Answer:
a) Electric potential = 853 V
b) Electron speed at point B, if at Point A, the speed were zero = 1.732 × 10⁷ m/s
Explanation:
For an electron moving in an electric field with potential V,
Work done = qV where q is the charge on the electron
And the Work done is equal to the change in kinetic energy of the electron
qV = m(v₂² - v₁²)/2
V = m(v₂² - v₁²)/2q
q = 1.602 × 10⁻¹⁹C
m = 9.11 × 10⁻³¹ kg
v₁= 10⁷ m/s
v₂ = 2 × 10⁷ m/s
Putting these values in for the variables and solving
V = 853 V
b) If the electron started from rest,
qV = mv²/2
v = √(2qV/m) =√((2 × (1.602 × 10⁻¹⁹) × 853)/(9.11 × 10⁻³¹)) = 1.732 × 10⁷ m/s
Answer:
a = 0.55 m/s/s
Explanation:
As the car accelerates in forward direction the string will make some angle with the vertical direction
So here horizontal component of the tension force will accelerates the dice in forward direction along with the car
Vertical component of the tension force will balance the weight of the dice
so we will have
here we know that
= angle made with the vertical by string
now divide the two equations
so we have
Here are the choices.
A. 200 m
<span>B. 31.25 m </span>
<span>C. 0.2 m </span>
<span>D. 5 m
</span>Pa = N/m²
<span>m² = N/Pa </span>
<span>12.5 kPa = 12500 Pa </span>
<span>m² = N/Pa </span>
<span>m² = 2500 / 12500 </span>
<span>m² = 0.2 </span>
<span>C. 0.2 m²
</span>
So the correct answer is 0.2m^2
<span>(A)
s{ t } = (¼)t² + 2t + 1 ... differentiate to get velocity versus time
v{ t } = (½)t + 2 ... linearly increases with time
(1)
Vavɢ{ a→b } = [ v{ b } + v{ a } ] ⁄ 2
Vavɢ{ 1→3 } = [ v{ 3 } + v{1 } ] ⁄ 2
Vavɢ{ 1→3 } = [ (½)(3) + 2 + (½)(1) + 2 ] ⁄ 2
Vavɢ{ 1→3 } = 3 m/sec
(II) and (III) are done the same way.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
(B)
v{ t } = (½)t + 2
v{1 } = (½)(1) + 2
v{1 } = 2.5 m/sec</span>
Answer:
im pretty sure its 10 m/s but its kinda hard sorry
Explanation: