Answer:
A). σ = 3.823 x 
/N-
B). 
C/
C). 
J
Explanation:
A). We know magnitude of charge per unit area for a conducting plate is given by
where, E is resultant electric field = 1.2 x 
V/m
is permittivity of free space = 8.85 x 
/N-
k is dielectric constant = 3.6
∴
= 3.6 x 8.85 x x 1.2 x
= 3.823 x /N-
B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by
C/
C).
Area of the plate, A = 2.5 
= 2.5 x 
diameter of the plate, d = 1.8 mm
= 1800 m
∴ Total energy stored in the capacitor
J
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
So the equation for angular velocity is
Omega = 2(3.14)/T
Where T is the total period in which the cylinder completes one revolution.
In order to find T, the tangential velocity is
V = 2(3.14)r/T
When calculated, I got V = 3.14
When you enter that into the angular velocity equation, you should get 2m/s
