A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended
1 answer:
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Answer:
Wavelength of light will be 632.73 nm
Explanation:
We have given width of the single slit
Angle is given as
We have to find the wavelength of light for first dark fringe
We know that for dark fringe it is given that
For first fringe n = 1
So
Answer:
a) The ball will clear the fence.
b) The distance between the ball and the top of the fence is 1.41 m.
Explanation:
The equation for the position of the ball is given by the vector "r":
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector of the ball at time t
v0 = initial velocity
x0 = initial horizontal position
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point.
When the ball reaches a horizontal distance of 116 m, the y-component of its vector position (r final in the figure, in red) is 0. Then:
r final = (116 m , 0)
Using the equations for the x and y-component of "r" we can calculate the total time of flight and the initial velocity:
x = x0 + v0 · t · cos α
116 m = 0 m + v0 · t · cos 45°
116 m / cos 45° · t = v0
Replacing v0 in the equation of the y-component, we can obtain the final time of flight:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 m = 0 m + 116 m / (cos 45° · t) · t · sin 45° - 1/2 · 9.8 m/s² · t²
0 m = 116 m - 4.9 m/s² · t²
-116 m / - 4.9 m/s² = t²
t = 4.87 s
Then, the initial velocity will be:
v0 = 116 m / cos 45° · t
v0 = 116 / cos 45° · 4.87 s
v0 = 33.7 m/s
a) Now, let´s find the y-component of the vector position when the x-component is 106 m (vector r in the figure). If the y-component plus 1.21 m is greater than 8.79, then, the ball will clear the fence.
Let´s find the time at which the position vector of the ball has an x-component of 106 m.
x = x0 + v0 · t · cos α
106 m = 0 m + 33.7 m/s · t · cos 45°
106 m / 33.7 m/s · cos 45° = t
t = 4.45 s
The y-component at that time will be:
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 0 m + 33.7 m/s · 4.45 s · sin 45° - 1/2 · 9.8 m/s² · (4.45 s)²
y = 9.01 m
Then, the height of the ball relative to the ground when the horizontal distance is 106 m (where the fence is) is 9.01 m + 1.21 m = 10.2 m. The ball will clear the fence.
b) The distance between the ball and the fence will be the height of the ball relative to the ground minus the height of the fence:
distance = 10.2 m - 8.79 m = 1.41 m
