This is an earth science question about Issac Newton’s laws and I need some major help

An 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55

Sindrei [870]

Answer:

a)Yes will deform plastically

b) Will NOT experience necking

Explanation:

Given:

- Applied Force F = 850 lb

- Diameter of wire D = 0.15 in

- Yield Strength Y=45,000 psi

- Ultimate Tensile strength U = 55,000 psi

Find:

a) Whether there will be plastic deformation

b) Whether there will be necking.

Solution:

Assuming a constant Force F, the stress in the wire will be:

stress = F / Area

Area = pi*D^2 / 4

Area = pi*0.15^2 / 4 = 0.0176715 in^2

stress = 850 / 0.0176715

stress = 48,100.16 psi

Yield Strength < Applied stress > Ultimate Tensile strength

45,000 < 48,100 < 55,000

Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.

6 0

3 years ago

Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

$\theta=39.49^{\circ}$

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0

3 years ago

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

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

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

3 years ago

Read 2 more answers

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm= $8\times 10^{-2} m$

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid= $8\times 10^{-2} m$

Magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Using the formula

$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

Where $\mu_0=4\pi\times 10^{-7}$

$N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns$

6 0

3 years ago

What is a newton equal to in terms of units of mass and acceleration

artcher [175]

1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .

1 N = 1 kg-m/s² .

It's a force equal to roughly 3.6 ounces.

7 0

3 years ago

Read 2 more answers