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3 years ago

Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?

Engineering

1 answer:

lianna [129]3 years ago

5 0

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents a crystal system which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have three axes which have only two of its axes equal in length.

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The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio

Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

$v=23.6m/s$

Explanation:

Velocity $v_c=18.0km/h$

Radius $r=21m$

initial velocity u $u=19=>5.27778$

Generally the equation for Angle is mathematically given by

$\theta=\frac{v_c}{2r}$

$\theta=\frac{18}{2*21}$

$\theta=0.45$

$\theta=25.7831 \textdegree$

Generally

Height of mass

$h=\frac{rsin\theta}{\theta}$

$h=\frac{21sin25.78}{0.45}$

$h=20.3m$

Generally the equation for Work Energy is mathematically given by

$0.5mv_0^2+mgh=0.5mv^2$

Therefore

$v=\sqrt{u^2+2gh}$

$v=\sqrt{=5.27778^2+2*9.81*20.3}$

$v=23.6m/s$

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3 years ago

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Which of the following are the same as 1545.5347

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2 years ago

An ECG has a scalar magnitude of 1 mV on lead II and a scalar magnitude of 0.5 mV on lead III. Calculate the scalar magnitude on

s2008m [1.1K]

Answer: the scalar magnitude on lead I is 0.5 mV

Explanation:

Given that;

scalar magnitude on lead II = 1 mV

scalar magnitude on lead III = 0.5 mV

the scalar magnitude on lead I = ?

we know that;

Lead I Voltage = LA - RA -----------let this be equation 1

where LA is left arm electrode and RA is right am electrode

Also

Lead II = LL - RA

where LL is the left leg of electrode

we substitute

1 mV = LL - RA ---------------------let this be equation 2

Again

Lead III = LL - LA

we substitute

0.5 mV = LL - LA ------------------let this be equation 3

now subtract equation 3 and 2

1 mV - 0.5 mv = LL - RA - (LL - LA)

0.5 mV = LL - RA - LL + LA

0.5 mV = -RA + LA

0.5 mV = LA - RA

now taking a look at our equation 1 ( Lead I Voltage = LA - RA )

hence, Lead I Voltage = LA - RA = 0.5 mV

Therefore the scalar magnitude on lead I is 0.5 mV

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3 years ago