All categories

3 years ago

How are we able to break wire by repeated bending

Physics

2 answers:

Arlecino [84]3 years ago

6 0

Answer:

We are able to break it because wire is elastic which means that it stretches. The more we stretch it the more the material comes off.

Explanation:

DIA [1.3K]3 years ago

3 0

Answer:

Repetitive bending cause metal fatigue which is responsible for the break of wire.

Explanation:

Metal fatigue is defined as the weakening of metal when cyclic load is applied on it. Due to metal fatigue cracks are grown in the structure and these cracks become the reason of metal breaking.

Moreover, If the repetitive bending is too fast, heat is produced. This heat changes the structure of atoms in the metal and make it more brittle. That’s why metals break due to repetitive bending.

You might be interested in

Convert 35 centimeters to inches

SIZIF [17.4K]

Answer:

13.7795276 Inches

Explanation:

5 0

3 years ago

A piece of plastic is uniformly charged with surface charge density eta1. The plastic is then broken into a large piece with sur

Amiraneli [1.4K]

C is the answer for this problem

3 0

3 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

3 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation: