<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:
where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
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Answer:
Explanation:
b ) First is concave lens with focal length f₁ = - 12 cm .
object distance u = - 20 cm .
Lens formula
1 / v - 1 / u = 1 / f
1 / v + 1 / 20 = -1 / 12
1 / v = - 1 / 20 -1 / 12
= - .05 - .08333
= - .13333
v = - 1 / .13333
= - 7.5 cm
first image is formed before the first lens on the side of object.
This will become object for second lens
distance from second lens = 7.5 + 9 = 16.5 cm
c )
For second lens
object distance u = - 16.5 cm
focal length f₂ = + 12 cm ( lens is convex )
image distance = v
lens formula ,
1 / v - 1 / u = 1 / f₂
1 / v + 1 / 16.5 = 1 / 12
1 / v = 1 / 12 - 1 / 16.5
= .08333- .0606
= .02273
v = 1 / .02273
= 44 cm ( approx )
It will be formed on the other side of convex lens
distance from first lens
= 44 + 9 = 53 cm .
magnification by first lens = v / u
= -7.5 / -20 = .375 .
magnification by second lens = v / u
= 44 / - 16.5
= - 2.67
d )
total magnification
= .375 x - 2.67
= - 1.00125
height of final image
= 2.50 mm x 1.00125
= 2.503mm
e )
The final image will be inverted with respect to object because total magnification is negative .
La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".
Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.
Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.
En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
Answer:
a) 600nm
b) 300nm
Explanation:
the path difference = 2t
t = thickness of the film
L' = wavelength of light in film = L/n
L = wavength of light in air
n = refractive index of glass
(a)
for destructive interference 2t = L'/2 = L/2n
L = 4*t*n
= 4*120*10^-9*1.25
L = 600 nm
(b)
for constructive interference 2t = L' = L/1.25
L = 2tn
= 2 × 1.25 × 120nm
= 300 nm
