Required value of initial speed of the bullet be ( 4M/m)√(gL).
Given parameters:
Mass of the bullet =m.
Mass of the bob of the pendulum = M.
speed of the bullet before collision = v
Speed of the bullet after collision = v/2.
Length of the pendulum stiff rod = L.
Let speed transmitted to the pendulum be u.
Using principle of conservation of momentum:
mv = Mu + mv/2
⇒ Mu = mv/2
⇒ u = (m/M)v/2
We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]
To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:
u = √(4gL)
⇒ (m/M)v/2 = √(4gL)
⇒ v =( 4M/m)√(gL).
Hence, minimum required speed of the bullet be ( 4M/m)√(gL).
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Answer:
t = 2.68 x 10¹⁴ years
Explanation:
First we need to find the amount of energy that Sun produce in one day.
Energy = Power * Time
Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)
Energy of Sun in 1 day = 3.32 x 10³¹ J
Now, the time required by the nuclear power generator, in years, will be:
Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J
3.32 x 10³¹ J = Power * Time
3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)
t = 3.32 x 10³¹ /1.24 x 10¹⁷
<u>t = 2.68 x 10¹⁴ years</u>
initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
here we know that
a = 9.8 m/s^2
t = 1.5 s
now from above equation
so it will hit the floor with speed 14.7 m/s
By Snell's law:
η = sini / sinr. i = 25, η = 1.33
1.33 = sin25° / sinr
sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.
r = sin⁻¹(0.3177)
r ≈ 18.52°
Option A.
God's grace.
