Explanation:
The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.
All of the following are non-renewable resources except
O natural gas
O oil
O minerals
O <em>water ✓ </em>
- <em>Water </em><em>is </em><em>a </em><em>renewable </em><em>source </em><em>because </em><em>evaporation </em><em>and </em><em>condensation </em><em>takes </em><em>place </em><em>everytime </em><em>on </em><em>our </em><em>planet</em>
Answer:
a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]
Explanation:
a)
The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.
![E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2060%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2010%20%5Bm%5D%5C%5CE_%7Bp%7D%3D60%2A9.81%2A10%5C%5CE_%7Bp%7D%3D5886%5BJ%5D)
b)
Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.
![E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5C5886%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B5886%7D%7B0.5%2A60%7D%20%7D%5C%5Cv%20%3D%2014%5Bm%2Fs%5D)
c)
The work is equal to
W = 5886 [J]
d)
We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.
![v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bo%7D%20%5E%7B2%7D-2%2Aa%2Ad%5C%5Cwhere%3A%5C%5Cd%20%3D%202.5%5Bm%5D%5C%5Cv_%7Bf%7D%3D0%5C%5Cv_%7Bo%7D%20%3D14%5Bm%2Fs%5D%5C%5CTherefore%5C%5Ca%20%3D%20%5Cfrac%7B14%5E%7B2%7D%20%7D%7B2%2A2.5%7D%20%5C%5Ca%20%3D%2039.2%5Bm%2Fs%5E2%5D)
By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.
m*g - F = m*a
F = m*a - m*g
F = (60*39.2) - (60*9.81)
F = 1763.4 [N]
Answer:
4 ohms
Explanation:
Current = Voltage/resistance
2 = 8/R
2R = 8
R = 4