How is a solution diluted?

Upon reaching a velocity of 100fps, the pilot of the airplane decides to abort the take off and applies brakes and stops the air

masha68 [24]

Answer:

5 ft/s²

Explanation:

u = Initial velocity = 100 ft/s

v = Final velocity = 0

s = Displacement = 1000 ft

a = Acceleration

From equation of motion

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-100^2}{2\times 1000}\\\Rightarrow a=-5\ ft/s^2$

$1\ ft/s^2=0.3048\ m/s^2\\\Rightarrow 5\ ft/s^2=5\times 0.3048\ m/s^2=1.524\ m/s^2$

The airplane's deceleration is 5 ft/s² or 1.524 m/s²

8 0

3 years ago

What Element is represented by the diagram?

pychu [463]

Answer:Beryllium

Explanation:

8 0

3 years ago

Read 2 more answers

A 12000 kg railroad car is traveling at a 2 m/s when it strikes another 10000 kg railroad is at rest. If the car lock together w

zimovet [89]

The final speed is $1.09m\frac{m}{s}$

Why?

To calculate the final speed, we need to know that we are working with an inelastic collision, it means that both bodies stick together after the colission. We can calculate the final speed of the bodies (if exists) using the following equation:

$m_{1}v{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{f}$

We are given the following information:

$m_{1}=12000kg\\v_{1}=2\frac{m}{s}\\m_{2}=10000kg\\v_{2}=0\frac{m}{s}(at rest)$

So, substituting and calculating, we have:

$m_{1}v{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{f}\\\\12000kg*2\frac{m}{s}+10000kg*0\frac{m}{s}=(12000kg+10000kg)*v_{f}\\\\v_{f}=\frac{24000\frac{kg.m}{s} }{22000kg}=1.09m\frac{m}{s}$

Hence, the final speed is 1.09 m/s.

Have a nice day!

5 0

3 years ago

5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up a

Svetach [21]

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

a)

The work done by a force is given by the equation

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the dispalcement of the object

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is $90^{\circ}$ ; this means that $cos 90^{\circ}=0$ , and therefore, the work done is zero:

$W=0$

b)

In this case, the box is pushed along the ramp. We have:

$F=mg=(50.0)(9.8)=490 N$ is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by $15.0^{\circ}$ , therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is: