Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
It slows the object down so it cannot move well and evetually the object cannot be pushed and farther
<h2>Answer:</h2>
<u>Distance covered is 6.9 meters</u>
<h2>Explanation:</h2>
Data given:
Work Done = 345 kJ = 345000 J
Force = 5 x 10 ^ 4 = 50000 N
Distance = ?
Solution:
As we know that
Work Done = Force applied x Distance covered
By arranging the equation we get
Work / Force = Distance covered
By putting the values
345000 / 50000 = 6.9
So distance covered is 6.9 meters
Answer:
2*10^-<em>5</em>
Explanation:
<em>B=</em><em>I</em><em>L</em>
<em>I=</em><em>B</em><em>/</em><em>L</em>
<em>I=</em><em>0</em><em>.</em><em>0</em><em>0</em><em>2</em><em>0</em><em>*</em><em>1</em><em>0</em><em>^</em><em>-</em><em>4</em><em>/</em><em>1</em><em>0</em>
<em>I=</em><em>2</em><em>*</em><em>1</em><em>0</em><em>^</em><em>5</em>
