Answer:
a)
down direction.
b)
3.82 µC
Explanation:
a)
Consider the motion of the positively charged bead in vertical direction
y = vertical displacement of charged bead = 5 m
a = acceleration of charged bead = ?
v₀ = initial velocity of bead = 0 m/s
v = final velocity of bead = 21.9 m/s
using the equation
v² = v₀² + 2 a y
inserting the values
21.9² = 0² + 2 a (5)
a = 47.96 m/s²
m = mass of the bead = 1 g = 0.001 kg
F = force by the electric field
Force equation for the motion of the bead in electric field is given as
mg + F = ma
(0.001) (9.8) + F = (0.001) (47.96)
F = 0.0382 N
Since the electric force due to electric field comes out to be positive, the electric force acts in down direction. we also know that a positive charge experience electric force in the same direction as electric field. hence the electric field is in down direction.
b)
q = magnitude of charge on the bead
E = electric field = 1 x 10⁴ N/C
Electric force is given as
F = q E
0.0382 = q (1 x 10⁴)
q = 3.82 x 10⁻⁶ C
q = 3.82 µC
Answer:
Launch velocity of the fleas = 1.80 m/s
Explanation:
Optimal launch angle is the angle that guarantees the maximum horizontal distance covered.
Optimum launch angle = 45°
The horizontal distance covered by a projectile is given as Range
R = (u² sin 2θ)/g
u = initial velocity of the projectile (flea) = ?
R = range = 33 cm = 0.33 m
θ = 45°
2θ = 2 × 45° = 90°
Sin 90° = 1
g = acceleration due to gravity = 9.8 m/s²
R = (u²/g)
u² = Rg = 0.33 × 9.8 = 3.234
u = 1.80 m/s
Hope this helps!!!
Answer is c that is he particles move perpendicular to the direction of the wave.
Answer:
x = 27.3 m
Explanation:
This is a projectile launching exercise, let's start by looking for the time it takes for the rock to reach the height of the window.
Let's use trigonometry to find the velocities of the rock
sin 40 = / v
cos 40 = v₀ₓ / v
v_{oy}= v sin 40
v₀ₓ = v cos 40
v_{oy} = 30 sin 40 = 19.28 m / s
v₀ₓ = v cos 40
v₀ₓ = 30 cos 40 = 22.98 m / s
we look for the time
= v_{oy}^2 - 2 g y
v_{y}^2 = 19.28 2 - 2 9.8 16 = 371.71 - 313.6 = 58.118
v_{y} = 7.623 m / s
we calculate the time
v_{y} = v_{oy} - gt
t = (v_{oy} - v_{y}) / g
t = (19.28 -7.623) / 9.8
t = 1,189 s
since the time is the same for both movements let's use this time to find the horizontal distance
x = v₀ₓ t
x = 22.98 1,189
x = 27.3 m
Answer:
Does shadow cover the Earth or only a portion of it?
Explanation: