An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

Answer 1
Answer:

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      \alpha=(L_(1)-L_(0))/(L_(0)(T_(1)-T_(0)))

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is 25*10^(-6)C^(-1)

Substitute,  L_(0)=17.400cm,T_(1)=100,T_(0)=20,\alpha=25*10^(-6)C^(-1)

                   25*10^(-6)C^(-1)  =(L_(1)-17.400)/(17.400(100-20))\n\n25*10^(-6)C^(-1)  = (L_(1)-17.400)/(1392) \n\nL_(1)=[25*10^(-6)C^(-1)  *1392}]+17.400\n\nL_(1)=17.435cm

Hence, The new length of aluminum rod is 17.435 cm.

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Answer 2
Answer:

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

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(a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ? 106 V/m). The plates are separated by2.98 mm and a potential difference of 5575 V is applied. (b) How close together can the plates be with this applied voltage without exceeding the breakdown strength?

Answers

Answer:

(a) 1.87×10⁶ V/m

(b) 1.12 mm closer

Explanation:

(a)

Electric Field = Electric potential/distance.

E = V/d ................... Equation 1

Where E = Electric Field, V = Electric potential, d = distance.

Given: V = 5575 V, d = 2.98 mm = 0.00298 m.

Substitute into equation 1

E = 5575/0.00298

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(b)

without exceeding the breakdown strength,

make d the subject of equation 1

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Analysis of the relationship between the fuel economy​ (mpg) and engine size​ (liters) for 35 models of cars produces the regression model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter​ engine, what does this model suggest the gas mileage would​ be?

Answers

Answer:

Gas mileage is 17.29

Explanation:

Given data:

The total number of the model is 35

The total size of the engine is 5 ltr

The regression model is given as

36.44 - 3.829* engine\ size

From the information given in question we have

Regression equation is : model- mpg = 36.44 - 3.829* engine\ size

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Explanation:

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect

Answers

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

v_s = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean f=1.01f'

From the Doppler effect equation we have

f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s

The velocity is -3.396 m/s

when f=0.99f'

f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s

The velocity is 3.465 m/s

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Answers

Answer:

a

Explanation:

a push or a pull that occurs when an object interacts with another object or field.

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If the velocity of a pitched ball has a magnitude of 41.0 m/s and the batted ball's velocity is 50.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answers

The change in momentum is (91 m/s) multiplied by the mass of the ball (which you neglected to mention).
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The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

Answers

Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth =  6 × 10²⁴ kg

r = Distance between Earth and Sun = 1.5* 10^(11)\ m

t = Time taken = 3 days

Acceleration is given by

a=(GM)/(r^2)\n\Rightarrow a=(6.67* 10^(-11)* 6* 10^(24))/((1.5* 10^(11))^2)\n\Rightarrow a=1.77867* 10^(-8)\ m/s^2

Velocity of the star

v=u+at\n\Rightarrow v=0+1.77867* 10^(-8)* 3* 24* 60* 60\n\Rightarrow v=0.00461031264\ m/s

The Sun's speed will be 0.00461031264 m/s