Solve for y11=y/3+7
1 answer:
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Answer:
i. The radius 'r' of the electron's path is 4.23 × m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r =
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 × ÷ 2.608 ×
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f =
= (1.6 × × 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 × ÷ 5.7263 ×
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
Answer:
Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
Hope this helped!
<h3>~AH1807</h3>The angular speed of the device is 1.03 rad/s.
<h3>What is the conservation of angular momentum?</h3>A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.
Using the conservation of angular momentum
Here, = the system's angular momentum before the collision
= 0 + mv
= (0.005)(450)(0.752)
= 1.692 kgm²/s
The moment of inertia of the system is given by
I = 2(M₁R₁² + M₂R₂²)+ mR₁²
= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²
= 1.6292 kgm²
Here, = Iω
So,
1.692 = 1.6292(ω)
ω = 1.03 rad/s
To know more about the conservation of angular momentum, visit:
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