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kogti [31]
4 years ago
6

A 3.7-mm-diameter wire carries a 20 A current when the electric field is 8.2×10−2 V/m .

Physics
1 answer:
agasfer [191]4 years ago
7 0

Answer:

   ρ = 4.4 10⁻⁸ Ω m

Explanation:

For this exercise let's start by finding the value of the resistance using Ohm's law

          V = I R

           R = V / I

   

The voltage is related to the electric field

          V = E L

           

let's substitute

          R = E L / I

          R = 8.2 10⁻² / 20  

          R = 4.1 10⁻³ L

now we can use the resistance relation

          R = ρ L / A

the area of ​​a circular wire is

          A = π r² = π d² / 4

         ρ = R π d² / (4 L)

let's calculate

        ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)

        ρ = 4.4 10⁻⁸ Ω m

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A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How
stepan [7]

Answer:

6 \times 10^4 \; \rm J.

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2, where

  • m is the mass of the car, and
  • v is the speed of the car.

The 2 \times 10^3\; \rm kg car would have an initial KE of:

\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J.

The 3 \times 10^3\; \rm kg car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum p of an object is equal to its mass m times its velocity v. In other words, p = m\cdot v.

Let the mass of the two cars be denoted as m_1 and m_2, and their initial speeds v_1 and v_2. Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as v_3.

Initial momentum of the two-car system:

\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

After the collision, both car would have a velocity of v_3 (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3.

Since momentum is conserved during the collision, the momentum of the system after the collision would also be 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}. That is: (m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}.

Solve for v_3:

\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}.

Hence, the total kinetic energy after the collision would be:

\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}.

The amount of kinetic energy lost during the collision would be:

\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}.

5 0
3 years ago
the potential energy of a 40kg cannon ball is 14000j. How high was the cannon ball to have this much potential energy?
noname [10]
We will use the formula p = mgh p is potential energy. m is mass of object in kg g is acceleration due to gravity (9.8m/s²) h is height of the objects displacement in meters. p = mgh → mgh = p → h = p / mg p is 14000j, m is 40kg and g is 9.8 m/s² h = 14000 / 40 × 9.8 → h = 1400 / 392 → h = 35.7 Therefore , the cannonball was 35.7 meters high .
7 0
4 years ago
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8_murik_8 [283]

Answer:

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4 0
3 years ago
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Explain why a man using a parachute falls through air slowly while a stone falls through air very fast
Korolek [52]

Answer:because the parachute is built so the wind will push up on it make the man/woman glide or fall slowly will a stone which has a lot of desentity

falls through the air faster do to is weight and shape .

Explanation:

3 0
3 years ago
A rocket accelerates upward from rest at a rate of 5 m/s^2. At a height of 1000m, the engines burn out and it coasts the rest of
laila [671]

Answer:

Explanation:

Given

acceleration of rocket(a)=5 m/s^2

At h=1000 m rocket burn out

v^2-u^2=2as

v^2-0=2\times 5\times 1000

v=\sqrt{10^4}=100 m/s

(b) time to reach v=100 m/s

v=u+at

100=0+5\times t

t=\frac{100}{5}=20 s

(c)Rocket maximum altitude

v^2-u^2=2as

here u=100 m/s

v=0

a=9.8 m/s^2

s=\frac{v^2}{2g}

s=\frac{100^2}{2\times 9.8}

[tex]s=510.204

Therefore maximum altitude=510.204+1000=1510.204 m

7 0
3 years ago
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