<span>v/2
This is an exercise in the conservation of momentum.
The collision specified is a non-elastic collision since the railroad cars didn't bounce away from each other. For the equations, I'll use the following variables.
r1 = momentum of railroad car 1
r2 = momentum of railroad car 2
x = velocity after collision
Prior to the collision, the momentum of the system was
r1 + r2
mv + m*0
So the total momentum is mv
After the collision, both cars move at the same velocity since it was non-elastic, so
r1 + r2
mx + mx
x(m + m)
x(2m)
And since the momentum has to match, we can set the equations equal to each other, so:
x(2m) = mv
x(2) = v
x = v/2
Therefore the speed immediately after collision was v/2</span>
Answer:
v = 4.10 10⁻³ m / s
Explanation:
For this exercise we will use Newton's second law where the force is magnetic
F -W = m a
As the field is directed to the north and the proton to the east, using the rule of the right hand the force is vertical upwards, the force balances the weight the acceleration is zero
F = W
q v B = m g
Let's calculate the speed
v = m g / (q B)
v = 1,673 10⁻²⁷ 9.8 / (1.6 10⁻¹⁹ 2.5 10⁻⁵)
v = 4.10 10⁻³ m / s
Strength of the magnetic field: 20 T
Explanation:
For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by
![\epsilon=BvL](https://tex.z-dn.net/?f=%5Cepsilon%3DBvL)
where
B is the strength of the magnetic field
v is the speed of the wire
L is the length of the wire
For the wire in this problem, we have:
(induced emf)
L = 0.20 m (length of the wire)
v = 3.0 m/s (speed)
Solving for B, we find the strength of the magnetic field:
![B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cepsilon%7D%7BvL%7D%3D%5Cfrac%7B12%7D%7B%280.20%29%283.0%29%7D%3D20%20T)
Learn more about magnetic fields:
brainly.com/question/3874443
brainly.com/question/4240735
#LearnwithBrainly
Answer:
a)ω₂=127.64 rad/s ( min)
ω₁= 333.32 rad/s ( max)
b) At d₂= 11.75 cm ,ω₂=127.64 rad/s
d₁=4.5 cm,ω₁= 333.32 rad/s
c)α = 7.1 x 10⁻³ rad/s²
Explanation:
Given that
r₂= 11.75 cm
r₁=4.5 cm
v= 7.5 m/s
a)
We know that
v=ω r
ω =Angular speed
r= radius
v= velocity
When d₂= 11.75 cm :
v=ω r
7.5 x 2 =ω₂ x 0.1175
ω₂=127.64 rad/s ( min)
![\omega=\dfrac{2\pi N}{60}](https://tex.z-dn.net/?f=%5Comega%3D%5Cdfrac%7B2%5Cpi%20N%7D%7B60%7D)
![127.64=\dfrac{2\pi N_2}{60}](https://tex.z-dn.net/?f=127.64%3D%5Cdfrac%7B2%5Cpi%20N_2%7D%7B60%7D)
N₂=1219.4 rpm
When d₁=4.5 cm :
v=ω r
7.5 x 2=ω₁ x 0.045
ω₁= 333.32 rad/s ( max)
![\omega=\dfrac{2\pi N}{60}](https://tex.z-dn.net/?f=%5Comega%3D%5Cdfrac%7B2%5Cpi%20N%7D%7B60%7D)
![333.32=\dfrac{2\pi N_1}{60}](https://tex.z-dn.net/?f=333.32%3D%5Cdfrac%7B2%5Cpi%20N_1%7D%7B60%7D)
N₁=3184.58 rpm
The average angular acceleration α given as
ω₁ = ω₂ + α t
333.32 = 127.64 + α x 8 x 60 x 60
α = 7.1 x 10⁻³ rad/s²