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3 years ago

Three consequences of unbalanced wheels on a motor vehicle

1 answer:

8 0

Unstable car and loss of grip on the road. A hazard to other vehicles and pedestrians as well as a very shaky car and risk of your wheels rolling off and riding on rims.

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A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe

LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, $\lambda = 500\ nm = 500\times 10^{- 9}\ m$

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, $\mu = 1.00028$

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

$d_{m} = \frac{2x}{\lambda }\times (\mu - 1)$

Now, substituting the appropriate values in the above formula:

$d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)$

$d_{m} = 41.77$ ≈ 42

6 0

3 years ago

A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when

lianna [129]

Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

= (50)(1000/3600) m/s

= 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

ΣF = mg sinθ-f = 0 ...... (2)

Substitute the equation (1) in equation (2), we get

ΣF = mgsinθ-μsN = 0

mgsinθ-μsmgcosθ =0

μs = sinθ/cosθ

tanθ = μs

θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma (Centripetal acceleration, $a = \frac {v^{2}}{r}$

Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$ ...... (4)

Dividing the equation (4) with equation (3),

[sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] = $\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$

From part (1), tanθ = μs

Then the above equation becomes

$\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

$r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$

= $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

= $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

= 163.3 m

5 0

3 years ago

A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the a

prisoha [69]

Answer:

W = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)

we know that maximum possible efficiency is given as

$\eta = 1- \frac{T_L}{T_H}$

$\eta = 1 - \frac{ 293}{368}$

$\eta = 0.2038$

rate of heat transfer is given as

$Q_H = \dot m C_p \Delta T$

$Q_H = 0.2 * 4.18 8(95 - 20)$

$Q_H = 62.7 kW$

Maximuim power is given as

$W = \eta Q_H$

W = 0.2038 * 62.7

W = 12.8 KW

3 0

3 years ago

Why do you suppose a value of 5 is used? Do you think other values might work?

Shtirlitz [24]

Answer:

YES

Explanation:

values other than five will work

3 0

3 years ago

The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity —

butalik [34]

Answer:

The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.

Explanation:

The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform. In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.

3 0

3 years ago

Three consequences of unbalanced wheels on a motor vehicle​

Three consequences of unbalanced wheels on a motor vehicle