4.266 m is the radius of the circular path the electron follows.
Given
Speed of electron (v) = 7.5 × 10⁶ m/s
Earth's Magnetic Field (B) = 1 × 10⁻⁵ T
We already know that
Mass of electron (m) = 9.1 × 10⁻³¹ kg
Charge on electron (q) = 1.6 × 10⁻¹⁹ C
According to the formula
Radius of circular path(r) = mass on electron × speed/ Charge × Magnetic field
Radius of circular path(r) = m × v/q × B
Put the values into the formula
r = 9.1 × 10⁻³¹ × 7.5 × 10⁶/ 1.6 × 10⁻¹⁹ × 10⁻⁵
On solving, we get
r = 4.266 m
Hence, 4.266 m is the radius of the circular path the electron follows.
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Kinetic energy is an
expression of the fact that a moving object can do work on anything it hits; it
quantifies the amount of work the object could do as a result of its motion. It
can be calculated as:
KE
= mv^2 /2
We
calculate the problem as follows:
KE
= mv^2 /2
400
= m(20)^2 / 2
<span>m
= 2 kg</span>
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
Answer:
Explanation:
<u>Relation between stress and Force:</u>
<u>Relation between stress and strain:</u>
Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:
So:
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½
)/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
