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3 years ago

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A steel railroad track has a length of 23 m when the temperature is 7◦C. What is the increase in the length of the rail on a hot

day when the temperature is 30 ◦C? The linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1 . Answer in units of m.

1 answer:

xxTIMURxx [149]3 years ago

7 0

Answer:

5.82 mm

Explanation:

length of the track, L = 23 m

Initial temperature, T = 7 °C

final temperature, T' = 30 °C

Coefficient of linear expansion, α = 11 x 10^-6 / °C

Let the increase in length of the track on the hot day is ΔL.

use the formula of the linear expansion

ΔL = L x α x ΔT

ΔL = L x α x (T' - T)

ΔL = 23 x 11 x 10^-6 x (30 - 7)

ΔL = 5.82 x 10^-3 m

ΔL = 5.82 mm

Thus, the increase in the length of the track is 5.82 mm.

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Salsk061 [2.6K]

Answer:

21.18 L

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 1 atm

Initial volume (V₁) = 18 L

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Final Volume (V₂) =?

With the application of the Boyle's law equation, we can obtain the final volume (i.e the new volume) of the gas as follow:

P₁V₁ = P₂V₂

1 × 18 = 0.85 × V₂

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3 years ago

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago

Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?

cluponka [151]

Answer:

Torque, $\tau=0i+0j-70k$

Explanation:

It is given that,

Force acting on the particle, $F=-10j\ N$

Position of the particle, $r=(7i+5j)\ m$

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

$\tau=r\times F$

$\tau=(7i+5j)\times (-10j)$

The cross product of vectors is given by :

$\tau=\begin{pmatrix}0&0&-70\end{pmatrix}$

or

$\tau=0i+0j-70k$

So, the torque on the particle about the origin $0i+0j-70k$ . Hence, this is the required solution.

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3 years ago

Pam has a mass of 42.4 kg and she is at rest on smooth, level, frictionless ice. Pam straps on a rocket pack. The rocket supplie

fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

I = Δp

I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v / 2x

a = 56.8 2/2 29.1

a = 55.43 m / s²

Let's look for the time you need to get this speed

v = v₀ + a t

t = v / a

t = 56.8 / 55.43

t = 1,025 s

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F t = m v- m v₀

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3 years ago

In attempting to pull a 1500 kg car out of a ditch, Al exerts a force of 200 N for 5 s, Bill exerts a force of 500 N for two s,

MArishka [77]

Answer:

Clyde will provide greater impulse

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We have given that Al exerts a force of 200 N for 5 sec

We know that impulse is given by, impulse = force ×time = 200×5 =1000 N-s

Bill exerts a force of 500 N for 2 sec

So impulse = 500×2 = 1000 N-s

Now the force exerted by Clyde 300 N for 4 sec

So impulse = 300×4 = 1200 N-s

From above calculation we can see that Clyde provide greater impulse than any other

4 0

3 years ago