The magnetic field at distance 'r' from the center of toroid is given by:
a).
b).
The distance is the radius add the cross section so:
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m