Please help me!! Need this done before the 40 min end
1 answer:
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Answer:
v₁₀ = 1.90 m / s
Explanation:
In this exercise we are given the maximum height data, with energy we can know how fast the body came out
Final mechanical energy, maximum height
= U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em=
½ m v² = m g h
v = √ 2gh
Now we can use the moment to find the speed with which objects collide
The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v
Initial before the crash
p₀ = M v₁₀ + 0
Final after the crash
= M v1f + m v
p₀ =
M v₁₀ = M + m v
As the shock is elastic the kinetic energy is conserved
K₀ =
½ M v₁₀² = ½ M ² + ½ m v²
Let's write the system of equations
M v₁₀ = M + m v
M v1₁₀² = M ² + m v²
We cleared v1f in the first we replaced in the second
= (M v₁₀ - mv) / M
M v₁₀² = M (M v₁₀ - mv)² / M² + m v²
M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²
v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
Let's substitute the value of v
v1₁₀= √ (2gh) (m +1) / (2M)
Let's calculate
v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)
V₁₀ = 7.668 (2.68) / 10.82
v₁₀ = 1.90 m / s
Answer:
b.
Explanation:
In case of Single Slit, diffraction will occur.
Then In Single slit Diffraction, width of central fringe is
where D= distance b/w screen and slit
a= slit width
\lambda = wavelength
Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as
On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.