Both bricks will hit the ground at the same time.
Falling vertically is always accelerating at 9.8 m/s² because of gravity.
Nothing that's happening horizontally has any effect on that.
The brick that happens to have some horizontal motion will
probably hit the ground way over there, but that will still be
at the same TIME as this one.
This is a perfect place to remind you of the old unbelievable story,
which I'll bet you heard before:
If you fire a bullet horizontally from a gun, and at the exact same
moment you DROP another bullet out of your hand next to the gun,
the two bullets will hit the ground at the same time ! Even though
they'll be far apart.
Horizontal speed has no effect on vertical behavior.
Answer:
7/150
Explanation:
The following data were obtained from the question:
Object distance (u) = 75cm
Image distance (v) = 3.5cm
Magnification (M) =..?
Magnification is simply defined as:
Magnification (M) = Image distance (v)/ object distance (u)
M = v /u
With the above formula, we can obtain the magnification of the image as follow:
M = v/u
M = 3.5/75
M = 7/150
Therefore, the magnification of the image is 7/150.
<h2>
Option 2 is the correct answer.</h2>
Explanation:
Elastic collision means kinetic energy and momentum are conserved.
Let the mass of object be m and M.
Initial velocity object 1 be u₁, object 2 be u₂
Final velocity object 1 be v₁, object 2 be v₂
Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s
Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M
Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J
Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M
We have
Initial momentum = Final momentum
24 = 3v₁ + 6M
v₁ + 2M = 8
v₁ = 8 - 2M
Initial kinetic energy = Final kinetic energy
96 = 1.5 v₁² + 18 M
v₁² + 12 M = 64
Substituting v₁ = 8 - 2M
(8 - 2M)² + 12 M = 64
64 - 32M + 4M² + 12 M = 64
4M² = 20 M
M = 5 kg
Option 2 is the correct answer.
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
