Answer: x ≈ 36.3 cm
Explanation:
Conservation of momentum during the collision
0.0340(120) + 1.24(0) = (0.0340 + 1.24) v
v = 3.2025 m/s
The kinetic energy of the block/bullet mass will convert to spring potential
½kx² = ½mv²
x = √(mv²/k)
x = √(1.274(3.2025²) / 99.0)
x = 0.363293... ≈ 36.3 cm
i think this is the incomplete page that you are showing but the answer is
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Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
Answer:
i = 0.477 10⁴ B
the current flows in the counterclockwise
Explanation:
For this exercise let's use the Ampere law
∫ B . ds = μ₀ I
Where the path is closed
Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.
From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.
We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field
Segment on the y axis
L₀ = (y2-y1)
L₀ = 3-0 = 3 cm
Segment on the point x = 2 cm
L₁ = 3-0
L₁ = 3cm
B L = μ₀ I
B 2L = μ₀ I
i = 2 L B /μ₀
i= 2 0.03 / 4π 10⁻⁷ B
i = 4.77 10⁴ B
The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise
We can solve the problem by using the first law of thermodynamics:
where
is the change in internal energy of the system
is the heat absorbed by the system
is the work done by the system on the surrounding
In this problem, the work done by the system is
with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is
with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find
