Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ = 
with the new dough
v = 
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v = 
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v = 
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v = 
v = 1 /√5 v₀
v = 0.447 v₀
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
from the center of the pattern. In the formula, m is the order of the minimum,
the wavelenght,
the distance of the screen from the slit and
the width of the slit.
In our problem, the distance of the first-order band (m=1) is
. The distance of the screen is D=86 cm while the wavelength is
. Using these data and re-arranging the formula, we can find a, the width of the slit:
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Pressure = Force/ Area = 3000/2 = 1500 pascal.
Answer: The five major fields of environmental science are social sciences, geosciences, environmental chemistry, ecology, and atmospheric sciences.
Explanation:
