Ok so if each side is 4.53 cm, we can multiply 4.53 x 4.53 x 4.53 to get the volume (since v= l x w x h). Density equals mass/volume, so
519 g/4.53 cm
114.57 g/cm^3 (since none of the units cancel)
Answer:
The weight lifter would not get past this sticking point.
Explanation:
Generally torque applied on the weight is mathematically represented as
T = F z
To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm
for z the perpendicular distance
So Elbow Torque is ![T_e= 4000 * 0.02](https://tex.z-dn.net/?f=T_e%3D%204000%20%2A%200.02)
![= 80Nm](https://tex.z-dn.net/?f=%3D%2080Nm)
To obtain the torque required we substitute 300 N for F and 30cm ![=\frac{30}{100} = 0.3 m](https://tex.z-dn.net/?f=%3D%5Cfrac%7B30%7D%7B100%7D%20%3D%200.3%20m)
So the Required Torque is ![T_R = 300 *0.3](https://tex.z-dn.net/?f=T_R%20%3D%20300%20%2A0.3)
![=90Nm](https://tex.z-dn.net/?f=%3D90Nm)
Now since
it mean that the weight lifter would not get past this sticking point
Work done is the distance a force acts over.
So, the work done here is 9.0N * 3.0m = 27 J
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest
Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest
THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding