Answer:
ELASTIC collision
kinetic energy is conservate
Explanation:
As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.
As the floor does not move the conservation of the moment
po = pf
-mv1 = m v2
- v1 = v2
So the speed with which it descends is equal to the speed with which it rises
Therefore the kinetic energy of the ball before and after the collision is the same
Answer:
42.87
Explanation:
Google
<u>OR</u>
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420/9.8 which equals 42.87.
<span>anwser will be
F = ma
where
F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet
Substituting appropriately,
F = 0.005a --- call this Equation 1
Next working equation is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel
Substituting appropriately,
326^2 - 0 = 2(a)(0.83)
a = 64,022 m/sec^2
the anwser will be
Substituting this into Equation 1,
F = 0.005(64,022)
F =320.11 Newtons
Hope this helps. </span><span>
</span>
Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²) = 25 joules
Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²) = 100 joules
Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²) = 250 joules
Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters
Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg
