Compare and contrast what happens to an object's motion when balanced or

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

3 years ago

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 7.0 nC. Using insulatin

maxonik [38]

Answer:

Block A will have a final charge of 3.5nC.

Explanation:

This is because at the point of contact with Block B, which is electrically positive, the electrons in Block A will be attracted to the excess 'unpaired' protons in block B. Hence, the electrons will flow into Block B causing unpaired protons to remain in Block A.

This process is called Charging by Conduction.

This charging process will continue until the charges are evenly distributed between both objects.

In case you're wondering, "<em>how's all this possible within a few seconds</em>?", remember that electrons travel very fast and so, this process is a rather rapid one.

6 0

3 years ago

Read 2 more answers

Why is it important to keep the muzzle of a firearm pointed in a safe direction, even though the firearm's safety is engaged?

Gnom [1K]

The correct answer is B. The safety only prevents you from pulling the trigger, but does not stop the pin from striking the primer. For example, if you drop the firearm, the pin may hit the primer and fire the firearm. It is always responsible to keep the firearm pointed in a safe direction so that if this happens, no consequences come out of it.

6 0

3 years ago

A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio

xxTIMURxx [149]

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

$v^2-u^2 = 2as$

substitute the values and solve for v

$v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s$

b) Use conservation of momentum:

Initial momentum of the system = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

3 0

3 years ago

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago