Answer:
radial acceleration is 41.8 m / s²
Explanation:
The acceleration for circular motion is
a = v² / r
They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations
x = vox t
y = v₀ₓ t - ½ g t2
Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)
x = v₀ₓ t
t = x / v₀ₓ
y = - ½ g t²
Let's replace and calculate the initial velocity on the X axis
y = - ½ g (x / vox)²
v₀ₓ = √ (g x² / 2 y)
v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]
v₀ₓ = 3.54 m / s
This is the horizontal velocity, but since it circle is in horizontal position it is also the velocity of the body at the point of rupture.
Now we can calculate the radial acceleration
a = v² / r
a = 3.54² / 0.300
a = 41.8 m / s²
Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.
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Determined by cross product or ( vector product )
Answer:
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
So, the wavelength of the infrared radiation is 
.
