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GaryK [48]
1 year ago
9

If we apply a potential difference of 4. 50 v between the ends of a wire that is 2. 50 m in length and 0. 654 mm in radius, the

resulting current through the wire is 17. 6. What is the resistivity of the wire material?.
Physics
1 answer:
tatyana61 [14]1 year ago
8 0

According to the resistance, the resistivity of wire material is 1.37 x 10¯⁷ Ωm.

We need to know about the resistance of conductive material to solve this problem. Resistance can be defined as the ratio of the voltage applied to current flow It can be written as

R = V / I

where R is resistance, V is voltage and I is current.

The resistance of the material depends on its resistivity and follows this equation

R = ρ . L / A

where ρ is resistivity, L is the length of the material and A is the surface.

From the question above, we know that

V = 4.5 V

L = 2.5 m

r = 0.654 mm = 6.54 x 10¯⁴ m

I = 17.6 A

Find the surface area

A = π . r²

A = π . ( 6.54 x 10¯⁴ )²

A = 1.34 x 10¯⁶ m²

Find the resistivity of the wire

R = V / I

ρ . 2.5 / (1.34 x 10¯⁶)= 4.5 / 17.6

ρ = 1.37 x 10¯⁷ Ωm

Find more on resistance at: brainly.com/question/17563681

#SPJ4

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While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

If f = 78 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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