Ask question

Ask question

All categories

2 years ago

10

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t

he Young's modulus of the material?

1 answer:

Salsk061 [2.6K]2 years ago

6 0

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

$stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2} \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\$

Now, we calculate the strain:

$strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain = 0.002\\$

Now, we will calculate the Young's Modulus (Y):

$Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\$

<u>Y = 4.775 x 10⁹ Pa = 4.775 GPa</u>

You might be interested in

A wave front has the form of a

podryga [215]

Your answer is "<span>surface of a sphere"

Hope this helps.</span>

3 0

2 years ago

Compare the collision between two baseballs and a catcher's mitt.

Nookie1986 [14]

The applied force is different for the two cases

The case A with a greater force involves the greatest momentum change

The case A involves the greatest force.

<h3>What is collision?</h3>

This is the head-on impact between two object moving in opposite or same direction.

The initial momentum of the two ball is the same.

P = mv

where;

m is the mass of each
v is the initial velocity of each ball

Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.

Thus, the final momentum of each ball will be different

The impulse experienced by each ball is different since impulse is the change in momentum of the balls.

J = ΔP

The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.

Thus, we can conclude the following;

The applied force is different for the two cases
The case A with a greater force involves the greatest momentum change
The case A involves the greatest force.

Learn more about impulse here: brainly.com/question/25700778

3 0

2 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

What do electric and magnetic fields have in common?

Nastasia [14]

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Magnetic field are a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are produced by moving electric charges. Everything is made up of atoms, and each atom has a nucleus made of neutrons and protons with electrons that orbit around the nucleus. Since the orbiting electrons are tiny moving charges, a small magnetic field is created around each atom.

Similarities between magnetic fields and electric fields: Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like pole repel unlike poles attract. Electric field points in the direction of the force experienced by a positive charge.

7 0

2 years ago