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g100num [7]
2 years ago
12

Compared to the resistivity of a 0.4-meter length of 1-millimeter-diameter copper wire at 0 degrees Celsius, the resistivity of

a 0.8-meter length of 1-millimeter-diameter copper wire at 0 degrees Celsius is: *
Physics
2 answers:
Ierofanga [76]2 years ago
4 0

Answer:

c) the same

Explanation:

Mazyrski [523]2 years ago
3 0

Answer:

Resistivity of both wires are same

Explanation:

Length of one wire,l_1=0.4 m

Diameter,d_1=1mm

Radius,r_1=\frac{d_1}{2}=\frac{1}{2}mm=0.5\times 10^{-3} m

1mm=10^{-3} m

l_2=0.8 m

d_2=1mm

r_2=0.5\times 10^{-3} m

Temperature in each case is same.

Area of each wire,A_1=A_2=A=\pi r^2=\pi (0.5\times 10^{-3})^2m^2

Resistivity is the property of material due to which it offers resistance to the flow of current.

Resistivity of material depends upon the temperature and material by which it is made.

It does not depends upon the length of object.

Therefore, the resistivity of both wires of different length  are same.

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At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n
Andreyy89
Below are the choices that can be found elsewhere:

 a. 268 kJ 
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have: 

<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>

<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>

<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>

<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>

<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>

<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>
7 0
2 years ago
A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample
Hatshy [7]

Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.        

3 0
2 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
2 years ago
Read 2 more answers
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
2 years ago
Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un
elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0
2 years ago
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