All meteorites come from inside our solar system. Most of them are fragments of asteroids that broke apart long ago in the asteroid belt, located between Mars and Jupiter. Such fragments orbit the Sun for some time–often millions of years–before colliding with Earth
We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.
Answer:
y and length is directly relation
Explanation:
Given data
A single-slit diffraction pattern is formed on a distant scree
angles involved = small
to find out
what factor will the width of the central bright spot on the screen change
solution
we know that for single slit screen formula is
mass ƛ /area = sin θ and y/L = sinθ
so we can say mass ƛ /area = y/L
and y = mass length ƛ / area .................1
in equation 1 here we can see y and length is directly relation so we can say from equation 1 that the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled
Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
Answer:
Explanation:
Resistivity and resistance are proportional and depends of the length and the cross-sectional area of the wire:
furthermore, the density is the mass divided by the volume, and the volume can be written as the area multiplyed by the length:
Now you have tw equations and two variables, so you can solve for each of them.
first, solve for A in both equations and replace them:
now replace this into any of the previous equiations:
If you assume the wire has circular cross-sectional area, then the area is:
solving for d:
replacing A and simplifying:
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