Answer:
F = 1958.4 N
Explanation:
By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side
so we have
![L_1A_1 = L_2A_2](https://tex.z-dn.net/?f=L_1A_1%20%3D%20L_2A_2)
![1.20(\pi 18^2) = L_2(\pi 5^2)](https://tex.z-dn.net/?f=1.20%28%5Cpi%2018%5E2%29%20%3D%20L_2%28%5Cpi%205%5E2%29)
![L_2 = 15.55 m](https://tex.z-dn.net/?f=L_2%20%3D%2015.55%20m)
so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m
So here the net pressure on the smaller area is given as
![P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)](https://tex.z-dn.net/?f=P%20%3D%20P_%7Batm%7D%20%2B%20%5Cfrac%7B12%2C000%7D%7B%5Cpi%20%280.18%29%5E2%7D%20%2B%20%5Crho%20g%20%281.2%20%2B%2015.55%29%20)
excess pressure exerted on the smaller area is given as
![P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)](https://tex.z-dn.net/?f=P_%7Bex%7D%20%3D%20%5Cfrac%7B12000%7D%7B%5Cpi%20%280.18%29%5E2%7D%20%2B%20800%289.81%29%2816.75%29)
![P_{ex} = 2.49\times 10^5 Pascal](https://tex.z-dn.net/?f=P_%7Bex%7D%20%3D%202.49%5Ctimes%2010%5E5%20Pascal)
now the force required on the other side is given as
![F = P_ex (area)](https://tex.z-dn.net/?f=F%20%3D%20P_ex%20%28area%29)
![F = (2.49 \times 10^5)(\pi (0.05)^2)](https://tex.z-dn.net/?f=F%20%3D%20%282.49%20%5Ctimes%2010%5E5%29%28%5Cpi%20%280.05%29%5E2%29)
![F = 1958.4 N](https://tex.z-dn.net/?f=F%20%3D%201958.4%20N)
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
![m_1v_1+m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
![m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
![(0.42)(21) = (90+0.42)v_f](https://tex.z-dn.net/?f=%280.42%29%2821%29%20%3D%20%2890%2B0.42%29v_f)
![v_f = 0.0975m/s](https://tex.z-dn.net/?f=v_f%20%3D%200.0975m%2Fs)
Therefore the velocity right after catching the ball is 0.0975m/s
Answer:
Explanation:
Let fuel is released at the rate of dm / dt where m is mass of the fuel
thrust created on rocket
= d ( mv ) / dt
= v dm / dt
this is equal to force created on the rocket
= 220 dv / dt
so applying newton's law
v dm / dt = 220 dv / dt
v dm = 220 dv
dv / v = dm / 220
integrating on both sides
∫ dv / v = ∫ dm / 220
lnv = ( m₂ - m₁ ) / 220
ln4000 - ln 2500 = ( m₂ - m₁ ) / 220
( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )
( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )
= 103.4 kg .
Answer:
Tension = 0.012 N
Explanation:
If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,
Tension = Weight
Tension = mg
where,
m = mass of spider = 1.27 g = 0.00127 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
Tension = (0.00127 kg)(9.8 m/s²)
<u>Tension = 0.012 N</u>