Wait a sec.. what is a diode? i would help if i knew what that was, sorry lol
Use the projectile motion equations
H = v^2 x sin^2(θ) ÷ 2g
t = 2 x v x sinθ ÷ g
R = v^2 x sin2θ ÷ g
Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:
![\sum F_y=m_1*a_m_i_n = T-m_1*g](https://tex.z-dn.net/?f=%5Csum%20F_y%3Dm_1%2Aa_m_i_n%20%3D%20T-m_1%2Ag)
If the package is barely lifted, that means that T=m_2*g; then:
![\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3Dm_1%2Aa_m_i_n%3Dm_2%2Ag-m_1%2Ag)
Solving the equation for a_mín, we have:
![a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2](https://tex.z-dn.net/?f=a_m_i_n%3D%28%28m_2-m_1%29%2Fm_1%29%2Ag%20%3D%20%28%2815kg-10kg%29%2F10kg%29%2A9%2C8%20m%2Fs%5E2%20%3D4%2C9%20m%2Fs%5E2)
Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey: ![\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20m_1%2Aa%20%5Crightarrow%20T-m_1%2Ag%3Dm_1%2Aa)
For the package: ![\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20m_2%2Aa%20%5Crightarrow%20m_2%2Ag%20-%20T%20%3D%20m_2%2Aa)
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package: ![\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20-m_2%2Aa%20%5Crightarrow%20T-m2%2Ag%3D-m_2%2Aa%20%5Crightarrow%20m_2%2Ag%20-T%3Dm_2%20%2Aa)
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
![m_1*a+m_1*g=m_2*g-m_2*a](https://tex.z-dn.net/?f=m_1%2Aa%2Bm_1%2Ag%3Dm_2%2Ag-m_2%2Aa)
Solving a, we have
![(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2](https://tex.z-dn.net/?f=%28m_1%2Bm_2%29%2Aa%20%3D%28m_2%20-%20m1%29%2Ag%5C%5C%5C%5Ca%3D%28%28m_2-m_1%29%2F%28m_1%2Bm_2%29%29%2Ag%20%5Crightarrow%20a%3D%28%2815kg-10kg%29%2F%2810kg%2B15kg%29%29%2A9%2C8%20m%2Fs%5E2%5C%5C%5C%5Ca%3D%202%2C0%20m%2Fs%5E2)
We can then replace this value of a in one for the sums of force and find the tension T:
![T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N](https://tex.z-dn.net/?f=T%20%3D%20m_1%2Aa%2Bm_1%2Ag%20%5Crightarrow%20T%3Dm_1%2A%28a%2Bg%29%5C%5C%5C%5CT%20%3D%2010kg%2A%282%2C0%20m%2Fs%5E2%2B9%2C8%20m%2Fs%5E2%29%20%5C%5C%5C%5CT%20%3D%20120%20N)
Speed = frequency * wavelength
<span>The statement is TRUE. Water does have potential energy at the top of a slope. The reason why is that potential energy is energy possessed by a body based on its position relative to a zero point. In this case, water at the top of the slope is at an elevation above ground (zero point). The energy is not kinetic (moving) energy since the water is not moving.</span>