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padilas [110]
3 years ago
11

Which are examples of perfectly in elastic collisions

Physics
1 answer:
sweet [91]3 years ago
8 0
Typical examples of inelastic collision are between cars, airlines, trains, etc.
For instance, when two trains collide, the kinetic energy of each train is transformed into heat, which explains why, most of the times, there is a fire after a collision. However, the momentum of the two trains that are involved in the collision remains unaffected. So, the trains collide with all their speed, maintaining their momentum, yet their kinetic energy is transformed into heat energy.
Another way to explain a train or a car collision is this: when the two trains or cars collide, they stick together while slowing down. They slow down because their kinetic energy is gradually lost. Still, they collide because they conserve their momentum.
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Neglecting air​ resistance, the distance s left parenthesis t right parenthesis in feet traveled by a freely falling object is g
ivann1987 [24]

Explanation:

It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet

The distance traveled by a freely falling object is given by :

s(t)=16t^2

16t^2=862

t=\sqrt{53.875}

t = 7.34 seconds

So, the object will take 7.34 seconds to fall to the ground from the top of the​ building. Hence, this is the required solution.

7 0
3 years ago
Is vectors and motion in plane same?
suter [353]

Answer:

<em>Plane Motion</em>

<em>Plane Motion One of the most common examples of motion in a plane is Projectile motion.</em>

8 0
2 years ago
Which evidence best supports the theory that the universe began with a massive explosion?
Ivanshal [37]
I think it's b. But I'm not sure
5 0
3 years ago
Read 2 more answers
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an
tensa zangetsu [6.8K]
The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
5 0
3 years ago
A Imagine you derive the following expression by analyzing the physics of a particular system: v2 + 2az. The problem requires so
Nuetrik [128]

Answer:

  z = 93.2 m

Explanation:

We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration

           v² = v₀² + 2 a d

If we make a term-to-term comparison with the expression obtained, they are equivalent

          u² = v² + 2 a z

From here we can clear the position

           2 a z = u² –v²

           z = (u² –v²) / 2 a

Let's calculate

For the speed to reduce the acceleration must be negative

         

         z = (0 - 21.8²) / 2(- 2.55)

         z = 93.2 m

7 0
3 years ago
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