I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.
I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball <u><em>drop</em></u>, at the same time and from the same height as the first ball.
Now I'll go ahead and answer the question that I've just invented:
Strange as it may seem, <em>both</em> balls hit the ground at the <em>same time</em> ... the one that's thrown AND the one that's dropped. The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.
And here's another example of the exact same thing:
Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else <em>drops</em> another bullet at exactly the same time, from a point right next to the end of the rifle barrel. I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.
Answer:
<h2>18 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 6 × 3
We have the final answer as
<h3>18 N</h3>
Hope this helps you
Answer:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
Answer:
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Explanation:
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