From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
<span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
</span><span> v = 120/3.6 = 33.33 m/s
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = <span>
146235.375 </span><span>kg/s
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span>
<span>4874025 </span></span><span>N
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
Answer:
a. 5.23 m/s² b. 44.23 N
Explanation:
a. What is the centripetal acceleration of the hammer?
The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.
Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.
So a = rω²
= 1.4 m × (3.74 rad/s)²
= 5.23 m/s²
b. What is the tension in the chain?
The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.
So, T = 8.45 kg × 5.23 m/s²
= 44.23 N
Answer:
true is the answer of the question
C is the correct answer, hope it helps
