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makvit [3.9K]
2 years ago
14

a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t

emperature at which the sphere could pass through the ring? alpha steel= 12×10^-6 and alpha brass= 20×10^-6​
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
6 0

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt  = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C  .

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3 years ago
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?​
Xelga [282]
<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

<h3>And,</h3>

30 cm = 0.3 m (distance)

<h3>So,</h3>

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=> 10 × 0.3

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\small \boxed{work \: done \:  by \: Ryan \:  = 3 \: Joules}

4 0
2 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

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Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

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Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

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u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

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2 years ago
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In Physics, 'work' has a very clear definition:

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'Work' has the units of Energy.

If you push against a shopping cart with 30 newtons of force, and
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3 years ago
A 50 n force is acting on a lever 1.5 m from the fulcrum balances an object 1m from the fulcrum on the other arm. what is the we
Angelina_Jolie [31]

Answer:

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Explanation:

hpoe this helps you.

4 0
2 years ago
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