In a cold pack, an ___________ reaction draws heat from the surroundings. A. exothermic B. endothermic C. exergonic
2 answers:
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Answer:
A) E = 145.6 N / C , B) y= 2,8 10-7 m with a downward direction
C) he shape of the trajectory of the two particles is to simulate a parabola,
D) F_{e} /F_{g} = 10³⁴
Explanation:
A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric
F = m a
- e E = m a
a = - e E / m
with the field directed downward, the acceleration is in the vertical upward direction.
We look for how much the electron moves with kinematics, in the x direction there is no acceleration,
x axis (parallel to plates)
x = v₀ t
t = x / v₀
y axis (perpendicular to plates)
y = y₀ + t + ½ a t²
Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known
y = ½ a t²
we substitute
y = ½ (e E /m) (x / v₀)²
y = ½ e x2 /m v₀² E
we look for the electric field
E = 2 m y v₀² / e x²
where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m
let's calculate
E = 2 9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)
E = 145.6 N / C
B) The electron is exchanged for a proton
Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates
y = ½ e x² / m v₀² E
y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)² 145.6
y = 28.4375 10⁻⁸ m
since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small
In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)
C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards
D) The force of gravity
= G m M / R²
electron
Between the electron and the positive charges of the conducting plate
F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²
F_{g} = 4.1 10⁻⁵¹ N
electric force
F_{e} = -e E
F_{e} = - 1.6 10⁻¹⁹ 145.6
F_{e} = 2.3 10⁻¹⁷ N
let's look for the reason between these two forces
F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹
F_{e} /F_{g} = 10³⁴
We see that the electric force is many orders of magnitude higher than the gravitational force.
Answer:
ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:
Where:
Now:
Where is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:
Solving for :
Changed to cm/s, we get:
b) The kinetic energy K is calculated as:
K =
where M is the mass and V is the velocity.
So, the initial K is:
And the final K is:
Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J