Given Information:
Radius = r = 0.5 m
Magnetic field = 1.0 T
Required Information:
Period = T = ?
Speed = v = ?
Kinetic energy = KE = ?
Answer:
Period = 0.13x10⁻⁶ seconds
speed = 24.16x10⁶ m/s
Kinetic energy = 12.11 MeV
Explanation:
(a) period
The time period of alpha particle is related to its orbital speed as
T = 2πr/v eq. 1
According to newton's law
F = ma
Force due to magnetic field is given by
F = qvB
qvB = ma
qvB = m(v²/r)
qB = mv/r
v = qBr/m eq. 2
substitute the eq. 2 in eq. 1
T = 2πr/qBr/m
r cancels out
T = 2π/qB/m
T = 2πm/qB
T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1
T = 0.13x10⁻⁶ seconds
(b) speed
From equation 1
T = 2πr/v
v = 2πr/T
v = 2π*0.5/0.13x10⁻⁶
v = 24.16x10⁶ m/s
(c) kinetic energy (in electron volts)
Kinetic energy is given by
KE = 0.5mv²
KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²
KE = 1.94x10⁻¹² J
since 1 electron volt has 1.602x10⁻¹⁹ J
KE = 1.94x10⁻¹²/1.602x10⁻¹⁹
KE = 12.11 MeV
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
It’s around the g force so it’s gonna be around 54 km/h
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
Answer:
A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15
∘
with the horizontal,the mass just being to slide what is the coffeficient of static friction between the block & the surface.
Explanation:
