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fenix001 [56]
4 years ago
14

Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. T

he length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 14.1 ∘C above room temperature?
Physics
1 answer:
Deffense [45]4 years ago
4 0

Answer:

0.00034 m

Explanation:

Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m

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Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i
Digiron [165]

Answer:

3.31m/s

Explanation:

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L = L_i_n_i + L_3_s

L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt

L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

Moment if inertia is

I = 2ml^2

I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

Angular speed

ω = L/I

= 38 / 5.75\\\\=6.61

The speed of each ball is

V = ωL

= 6.61\times0.5\\\\= 3.31m/s

7 0
3 years ago
Should the shape of an object be considered a property of the material?
solniwko [45]
No
For example a rock was broken into one big and one little piece. The properties of these 2 pieces are still the same even though they have different shapes.
7 0
3 years ago
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vlada-n [284]
Protons
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7 0
4 years ago
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Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg
Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C

Calculating the common potential:

\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\

Calculating the charge after redistribution:

When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2        

\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\  \\  \to               q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C

6 0
3 years ago
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