<u>Answer:
</u>
Cat has 2.02 seconds to right itself.
<u>Explanation:
</u>
Initial height of cat from ground = 20 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8
, we need to calculate time when s = 20 meter.
Substituting
So, cat has 2.02 seconds to right itself.
Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
Answer:
8 time increase in K.E.
Explanation:
Consider Mass of truck = m kg and speed = v m/s then
K.E. = 1/2 ×mv²
If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then
(K.E.)₀ = 1/2 ×2m(2v)²
(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.
Answer:
The deceleration is
Explanation:
From the question we are told that
The distance of the car from the crossing is 
The speed is 
The reaction time of the engineer is 
Generally the distance covered during the reaction time is

=> 
=> 
Generally distance of the car from the crossing after the engineer reacts is
=>
=> 
Generally from kinematic equation

Here v is the final velocity of the car which is 0 m/s
So

=>