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Paladinen [302]
2 years ago
5

The resistivity of iron is 1 x 10 -7 Ωm. The resistance of a iron wire of particular length and thickness is 1 ohm. If the lengt

h and the diameter of wire both are doubled, then the resistivity in Ωm will be
Physics
1 answer:
Tatiana [17]2 years ago
7 0

Answer:

2×10⁻⁷ Ωm

Explanation:

From the question,

R = 4ρL/πd².................... Equation 1

Where R = Resistance of the wire, ρ = resistivity of the wire, L = length of the wire, d = diameter of the wire.

Therefore,

ρ = Rπd²/4L............. Equation 2

Given: R = 1 ohm, ρ = 1×10⁻⁷Ωm

1×10⁻⁷ = πd²/4L.................... Equation 3

If the length and the diameter are doubled, and the resistance remaining constant

ρ' = π(2d)²/4(2L)

Where ρ'  = new resistivity

ρ' = 4πd²/8L

ρ' = 2πd²/4L = 2ρ

ρ' = 2(1×10⁻⁷)

ρ' = 2×10⁻⁷ Ωm

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Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i
fiasKO [112]

Answer:

The algebraic equation is:

F_{v} =\frac{m_{b}v_{b}^{2}   }{R} -m_{b} g

Explanation:

Given information:

mb = book's mass

vb = tangential speed

R = radius of the path

Question: Derive an algebraic equation for the vertical force, Fv = ?

To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:

F_{v} =\frac{m_{b}v_{b}^{2}   }{R} -m_{b} g

The variables were defined above and g is the gravity.

4 0
2 years ago
A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra
PIT_PIT [208]

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

4 0
2 years ago
The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?
Klio2033 [76]
The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
n= \frac{c}{v}
where the speed of light in vacuum is c=3 \cdot 10^8 m/s. The speed of light in benzene is v=2.00 \cdot 10^8 m/s, so we can use the previous relationship to find the refractive index of benzene:
n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5
7 0
2 years ago
How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by th
ololo11 [35]

Answer:

1.0\cdot 10^{41} times

Explanation:

First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.

Age of the universe:

T=100,000,000,000,000,000s = 1.0\cdot 10^{17} s (1 followed by 17 zeroes)

Lifetime of the top quark:

\tau = 0.000000000000000000000001s = 1.0\cdot 10^{-24} s (we moved the decimal point 24 places to the right)

Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

r = \frac{T}{\tau}=\frac{1.0\cdot 10^{17} s}{1.0\cdot 10^{-24} s}=1.0\cdot 10^{41}

6 0
2 years ago
A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
svetlana [45]

Answer:

7 deg

Explanation:

m = mass of the rod = 1.4 kg

W = weight of the rod = mg = (1.4) (9.8) = 13.72 N

k_{L} = spring constant for left spring = 59 Nm^{-1}

k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

L = length of the rod = 0.75 m

\theta = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

3 0
2 years ago
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