Please some one help! I am about to go stir crazy. I am really struggling to answer a the following question...describe how force, momentum, Angular momentum, kinetic and potential energy are linked with respect to mass, acceleration and velocity.
Its probably really easy but science is not my...
I understand that angular velocity is technically not a vector so does that mean the cross product of the radius vector and the angular velocity vector, the tangential vector, is also not a vector?
Let there be a track 450,000 km long and a rocket 300,000 km long with a laser attached to the bottom of it's back end with a clock beside it, and a second synchronized clock attached to bottom of its front end. Both clocks were also synchronized with a track clock while the rocket was parked...
Firstly I would like to start with solving the problem with energy conservation principle which most solutions to the question show.
-Gmm/r= 1/2 mv^2 +1/2mv^2 -Gmm/2R
Where m= mass of planet
r= initial seperation
v= final velocity.
R=...
Refractive index is a function of velocity in the medium. How is this related to deviation angle inside the medium? I am not asking for the known formula, but for the mechanism behind it.
Hi, I am trying to determine the velocity of the particle with the mass m coming out of the acclerator.
I tried writing :
Ep(i) + Ec(i) = Ep(f) + Ec(f)
Ec(f) = Ec(i) - Ec(f)
But at this step, I'm no longer sure how to express Ep with V because :
In my textbook, it's written :
Ep = 0,5...
Hello, I've made a SPK file for asteroid 7482 (1994 PC1) with Horizon. I wan't to change the initial velocity with cspice, because I want to know where it will be in a future time at the speed changed. (now I'm using Newtons calc but is slow an error increases with time). This is for calculating...
Find the question and its solution below;
Ok i realized that we could also use cosine rule here, in my approach i considered the sketch below;
##V_b= 18,125- (2×50×125×cos 135)##
##V_b=164.2##
To find direction, i used sine rule;
##\frac {125}{sin α}##=##\frac {164.2}{sin 135}##
##α=32.56##...
The variation with time t of the acceleration a of an object is shown
What is the change in velocity of the object from ##t=0## to ##t=6##?
A. ##6ms^{-1}##
B. ##8ms^{-1}##
C. ##10ms^{-1}##
D. ##14ms^{-1}##
So apparently the answer is B, which I am having trouble reconciling.
Using methods...
Figure shows a locus of the figure axis of a symmetrical top on a unit sphere such that
##\dot{\theta}=\dot{\psi}=0## at the upper bounding circle. Where
##{\theta}## is the polar angle and ##{\psi}## is the azimuthal angle.
Suppose the figure axis is at the upper circle, since...
Adopt the speed of light equals one.
Calls ##cos = c##, ##sin = s##
$$ux' = \frac{v-uc}{1-uvc}$$
$$uy' = \frac{us}{\gamma(1-uvc)}$$
$$tan \theta' = uy' / ux' = \frac{us}{\gamma(v-uc)}$$
So that's basically my solution. The problem is: The answer is ##\frac{us}{\gamma(v+uc)}##. Now, i can't...
Sufficiently pressurised (difference between inlet and release pressure is enough to create supersonic flow) gaseous fluid is being released through a convergent-divergent nozzle. And it's a known fact that if pressure difference is sufficient, a convergent-divergent nozzle can release gaseous...
Here's a picture of the question:
This is a Khan Academy question and although I could just click on hint to find out what the answer is, I think it would be helpful to still ask this here before looking at the answer over there, so that I know what I did wrong. Thank you in advance to anyone...
My line of thinking is as follows:
\omega_{PQ} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}
Similarly for rod ##QR##
\omega_{QR} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}
Is my reasoning correct?
Here's where I got the questions:
These are from a worksheet I downloaded online: Answer Key
The answer key says that the answer to the first question is 500J and for the next question it's 433J.
It says constant speed though, so I don't understand why the answers aren't zero. I get how they...
First case, descends with the wheel:
mgh = .5(I)(w^2) ———- GPE converted to wheel energy
w = .1095. ———- rotation result is .1095
Second case, allow to free fall and impulse:
mgh = .5(m)(v^2). ———- GPE converted to kinetic energy
v = 7.746 ———-...
Hello, guys. Interesting riddle here.
I have no idea how to solve it. Tried different methods, but point is answer is always wrong,
exact answer Downriver, at an angle of 53.13(degree) to the bank.
That exercise is from
"Pohl’s Introduction to Physics"
I(i)w(i)= I(f)w(f)
I(i)= 1.08 x 10-3 kg·m2
w(i)= 0.221 rad/s
I(f)= mr^2 + I(i) = (5 x 10^-3)(.138)^2 + (1.08 x 10^-3)
(1.08 x 10-3)(.221) = ((1.08 x 10^-3)+9.22 x 10^-5))w(f)
w(f) = (2.3868 x 10^-4)/(0.00117522)
w(f)= 0.203094 rad/s
This is my attempt; however, I cannot seem to get it...
Distance:
substitute t=5 into x=3e^(0.4t)
22.17m
Velocity:
v=dx/dt
=1.2e^0.4t____(1)
Sub t=5 back into (1)
v= 8.867m/s
Acceleration:
a=dV/dt
=0.48e^0.4t____(2)
sub t=5 back into (2)
a= 3.547 m2/s
I am not sure if i am doing this right on dx/dt and dv/dt
Where exactly have I gone wrong? I think it is the part where I assume that the person gains the deceleration of the car, but I have no other way to proceed in this case. Also please only use the equations that I have posted below, and it would help if you would not use the equation for...
I am not sure what form of mass conservation to use to solve the above problem from An Introduction to Combustion by Stephen Turns. Can anyone explain what form of mass conservation applies to a sphere in this context?
He explain escape velocity in example where rocket goes straight up,isnt escacpe velocity ,velocity where centrifugal forces and gravity are equal,so refers only when rocket going in circle/orbit?
Can rocket really leave Earth in straight line like he show in video once reach this velocity and...
assuming initial velocity is 0 and we have the value for acceleration I'm unsure how to still use any of those equations because you must have a time value at least or a final velocity
I am having trouble with part b). I found the answer to part a) is -0.4 and for part b) the mark scheme says that the velocity of the 2nd bag is 1.6m/s but how do I find that out? If someone could perhaps show step-by-step of part b)?
Thank you!
I am currently doing an assignment on nuclear power and in the turbine, the steam is moving from pressure of 6Mpa to 0.008 Mpa. is there any way to work out the velocity of the steam when moving between these pressure differences?
A point mass is moving at speed v, on a horizontal plane, until it reaches an incline. Immediately after just climbing up the incline, its speed remains at v, but its direction changes. How does this happen?
Q2: Now, I drop a point mass such that it falls vertically downward onto a fixed ramp...
We can read: "The velocity dependence of the stopping power, increasing with decreasing velocity, is obvious from Fig.4".
I know why the stopping power depends on velocity as Bethe equation states, but I do not know how I can observe that dependence on a Bragg curve.
A rocket of initial mass m0 is launched vertically upwards from the rest. The rocket burns fuel at the constant rate m', in such a way, that, after t seconds, the mass of the rocket is m0-m't. With a constant buoyancy T, the acceleration becomes equal to a=T/(m0-m't) -g. The atmospheric...
I am trying to determine if there is a set of formula to calculate if a cabinet will when subjected to a force on it's upper edge for an angular distance (not a duration) will gain enough momentum to continue over the balance point and fall over or whether the opposing moment due to mass and...
I am a new tutor working with students in 11th grade physics in New York City. I am required to keep notes internally on the sessions and the standards that I cover in them; after noticing that the physical science standards don’t include mentions of velocity or displacement as a content area, I...
I just have a question that could you guys make an equation that expresses the terminal velocity based on followed condition?
- When diameter increase, velocity decrease
- velocity should change depending on both cylinder and sphere's diameter
- We know every variable
- The sphere is in...
For a particle moving in a straight line, if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
I am told the above statement is true.
If I look at the equations
v = dx/dt
a = dv/dt
It looks like if the velocity is zero for a time...
Let D be the opposite corner. In the CM frame, A moves towards CM, D moves towards CM as well. The other two corners (let them be B and C) moves away from the corner at ##v##. Then
##v_A \cos \theta/2 = v \sin \theta/2##
##v_{CM} = \frac{j}{4m}##
This is where it seems like the problem is...
So far I have:
The velocity of the belt will be the same for pully A and D, so we can calculate the angular velocity of pulley D:
## V_A = V_B ##
## \omega_A r_A = \omega_D r_D ##
## ((20*3)+40)(0.075) = \omega_D (0.025) ##
## \omega_D = 300 Rad/s ##
My next step was to determine the angular...
Hello!
I have done this problem :
vf^2 = (4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3)
so vf= sqrt((4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3))
I get vf = 4.7 x 10^5 m/s
However, the textbook solutions says vf = 8.7x10^5 m/s.
Where did I go wrong?
Thank you for any help! :)
Hi there,
I am solving a problem which requires me to use Bernoulli's equation and I have across a hint that says that "velocity at the reservoir is negligible because the reservoir is large" and I do not understand what that means exactly. How is velocity at a reservoir negligible?
I first used differentials to find an equation for the displacement.
$$6.0t^2+5.0t=13$$
and using the quadratic formula I got time ##t=1.1##
I then got ##V_x## from ##6.0t+5.0=v_x=11.6##
and ##V_y## from ##7.0t=v_y=7.7##
The I got v from ##V=\sqrt {V_x^2 + Vy^2} ##
$$\sqrt {11.6^2 + 7.7^2}...
Hi all,
It has been some time since I've done physics. I wish to model some projectile motion of a lure being cast from a fishing rod. The setup is very similar to that of a trebuchet.
The fishing rod - we'll assume a perfectly rigid beam - is rotating about a fixed axis. I can calculate...
Hi everyone :)!
I resolve this problem with components method and trigonometry method.
My results with components method its okay, but i can´t obtain the correct VE velocity.
Im sure that the problem its in the angles, but i don't know how to fix it.
The correct answers:
-Angular velocity...
I'm trying to understand what the time dilation looks like when moving through space and then approaching a planet's gravity field. So I have the broad understanding that if you are moving near the speed of light in a spaceship, your clock ticks normal but the clocks on other stationary objects...
In the figure below Ball 2(B2) is hanging and attached to a square lever. B1 drop from gravity and hit the lever as shown. Calculate velocity of B1 and B2 with gravity influence right after collision. The mass of B1 and B2 is identical.
Using energy equation E1=E2, therefore VB2 final = VB1...
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 :)
Hi! everyone! ;)
I have a problem with the development of this problem.
I need to resolve it with 2 procedures: trigonometry and instant centers. My advance can be see in the next image:
The instant centers procediment its (1) up and trigonometry procediment its (2) down.
I know that the...
i have tried V=Vi - (g*t) with an answer equals to 4
as well as
d=Vi(t) + 1/2 a t^2 with an answer equals to 19
my final asnwer was 1.43 and it was still wrong
i used formula a=v/t
v=14
t=10
i divided 14/10 to get 1.4
Here's what I did so far.
The velocity of the first car is ##v = v_0 +at##
Frame of reference S = the road
Frame of reference S' = the second car
thus, v' is the speed of the first car in the frame of reference S' and v the speed in the frame of reference S.
Here's what make me doubt.
The...