Answer:
B = 7.6 T direction of + x
Explanation:
For the proton beam to continue in the same direction the electric and magnetic forces must be equal
= 0
= F_{e}
Fm = q E
The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y
The magnetic force is
F_{m} = q v x B = q v B sin θ
θ = 90
B = q E / q v
B = E / v
B = 800/105
B = 7.6 T
To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.
Thumb goes in the direction of the + z axis
Palm in the direction of +y
Fingers point in the direction of + x
Newton has 3 Laws specifically The Three Laws of Motion
Answer:
The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Explanation:
The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.
Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.
From the law of conservation of energy, U₁ + K₁ = U₂ + K₂
So, kQ²/d + 0 = kQ²/3d + K
K₂ = kQ²/d - kQ²/3d = 2kQ²/3d
So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Answer:
170N
Explanation:
First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
