Answer:
0.2cm towards the retina.
Explanation:
the focal length of the frog eye is
(1/f) = (1/10) + (1/0.8)
f = 0.74cm
Since the distance of the object is 15cm Hence
(1/0.74) = (1/15) + (1/V)
V = 0.78cm
Therefore the distance the retina is to move is
0.78cm - 0.8cm = 0.02cm towards the retina.
Elastic potential = 1/2 x constant x square of compression lenght
So it's 360 N/m
Answer:
The moment (torque) is given by the following equation:
![\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)](https://tex.z-dn.net/?f=%5Cvec%7B%5Ctau%7D%20%3D%20%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%5C%5C%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C%5Ek%5C%5Cr_x%26r_y%26r_z%5C%5CF_x%26F_y%26F_z%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C3k%5C%5C0.23%260.04%260%5C%5C150%26260%260%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5C%5Ek%28%280.23%2A260%29%20-%20%280.04%2A150%29%29%20%3D%20%5C%5Ek%20%2853.8~Nm%29)
Explanation:
The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.
Formula for height
<span> r(t) = a/2 t² + v₀ t + r₀
</span><span> where
</span><span> a = acceleration = -32 ft/sec² (gravity)
</span><span> v₀ = initial velocity
</span><span> r₀ = initial height
</span><span> r(t) = -16t² + v₀ t + r₀
</span> <span>Tomato passes window (height = 450 ft) after 2 seconds:
</span><span> r(2) = 450
</span><span> -16(4) + v₀ (2) + r₀ = 450
</span><span> r₀ = 450 + 64 - 2v₀
</span><span> r₀ = 514 - 2v₀
</span><span> Tomato hits the ground (height = 0 ft) after 5 seconds:
</span><span> r(5) = 0
</span><span> -16(25) + v₀ (5) + r₀ = 0
</span> r<span>₀ = 16(25) - 5v₀
</span><span> r₀ = 400 - 5v₀
</span><span>
r₀ = 514 - 2v₀ and r₀ = 400 - 5v₀
</span> <span>514 - 2v₀ = 400 - 5v₀
</span><span> 5v₀ - 2v₀ = 400 - 514
</span> <span>3v₀ = −114
</span><span> v₀ = −38
</span><span> Initial velocity = −38 ft/sec (so tomato was thrown down)
</span><span> (initial height = 590 ft) </span>
Answer:reactant,active site,enzyme below,substrate,products
Explanation:
