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Allushta [10]
2 years ago
5

A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through th

e wire, it heats up and melts ice and snow on the window. For one window the wire has a total length of 12.2 m, a width of 1.8 mm, and a thickness of 0.11 mm. The wire is connected to the car's 12.0 V battery and draws 7.5 A. What is the resistivity of the wire material?
Physics
2 answers:
babunello [35]2 years ago
7 0

Answer: 2.6*10^-8 Ωm

Explanation:

given

Length of the wire, l = 12.2 m

width of the wire, w = 1.8 mm

thickness of the wire, T = 0.11 mm

Potential difference of the battery, v = 12 V

Current in the battery, I = 7.5 A

Remember, Ohms law says, V = IR.

So that, R = V/I

R = 12/7.5 = 1.6 Ω

Resistivity if a material is

ρ = RA/l

ρ = [1.6 * 1.8*10^-3 * 0.11*10^-3] / 12.2

ρ = (3.168*10^-7) / 12.2

ρ = 2.596*10^-8

Therefore, the resistivity of the wire is 2.60*10^-8 Ωm

Setler [38]2 years ago
7 0

Answer:

Answer is given below.

Explanation:

resistance = voltage / current

R = V/I

= 12 / 7.5 = 1.6 ohm

R = pL/A

where p = resistivity

L = length , A = cross sectional area

1.6 = p*12.2 / (1.6*10^-3*0.11*10^-3)

p = 2.3*10^-8 ohm - m

so the resistivity of the material is 2.3*10^-8 ohm - m

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
Blizzard [7]
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

R=8031.38m   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0
2 years ago
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
Why is thesize saved prior to entering the for loop? 2. what is the running time of removefirsthalf if lst is an arraylist? 3. w
Mrac [35]
<span>After entering the loop, it should use the correct list size and the loop will be affected if the remove call changes the size of the list. If lst is an Arraylist the running time of removefirsthalf is O (n^2). So when the beginning is removed the next element will move forward. If lst is a LinkedList which is a dynamic structure the running is O (n) for removefirsthalf</span>
8 0
2 years ago
Which changes in an electric motor will make the motor stronger? Check all that apply.
ollegr [7]
To make the motor turn faster we can:
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A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0
2 years ago
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