Answer:
The final velocity of the ball is 7m/s
Explanation:
M1=8kg, V1 =10m/s
, M2=2kg
, V2=-5m/s
initial momentum before collison
m1v1+m2v2
=8×10 +2×(-5) =80-10 = 70kg m/s
final momentum after collison
=(m1+m2)×v
=(8+2)×v
=10v
According to the law of conversion of momentum
initial momentum =final momentum
70=10v
10v=70
v=70/10
v=7m/s
Answer:
That insane it might be true because a planet sometimes quoted to be an Earth 2.0 or Earth's Cousin based on its characteristics; also known by its Kepler Object of Interest designation KOI-7016.01) is an exoplanet orbiting the Sun-like star Kepler-452 about 1,402 light-years (430 pc) from Earth in the constellation Cygnus.
Explanation:
Answer:
Explanation:
L 1: front radius 950 mm, rear radius 2700 mm, refractive index 1.528;
We shall use lens maker's formula , that is
1/f = (μ-1) ( 1/R₁ - 1/R₂) , μ is refractive index of the lens , R₁ and R₂ are radius of curvature of front and rear curved surface.
1/f₁ = (1.528-1)( 1/950 + 1/2700)
f₁ = 1331 mm
L2: front radius 535 mm, rear radius 500 mm, refractive index 1.550.
1/f₂ = (1.550-1)( 1/535 + 1/500)
f₂ = 470 mm
largest angular magnification possible
= f₁ /f₂
= 1331 / 470
= 2.83 ( approx )
Length between two lenses
=1331 +470
= 1801 mm
= 1.8 m Ans
Answer:
The answer is (C) E= 1/2 kx^2
Explanation:
Answer:
a)y = 485 m
, v = 220 m / s
, b) y = 2954.39 m
, c) t_total = 51 s
,
d) v = 240.59 m / s
Explanation:
a) We can use vertical launch ratios for this exercise
the speed of the rocket the run out the fuel is
v = v₀ + a t
the rocket departs with initial velocity v₀ = 0
v = a t
v = 55 4
v = 220 m / s
the height at this point is
y = y₀ + v₀t + ½ a t²
y = y₀ + 1/2 a t²
y = 45 + ½ 55 4²
y = 485 m
b) the maximum height of the rocket is when its speed is zero
for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m
v² = v₀´² + 2 g (y-y₀´ )
0 = v₀´² + 2 g (y-y₀´ )
y = y₀´ + v₀´² / 2g
y = 485 + 220 2/2 9.8
y = 2954.39 m
c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)
let's calculate the rise time
v = v₀´- g t
v = 0
t₂ = v₀´ / g
t₂ = 220 / 9.8
t₂ = 22.45 s
Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor
y = y₀´´ + v₀´´ t - ½ g t²
0 = y₀´´ - ½ g t²
t = √ (2 y₀´´ / g)
t = √ (2 2954.39 / 9.8)
t = 24.55 s
the total flight time is
t_total = t₁ + t₂ + t₃
t_total = 4 + 22.45 + 24.55
t_total = 51 s
d) the veloicda right now
v = vo + g t
v = 9.8 24.55
v = 240.59 m / s