Rachel and Sarah are on a bus travelling at 5 mph past John who is standing on the sidewalk. Rachel then throws a ball

Which are examples of perfectly in elastic collisions

sweet [91]

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.

8 0

3 years ago

man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h

-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

L₀ = I₁ w₁

final moment. After releasing the bricks

$L_{f}$ = I₂W₂

L₀ = L_{f}

I₁ w₁ = I₂ w₂

w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

let's calculate

w₂ = 6.0/2.0 7.54

w₂ = 22.6 rad/s

3 0

3 years ago

- A thin film of oil * (n = 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place whe

Darya [45]

Answer:

There will be a phase change at the first interface and no phase change at the second interface:

If the film is 1/4 wavelength thick this restriction will hold

The wavelength of the light in oil is 545 nm / 1.45 = 376 nm

376 nm / 4 = 94 nm

"D" is correct

4 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

What are the top three osha cited ladder violations

bonufazy [111]

Answer:

Explanation:

1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS

2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671

3. SCAFFOLDING (29 CFR 1926.451). 2,813

6 0

3 years ago