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Svetlanka [38]
3 years ago
12

The 8 mm diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 20 mm. If

the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the materials at A is rigid. Eal = 70 GPa, Emg = 45 GPa.
Physics
1 answer:
dlinn [17]3 years ago
4 0

Answer:

Strain on bolt= 159.15MPa

Strain on sleeve = 39.79MPa

Stress on bolt= 0.00227mm/km

Stress on sleeve= 0.000884mm/mm

Explanation: Given : do= 20mm , di= 12mm , p= 8 × 10^3N, Eol= 70Gpa , Emg = 45Gpa

Normal stress = P/Ab

Ab= pi/4 ×d^2=3.142/4 × (8 × 10^-3)^2

Ab = 5.024× 10^-5m^2

Stress of bolt= 8×10^3/5.024×10^5=159.15Mpa

Eb= stress on bolt= 159.15×10^6/70×10^9

Eb= 0.00227mm/mm

For the sleeve

Normal stress= P/As As= pi×d^2/4 =3.142/4× ((20×10^-3)-(12×10^-3))^2

As = 5.024× 10^-5m

Stress on sleeve = 8×10^3/5.024×10^-5)

= 39.79MPa

Es= 39.79/ 45×10^9 =0.000884mm/mm

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2 years ago
2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in
aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

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6 0
1 year ago
You place a box weighing 200.2 N on a inclined plane that makes a 37.1 angle with the horizontal. Compute the component of the g
salantis [7]

Answer:

141.56 N.

Explanation:

Data given:

Weight of the box= 200.2 N

Angle with the horizontal= 37.1°

Solution;

Gravitational force on the box, F_g= weight of the box

                                                                           = 200.2 N

Component of gravitational force along plane = F_g * sin( ∅ )

                                                                                   = W * (sin∅)

                                                                                   = (200.1) * sin (37.1°)

                                                                                   = 141.56 N

8 0
2 years ago
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slava [35]
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3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
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