Answer:
A. negative; virtual.
Explanation:
A concave lens always forms a virtual erect and diminished image. The reason why is that the image is actually formed by the intersection of virtually extended refracted rays.
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Answer:
Given a tube of diameter d, = 3cm = 0.03m
Pressure Balance
Mercury pressure at the tube bottom Pₓ = Pa + ρgh
where
Pa = Atmospheric pressure = 101kpa
ρ = Density of mercury = 13,546kg/m3
g = acceleration due to gravity
h = height of the tube?
Given
Bottom pressure in excess of the atmospheric pressure = 48kPa = Pₓ - Pa
Therefore, 48kPa = ρgh
h = 48(kN/m2)/ρg
h = 48,000kgms⁻²m⁻²/(13546kgm⁻³ x 9.81ms⁻²)
h = 0.36m
the tube is 36cm tall
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
A convex lens<span> makes </span>light<span> rays </span><span>converge </span>
