alright got it now, thanks a lot for your help.
i was about 80% through typing out my reworking to see if you could check it when i accidentally clicked the back button on my mouse and lost it all, and i really can't put myself through typing it again lol.
1. Homework Statement
Three point charges lie along the x axis. A positive charge q1=16.0mC is at x=2.00m. Another positive charge q2=9.00mC is at the origin.
Where should we put a third charge so the resultant force acting on it is zero?
What should be the sign of the charge?
2...
1. Homework Statement
A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle theta from the horizontal. There is a uniform, vertical magnetic field vecB at all points (produced by an arrangement of magnets not shown in the...
ΔB = μo/4π(IΔLsinθ)/R^2
so I = 27
θ = 45
ΔL = 2mm???
and R can be found using trig.
so what is μ and o??
soz just really lost. any help would be great.
1. Homework Statement
A wire carrying a 27.0A current bends through a right angle. Consider two 2.00mm segments of wire, each 3.00cm from the bend.
Find the magnitude of the magnetic field these two segments produce at point P, which is midway between them.
2. Homework Equations...
r = mv/qb
0.008276 = (1.67x10^-27 x 1200) / (1.602×10^−19 x b)
(1.602×10^−19 x b) = (1.67x10^-27 x 1200) / 0.008276)
(1.602×10^−19 x b) = 2.42146x10^-22
b = 1.5115x10^-3
would i be correct in saying that using the arc length formula:
pi/2 radians is equivalent to 1.3cm
θ = s / r
r = s / θ
= 1.3 / pi/2
r = 0.8276 cm
t = 0.013m / 1200m/s
= 1.083x10^-5 s
then w = θ / t
= (pi/2) / 1.083x10^-5
= 145041.2121
w = w0 + at
145041.2121 = 0...
yeh i was about to ask how you knew the radius. ok i'll work on those and get back to you in a few mintues.
**digs deep into the memory bank for equations**
ok so:
F = qv x B
B = F/qv
= F / (1.602 × 10^−19) x 1.2 (charge of a proton x velocity of proton)
F = ma?
= 1.67x10^-27 x a (mass of a proton x acceleration)
however, as:
m(v/t) / q(v)
= m(1/t) / q
= (1.67x10^-27 x (1/t)) / 1.602 × 10^−19 (where t = distance / velocity??)
1. Homework Statement
A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction. The beam travels a distance of 1.30 cm while in...
hey all,
i've recently been given an assignment of special relativity, the question is as follows:-
Topic B: SPECIAL RELATIVITY
Write a short report on any experiment, published in the literature (ie journals), which proves or depends on a relativistic effect. Include essential experimental...
similar question...
A toy gun uses a spring to project a 5.3 g soft rubber sphere. The spring constant k = 8.0 N m-1. The barrel of the gun is 15 cm long and there is a constant frictional force of 0.032 N between the barrel and the projectile. If the spring is compressed 5 cm what is the speed...
1. Homework Statement
The rotor (flywheel) of a toy gyroscope has mass 0.140 kilograms. Its moment of inertia about its axis is 1.20x10^{-4} kilogram meters squared. The mass of the frame is 0.0250 kilograms. The gyroscope is supported on a single pivot with its center of mass a horizontal...
a)
try the formula u^2 = v^2 + 2as
where u=initial velocity, v=final velocity, a=acceleration (will be a negative), and s=the distance traveled.
b)
try using v = u + at
using the final velocity that you found in section a.
t will give you time.
1. Homework Statement
A spring-loaded toy gun is used to shoot a ball of mass m= 1.50kg straight up in the air, as shown in the figure. The spring has spring constant k = 667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the...
horizontal component:
-Tacos60+Tbcos45=0
Tacos60=Tbcos45
Ta=Tb(cos45/cos60)
=Tb(2^.5)
vertical component:
-Tasin60+Tbsin45-Tcsin90=0
-Tb(2^.5)(sin60)+Tbsin45=w
0.707Tb-((2^.5)(sin60))Tb=w
Tb=-w/0.5176
which i put in and was wrong. so i misc decided to make the a component positive and put...
sweet thanks.
got them both right.
now need to try for part b :p
do i apply the same rules or does it change up since one cord is coming from the side now?
ok so say i take the first of the 2 options:
x component:
Ta+Tb+Tc=0
Ta+Tb=0 (as there is no x direction of Tc)
-Tacos30+Tbcos45=0
Tacos30=Tbcos45
Ta=Tbcos45/cos30
y component:
Tasin30+Tbsin45-Tcsin90=0 (Tc is in the negative direction?)
Tb(cos45/cos30)(sin30)+Tbsin45=w (subbing Tc for w...
ok so i'm trying to follow an example from my textbook, but it's only dealing with 2 strings so that is probably where i am getting stuck.
Ta+Tb+Tc+w=ma
as net force is 0, ma = 0.
then into x and y components:
x:
Tax+Tbx+Tcx+wx=0
-Tacos30+Tbcos45+Tccos90+0=0
y:
Tay+Tby+Tcy+wy=0...