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GarryVolchara [31]

2 years ago

15

Some iron has a coefficient of linear expansion of 12 x 10-6 / o K . A 100 mm length of iron piping is heated through 20 K. The

pipe extends by:

1 answer:

VLD [36.1K]2 years ago

8 0

Given that the linear expansion of the pipe is 12×10¯⁶ K¯¹, the pipe extends by 0.024 mm

<h3>Data obtained from the question </h3>

Original length (L₁) = 100 mm
Change in temperature (ΔT) = 20 K
Coefficient of expansion (α) = 12×10¯⁶ K¯¹
Change in length (ΔL) =?

<h3>How to determine the new diameter </h3>

α = ΔL / L₁ΔT

12×10¯⁶ = ΔL / (100 × 20)

12×10¯⁶ = ΔL / 2000

Cross multiply

ΔL = 12×10¯⁶ × 2000

ΔL = 0.024 mm

Learn more about linear expansion:

brainly.com/question/23207743

#SPJ1

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which statements about velocity are true? check all that apply. a. for velocity, you must have a number, a unit, and a direction

satela [25.4K]

a). for velocity, you must have a number, a unit, and a direction.
Yes. This one isn't bad. The 'number' and the 'unit' are the speed.

b). the si units for velocity are miles per hour.
No. That's silly.
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity.

c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.

d). to calculate velocity, divide the displacement by time.
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3 0

3 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago

Which sentence describes an example of sublimation?

k0ka [10]

answer C is the correct one

7 0

3 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

BaLLatris [955]

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>

4 0

4 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

3 years ago