I wanted to make sure these were the correct equations to use for they were not given. I had to research to get this far and only had the below statement to go off.
The problem is an object with initial velocity = "0." It travels 0.495m in 0.1625sec and weights 8.4 kg
or
Initial velocity (u) =...
I tried to write the data I understood from the image:
y0=160m
yf=0
x0=0
x1=192m
I tried to express the total change in time using the position over time equation on the Y direction:
y(t)=y0+v0y*(t2-t1)-0.5a(t2-t1)^2
but then I stuck with 2 variables and didn't know what to do
any help?
Givens:
Vyi=12.5 m/s
Vyf=-12.5 m/s (at the same horizontal level)
ay=-9.81 m/s^2
Δy= zero m (as the displacement on the y-axis, when the projectile reaches the same horizontal level, is zero m)
Δt=?
When I use
Δy=[(vyi+vyf)/2]*Δt
I get the time as undefined.
Δt= 2Δy/(vyi+vyf)
= 2*0 m/(12.5...
I was able to solve for the velocity of MB and got my answer as 4.47m/s.
The main issue right now for me is how to get the angles. I'm really confused on what most people have been posting as we didn't get a groundwork on this topic and so most of the basics I had them self taught.
So far I...
I have tried this problem over 4 times and keep getting the exact same answer, which is incorrect. Could someone please help by pointing out what I did wrong, or letting me know if I am actually on the completely wrong path?
Thanks!
Have a wonderful day :)
Derek Muller is the Veritasium YouTube channel host and he had a wager against UCLA professor Alexander Kusenko on whether a unpowered car going downwind can go faster than the wind.
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Figure 1. A map of the crime scene.
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Details of Question:
ds/dt= v which becomes ds=v dt, where s=displacement, t =time, and v=velocity
Then we can integrate both sides of this equation, and do a little algebra, and turn the above equation into:
s − s0 = v0t + ½at2
My main question is about the integration of...
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Part A)
So, I set the two positions equations of A and B equal to each other since the position has to be the same.
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Magnitude of acceleration of system:
a = (4.59kg - 1.71kg)(9.81N/kg)/(4.59kg + 1.71kg)
= 4.48 m/s^2
Velocity of lighter mass when heavier one hits the ground:
vf^2 = vi^2 + 2ad
= 0 + 2(4.48m/s^2)(2.60m)
vf = 4.83 m/s [up]
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None[/B]The Attempt at a Solution
None[/B]
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Vf=Vi(a)(t)
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Vf^2=Vi^2+2(a)(Δd)
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I...
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[/B]
Average...
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I recognize...
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Find the initial velocity v0x such that the particle will have the same x-coordinate at time t= 4.07 s as it had at t=0.
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Why 's' in all kinematics equation is displacement.
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Ok I have the answers to parts a and b, which I am 100% confident in since they are simple computations
Question is in attachement
a)74.9 rad/s
b)168.08 rad
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Two points P and Q move in a straight line AB.The point P starts from A in the direction AB with velocity and acceleration http://latex.artofproblemsolving.com/b/b/2/bb2c93730dbb48558bb3c4738c956c4e8f816437.png.At the same instant of time Q starts from B in the direction of...
x = x0 + v0 * t + 1/2 * constant acceleration * t^2
So this is supposed to be very very simple physics, but I still feel like there's a part of this equation I don't fully understand. The first term is the initial position of the body at t=0. The second term is the initial velocity at t=0...