Answer:
Explanation:
In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…
Answer:
C/100 = (F-32) / 180
or, C/5 = (F-32)/9
Explanation:
relation between any two scales is given by:
(X- lower fixed point ) / (upper fixed point -lower fixed point)
where X is any temperature
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>
Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
or
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump:
