Answer:
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg
velocity of meteor v=40km/s \approx 40000 m/s
Kinetic Energy of Meteor
K.E.=\frac{mv^2}{2}
K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}
K.E.=2\times 10^6 J
Kinetic Energy of Car
=\frac{1}{2}\times Mu^2
=\frac{1}{2}\times 900\times u^2
\frac{1}{2}\times 900\times u^2=2\times 10^6
900\times u^2=4\times 10^6
u^2=\frac{4}{9}\times 10^4
u=\frac{2}{3}\times 10^2
u=66.67 m/s
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
Answer:
1.549 m
Explanation:
Given:
The radius of the circular board, r = 2 m
The probability of hitting the red is given as 0.6
Now, this probability of hitting the red can be conclude as
0.6 = (Area of red)/ (Total area of the board)
Total area of the board = πr² = π × 2²
let the radius of the red area be R
thus, area of red circle, = πR²
on substituting the value of the area, we have
0.6 = (πR²)/ (π × 2²)
or
R² = 2.4
or
R = 1.549 m
Thus, the radius of the red circle is 1.549 m
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
