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Tema [17]
3 years ago
14

Why do engineers place tolerances on dimensions?

Physics
1 answer:
ser-zykov [4K]3 years ago
4 0
<span>Engineers place tolerance on dimensions since the when engineers use tolerance, buildings and constructions projects like tall towers, roads, and bridges become more reliable, strong and stable. In harnessing tolerance, the end product that is made by the engineers, specifically buildings like schools, hospitals, malls and skyways are sturdy and dependable.</span>
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Which answer choice provides the best set of labels for Wave A and Wave B?
hodyreva [135]
A has less energy and lower frequency, while B has greater energy and higher frequency.
3 0
4 years ago
Read 2 more answers
The length of a 100 mm bar of metal increases by 0.3 mm when subjected to a temperature rise of 100°C. The coefficient of linear
Juli2301 [7.4K]

Answer:

α = 3×10^-5 K^-1

Explanation:

let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.

then, the coefficient of linear expansion is given by:

α = ΔL/(ΔT×L)

   = (0.3×10^-3)/(100)(100×10^-3)

   = 3×10^-5 K^-1

Therefore, the coefficient of linear expansion is 3×10^-5 K^-1

5 0
4 years ago
How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat
viva [34]

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

8 0
3 years ago
It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati
Soloha48 [4]

Answer:  47.6 m/s

Explanation:  Please see attached for the calculation and formula.

5 0
3 years ago
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with
Dvinal [7]

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

6 0
4 years ago
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