In the laboratory you will find three bottles with the following labels:

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

3 years ago

Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an

Svet_ta [14]

Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
(velocity)^2 = 10.78
velocity = 3.28 meters/sec

8 0

4 years ago

Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev

masha68 [24]

Answer:

The speed in the smaller section is $43.2\,\frac{m}{s}$

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

$\Delta Q=0$ (1)

With Q the flux of water that is $Av$ with A the cross section area and v the velocity, so by (1):

$A_{2}v_{2}-A_{1}v_{1}=0$

subscript 2 is for the smaller section and 1 for the larger section, solving for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$ (2)

The cross section areas of the pipe are:

$A_{1}=\frac{\pi}{4}d_{1}^{2}$

$A_{2}=\frac{\pi}{4}d_{2}^{2}$

but the problem states that the diameter decreases 86% so $d_{2}=0.86d_{1}$ , using this on (2):

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}$