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Lelu [443]
3 years ago
7

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 40.0°C greater th

an when they were laid? Their original length is 41.0 m.
Physics
1 answer:
ella [17]3 years ago
8 0

Answer:

19.68 × 10⁻³ m

Explanation:

Given;

Original Length, L₁ = 41.0 m

Temperature Change, ΔT = 40.0°C

Thermal Linear expansion of steel is given to be, ∝_{steel} = 12 × 10⁻⁶ /°C

   Generally, Linear expansivity is expressed as;

                               ∝ = ΔL / L₁ΔT

Where

∝ is the Linear expansivity

ΔL is the change in length, L₂ - L₁

L₂ is the final length

L₁ is the original length

ΔT is the change in temperature θ₂ - θ₁ (Final Temperature - Initial Temperature)              

From equation of linear expansivity

                         ΔL = ∝_{steel}L₁ΔT

                         ΔL = 12 × 10⁻⁶ /°C × 41.0 m × 40.0 °C

                         ΔL = 19.68 × 10⁻³ m

                         ΔL = 19.68 mm

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raketka [301]

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast f=99.4MHz=99.4\times 10^4Hz

The speed of light c=3\times 10^8m/sec

The distance of the antenna to receive a minimum signal from the station is given by d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m

So the distance of the antenna from the station will be 3.018 m

5 0
3 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0
2 years ago
A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the
Nezavi [6.7K]

We use the equation of motion for vertical component,

s_{y} = u_{y} t+\frac{1}{2} gt^2.

Here, s_{y}   is displacement of bullet, u_{y}  is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

s_{y} =\frac{1}{2}gt^2

Given, s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m

Substituting the values, we get time of flight

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4 0
3 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016 Na+ ions arrive at the negative electrode and 3.
sladkih [1.3K]

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = \frac{q_{1}}{t} = \frac{0.004288}{1} = 0.004288 A

i₂ = Current due to Cl⁻ ions = \frac{q_{2}}{t} = \frac{0.006272}{1} = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

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