Answer:
q=1.4*10^{-9}C
Explanation:
Given data:
charge on ruler = -14μC
Mass of tissue is 5 g
To Know the minimum charge, equate electrostatic force to weight
we have F = W
so
putting all value in equation,
solving for q
or q=1.4*10^{-9}C
A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050
1 answer:
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Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum
Substitute the values
Magnitude of velocity of stone
=
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction
=
=33.3 degree below the horizontal
(b)
Initial kinetic energy
Final kinetic energy
=
Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.
(A) We can solve the problem by using Ohm's law, which states:
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
and 
, therefore we can rearrange Ohm's law to find the current through the device:
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
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.