1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Svet_ta [14]
3 years ago
12

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

7.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.7 A , while at 92.0 ∘C it reads 17.3 A . You can ignore any thermal expansion of the rod.
Part A

Find the resistivity and for the material of the rod at 20 ∘C.

Part B

Find the temperature coefficient of resistivity at 20∘C for the material of the rod.

2) An 18-gauge copper wire (diameter 1.02 mm) carries a current with a current density of 1.90×106 A/m2 . Copper has 8.5×1028 free electrons per cubic meter.

Part A

Calculate the current in the wire.

Express your answer using two significant figures.

Part B

Calculate the drift velocity of electrons in the wire.

Express your answer using two significant figures.
Physics
1 answer:
bija089 [108]3 years ago
8 0

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

You might be interested in
Please help me......
Novosadov [1.4K]
The first one, Climate
6 0
3 years ago
Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of coppe
alexandr402 [8]

Answer:

0.699 L of the fluid will overflow

Explanation:

We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C

and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C

Substituting these values into the equation, we have

ΔV = V₀β(T₂ - T₁)

= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)

= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L

= 0.698832 L

≅ 0.699 L = 0.7 L to the nearest tenth litre

So, 0.699 L of the fluid will overflow

6 0
3 years ago
In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v
dangina [55]
Volume of gold in the phone = 10 cm^3
                                              = 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
                                        = 0.193 kg 
So
0.193 kg = 0.193 * 2.2 pounds
               = 0.43 pounds
I think there is something wrong with the options given in the question.</span>
7 0
3 years ago
To lower the voltage from a "120 V" outlet, what is used (first, at least) in most situations
nasty-shy [4]

Answer:

B. Transformer

Explanation:

A transformer is a device that is used to either raise or lower voltages and currents in an electrical circuit. In modern electrical distribution systems, transformers are used to boost voltage levels so as to decrease line losses during transmission. It basically trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice versa.

4 0
3 years ago
Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
Other questions:
  • Why is a small crescent of light often observed on the moon when it is exactly in the new phase?
    13·1 answer
  • The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗
    9·1 answer
  • Help please!!!!!!!!!!
    8·1 answer
  • Many fungi are decomposers. true or false ?
    5·1 answer
  • Describe two examples of activities that you do where you depend on friction, and would not be able to do them without friction.
    8·1 answer
  • The greater the amplitude of a wave, the greater its
    8·1 answer
  • Two small plastic spheres each have a mass of 1.2 g and a charge of -56.0 nC . They are placed 3.0 cm apart (center to center).
    11·1 answer
  • 2. How much work does gravity do in causing a 6 kg hammer to fall to the ground from a
    9·1 answer
  • A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards across a friction-free surface. determine the accelera
    7·2 answers
  • 14. If 100 grams of sodium nitrate are dissolved in 100 grams of water
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!