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3 years ago

An iron vat is 11 m long at room temperature (20°C). How much longer is it when it contains boiling water at 1 atm pressure?

Physics

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

The change in length is 0.01056 m.

Explanation:

Given that,

Length = 11 m

Temperature = 20°C

Pressure = 1 atm

Boiling temperature = 100°C

We need to calculate the length

Using formula of change in linear expansion

$\dfrac{\Delta L}{L}=\alpha_{l}\Delta T$

$\Delta L=L\times\alpha_{l}\times(T_{f}-T_{i})$

Put the value into the formula

$\Delta L=11\times12\times10^{-6}\times(100-20)$

$\Delta L=11\times12\times10^{-6}\times80$

$\Delta L=0.01056\ m$

Hence, The change in length is 0.01056 m.

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t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s

The distance traveled during the acceleration phase is
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3 years ago

Consider two planets in space that gravitationally attract each other if the mass of one of them stays the same and the mass of

irga5000 [103]

Answer:

D. half as much

Explanation:

let m and M be the mass of the planets and r be the distance between them.

then: the force of attraction between them is given by,

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= (1/2)G×m×M/(r^2)

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3 years ago

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djyliett [7]

Answer:

In an a.c generator, slip rings are used.

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Explanation:

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3 years ago

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A 10 kg runaway grocery cart runs into a spring with spring constant 250 n/m and compresses it by 60 cm. what was the speed of t

andrew-mc [135]

The initial kinetic energy of the cart is
$K= \frac{1}{2} mv^2$ (1)
where m is the mass of the cart and v its initial velocity.

Then, the cart hits the spring compressing it. The maximum compression occurs when the cart stops, and at that point the kinetic energy of the cart is zero, so all its initial kinetic energy has been converted into elastic potential energy of the spring:
$U= \frac{1}{2}kx^2$
where k is the spring constant and x is the spring compression.

For energy conservation, K=U. We can calculate U first: the compression of the spring is x=60 cm=0.60 m, while the spring constant is k=250 N/m, so
$U= \frac{1}{2}kx^2= \frac{1}{2}(250 N/m)(0.60 m)^2=45 J$

So, the initial kinetic energy of the cart is also 45 J, and from (1) we can find the value of the initial velocity:
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3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

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3 years ago