Ask question

Ask question

All categories

2 years ago

5

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

he train is initially moving at a speed of 54 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

1 answer:

coldgirl [10]2 years ago

7 0

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

You might be interested in

A vase of flowers is stationary on a table. which of the following statements best explains why the vase is stationary?

goldenfox [79]

D. The table pushes up on the vase with the same amount of force as gravity pulling it down.

7 0

2 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

If a reaction occurs when a piece of metal is placed in a solution, what can you conclude about the solution?

Tju [1.3M]

You can conclude that the solution is probably acidic because bases rarely react with metals.

5 0

2 years ago

How much time would it take for an object to fall 4.7 meters

levacccp [35]

Answer:

4.7 is 10 as much as the number 0.47.

If you multiply 0.47 x 10 it will equal 4.7

Explanation:

4 0

2 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$

$m_1u_1 +m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

$m_2u_2 = (m_1+m_2)v_f$

Rearranging to find the final velocity

$v_f = \frac{m_2u_2}{ (m_1+m_2)}$

$v_f = \frac{ 12500*7.8}{ 12500+7430}$

$v_f = 4.8921ft/s$

The expression for the impulse received by the first car is

$I = m_1 (v-u)$

$I = \frac{W}{g} (v-u)$

Replacing,

$I = \frac{12500}{32.2}(4.89-7.8)$

$I = -1129.65lb\cdot s$

The negative sign show the opposite direction.

7 0

2 years ago