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fgiga [73]
3 years ago
9

The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards

the center of the path. c. perpendicular to the transverse component of acceleration d. all of the above
Engineering
1 answer:
lesya [120]3 years ago
7 0

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

a_r = \frac{V^2}{R}

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

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Write an expression that evaluates to true if the value of variable lastName is greater than the string Dexter.
gregori [183]

Answer:

lastName.compareTo("Dexter")>0

Explanation:

The expression that evaluates to true if the value of variable lastName is greater than the string Dexter is; lastName.compareTo("Dexter")>0.

4 0
3 years ago
A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted
alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - D^{2}/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - U^{2}/Uf)

2100 = 240 ( U - U^{2}/55)

Solve for U

U = 44m/h

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3 years ago
An air mass usually forms in an area of____.
hoa [83]
I think it’s b hopefully it is
3 0
3 years ago
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A metal having a cubic structure has a density of , an atomic weight of , and a lattice parameter of Å One atom is associated wi
Veronika [31]

Answer:

Explanation:

Answer: The crystal structure of the metal is BCC

Explanation:

we first calculate the volume of the unit cell.

Volume of unit cell= (a°)^3.

The lattice parameter here is a°.

Substitute (6.13 * 10^-8)cm for a°.

Volume of unit cell = (6.13 * 10^-8)^3 = 2.3034 * 10^-22 cm^3/cell.

To determine the crystal structure we use

Density (p) = {(Number of atoms per cell) (Atomic mass)} / {(volume of unit cell)(Avogrado constant)}.

Substitute 1.892g/cm^3 for p (6.02*10^23) atoms/mol for Avogrado constant 1.3921g/mol.

For atomic mass and (2.3034 * 10^-22) cm^3/cell for unit cell.

1.892g/cm^3 = {(Number of atoms per cell) (1.3291g/mol)} / {(2.3034 * 10^-22) (6.02 * 10^23 atoms/mol)}.

Changing the subject of formula we have :

Number of atoms per cell = {(2.3034 * 10^-22) * (6.02 * 10^23) * 1.892} / 132.91

Number of atoms per cell = 2.

Since the number of atoms per cell is 2, :. the crystal structure of metal is BCC.

Note: p = density

a° = a subscript o

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3 years ago
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ANSWER:
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