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scoundrel [369]
3 years ago
15

A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in

m/s^2.
Engineering
1 answer:
DochEvi [55]3 years ago
7 0

Answer:

a=5.515\frac{m}{s^{2} }

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

1 mile=1609.34 mts.

1 hora=3600 segundos

Performing the operations

60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}

Now, we will use the acceleration formula

a=\frac{v}{t}

Where v = speed and t = time

Substituting the values ​​of t=5.2s

a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }

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Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.
gayaneshka [121]

Answer / Explanation:

On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.  

WHILE,

True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.

5 0
3 years ago
While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently
musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

4 0
3 years ago
A minivan starts from rest on the road whose constant radius of curvature is 40 meters and whose bank angel is 10 degrees. the m
zimovet [89]

Based on the calculations, the magnitude (a) of it's total acceleration is equal to 2.71 m/s².

<u>Given the following data:</u>

  • Angle of inclination = 10°.
  • Radius of curvature, r = 40 meters.
  • Acceleration of the minivan, A = 1.8 m/s².
  • Initial velocity, u = 0 m/s (since it's starting from rest).
  • Time, t = 5 seconds.

<h3>How to determine the magnitude (a) of it's total acceleration?</h3>

First of all, we would determine the final velocity of the minivan by applying the first equation of motion as follows:

V = u + at

V = 0 + 1.8 × 5

V = 9 m/s.

Next, we would calculate the centripetal acceleration of this minivan:

Ac = V²/r

Ac = 9²/40

Ac = 2.025 m/s².

Now, we can determine the magnitude (a) of it's total acceleration:

a = √(Ac² + A²)

a = √(2.025² + 1.8²)

a = 2.71 m/s².

Read more on acceleration here: brainly.com/question/24728358

#SPJ1

8 0
2 years ago
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su
Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0
4 years ago
Read 2 more answers
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

4 0
3 years ago
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