The acceleration of one of those bugs is equal to 305mi/s.
<h3>Acceleration calculation</h3>
To calculate the insect's acceleration, the action and reaction force of the impact must be considered.
As the insect will hit the helmet, the force it hits is the same force it receives, so we can make the following expression:
<em>Speed has been converted to miles per second</em>
So, the acceleration of one of those bugs is equal to 305mi/s.
Learn more about acceleration calculation: brainly.com/question/390784
Answer:
a ) 2.13 X 10⁶ m/s .
b ) 1.28 X 10⁶ m/s
Explanation:
When electrons are repelled by negative plates and attracted by positive plates , it will increase their kinetic energy.
Increase in their energy = 4.15 eV
= 4.5 X 1.6 X 10⁻¹⁹ J
= 6.64 x 10⁻¹⁹ J
Initial kinetic energy
= 1/2 mv²
= 1/2 x 9.1 x 10⁻³¹ x ( 1.76 x 10⁶)²
= 14.09 X 10⁻¹⁹ J
Total energy
= 6.64 x 10⁻¹⁹+14.09 X 10⁻¹⁹
= 20.73 x 10⁻¹⁹
If V be the increased velocity
1/2 m V² = 20.73 X 10⁻¹⁹
.5 X 9.1 X 10⁻³¹ V² = 20.73 X 10⁻¹⁹
V = 2.13 X 10⁶ m/s .
b ) When electrons are released from positive plate , their speed are reduced because of attraction between electrons and positively charge plates.
Initial kinetic energy
= 14.09 x 10⁻¹⁹ J (see above )
reduction in kinetic energy
= 6.64 x 10⁻¹⁹ J ( See above )
Total energy with electron -
= 14.09 x 10⁻¹⁹ - 6.64 x 10⁻¹⁹
= 7. 45 x 10⁻¹⁹ J .
If V be the energy of electrons reaching the negative plate,
1/2 m V² =7. 45 x 10⁻¹⁹
V = 1.28 X 10⁶ ms⁻¹
Answer:
7.6427m/s
Explanation:
Given:
#Applying the conservation of momentum along the x-axis:
#And along y-axis:
#Solving for 
:
#By substitution in the x-axis equation:
Hence the original speed of the ball before impact is 7.6427m/s
Answer:
c.) 36E
Explanation:
The magnitude of the electric field is given by the expression
(1)
where k is the Coulomb's constant, q is the charge that generates the field, and d is the distance from the charge.
In this problem, we have that the magnitude of the field at a distance d is 4E, so we can rewrite the previous equation as
Now we want to determine the electric field at a distance of 
away. Substituting into (1), we find
(2)
We also know that
(3)
So combining (2) with (3), we find a relationship between the original field and the new field:
Answer:
(a) Maximum current through resistor is 1.43 A
(b) Maximum charge capacitor receives is 
.
Explanation:
(a)
In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by
Here, 
is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).
The maximum current is given by
<em>V</em> is the emf of the battery and 
is the effective resistance.
In this question, = 10.0 Ω + 25.0 Ω = 35.0 Ω
(b) The maximum charge is given
<em>Q</em> = <em>CV</em>
where <em>C</em> is the capacitance of the capacitor
