Question

A brass plug is to be placed in a ring made of iron. At 15∘C, the diameter of the plug is 8.864 cm and that of the inside of the ring is 8.855 cm .

They must both be brought to what common temperature in order to fit?

Answer #1

A brass plug is to be placed in a ring made of iron. At 15∘C,
the diameter of the plug is 8.900 cm and that of the inside of the
ring is 8.889 cm .
They must both be brought to what common temperature in order to
fit?

A brass plug is to be placed in a ring made of iron. At 16 ?C,
the diameter of the plug is 8.793 cm and that of the inside of the
ring is 8.778 cm .
They must both be brought to what common temperature in order to
fit?
What if the plug were iron and the ring brass?

A brass plug is to be placed in a ring made of iron. At 20 ∘C,
the diameter of the plug is 8.735 cm and that of the inside of the
ring is 8.722 cm .
A) They must both be brought to what common temperature in order
to fit?
B) What if the plug were iron and the ring brass?

A brass plug is to be placed in a ring made of iron. At 10 ∘C,
the diameter of the plug is 8.741 cm and that of the inside of the
ring is 8.729 cm .
1.They must both be brought to what common temperature in order
to fit?
2.What if the plug were iron and the ring brass?

A brass ring with inner diameter 2.00 cm and outer diameter 3.00
cm needs to fit over a 2.00-cm-diameter steel rod, but at 20
degrees Celsius the hole through the brass ring is 40 μm too small
in diameter.
To what temperature, in degrees Celsius, must the rod and ring
be heated so that the ring just barely slips over the rod?

A steel rod is 3.000 cm in diameter at 33.0 °C. A brass ring has
an interior diameter of 2.992 cm at 33.0 °C. At what common
temperature will the ring just slide onto the rod? The coefficients
of linear expansion of steel and brass are 1.100×10-5 , 1.900×10-5
/°C, respectively, answer in °C.

A steel rod is 2.689 cm in diameter at 35.00°C. A brass ring has
an interior diameter of 2.684 cm at 35.00°C. At what common
temperature will the ring just slide onto the rod? The linear
expansion coefficient of steel is 11.00 × 10-6 1/C°. The
linear expansion coefficient of brass is 19.00 × 10-6
1/C°.
4sig figs

At 18.10 ∘C a brass sleeve has an inside diameter of 2.22063 cm
and a steel shaft has a diameter of 2.22333 cm. It is desired to
shrink-fit the sleeve over the steel shaft.
A.) To what temperature must the sleeve be heated in order for
it to slip over the shaft?
B.) Alternatively, to what temperature must the shaft be cooled
before it is able to slip through the sleeve?

A brass ring of diameter 10.00 cm at 15.7°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 15.7°C. Assume
the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to
separate the two metals?
°C
Is that temperature attainable?
Yes No
(b) What if the aluminum rod were 10.04 cm in diameter?
°C
Is that temperature attainable?

A brass ring of diameter 10.00 cm at 17.8°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 17.8°C. Assume
the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to
separate the two metals?
°C
(b) Is that temperature attainable?
Yes or No
(c) What if the aluminum rod were 9.16 cm in diameter?
°C
(d) Is that temperature attainable?
Yes or No

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