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Andre45 [30]
3 years ago
12

A hole is to be drilled in the plate at A. The diameters of the bits available to drill the hole range from 12 to 24 mm in 3-mm.

increments. Determine: (a) the diameter d of the largest bit that can be used if the allowable load at the hole is not to exceed that at the fillets.

Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

The maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.

Explanation:

As the question is not complete, the complete question is found online and is attached herewith.

From the given data

Allowable stress is given  as \sigma_{all}=145 MPa

Diameters of the bits are 12 mm, 15 mm, 18 mm, 21 mm, 24 mm

For the fillet, the dimensions are given in the attached question which are as

radius rf=9 mm

Small diameter d=75 mm

Large Diameter D=112.5 mm

The ratios of the diameters are calculated as

D/d=1.5

rf/d=0.12

So for these values and from the table, the value of K is found as 2.1

The area is given as

A=d*t

where d=75 mm, t=12 mm

A=75*12=900x10^-6 m2

Now the value of the average stress is given as

\sigma_{av}=\dfrac{\sigma_{max}}{K}\\P=\sigma_{av}A\\P=\dfrac{\sigma_{max} A}{K}\\P=\dfrac{145 \times 10^6 *900\times 10^{-6}}{2.1}\\P=62.1 kN

So for the various holes the value of r is the radius of the hole so

D=112.5 mm as given above whereas

d=D-2r

so the areas, value of k and respective values of P and σ are calculated for each radii and are attached with the solution.

As seen from the attached values, it is observed that the maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.

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URGENT What does Newton's third law say about why momentum is conserved in collisions?
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Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.For such a collision<span>, the forces acting between the two objects are equal in magnitude and opposite in direction</span>
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A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co
Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = -F_Dd

Where:

d = Distance traveled = 450 m

Therefore,

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd and

v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh  -F_Dd}{ \frac{1}{2}m}  = v_1^2 + 2gh -\frac{   2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0
3 years ago
One object is thrown vertically upward with an initial velocity of 100 m/s and
gavmur [86]

We have that for the Question it can be said that The maximum height reached  by the first <em>object</em> will be  100 times  that of the other.

  • (H_{max})_1=100*(H_{max})_2

From the question we are told

One object is thrown vertically upward with an initial velocity of 100 m/s and  another object with an initial velocity of 10 m/s. The maximum height reached  by the first object will be

that of the other.

a. 10,000 times

b. none of these

<em>c. </em><em>1000 times</em>

d. 100 times

<em>e.</em><em> 10 times</em>

Generally the equation for the velocity is mathematically given as

v=\frac{d}{t}\\\\Where\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{(V_1)^2}{(v_2)^2}\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{10000}{(10}\\\\

(H_{max})_1=100*(H_{max})_2

Therefore

The maximum height reached  by the first <em>object</em> will be  100 times  that of the other.

(H_{max})_1=100*(H_{max})_2

For more information on this visit

brainly.com/question/23379286

8 0
2 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

u = 27.7 m/s --> initial velocity of the car

\mu=0.804 --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

F=-\mu mg

where m is the car's mass and g=9.81 m/s^2 is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2

And the negative sign means it is a deceleration.


2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

v^2 - u^2 = 2aS

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m


3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

d_t = ut=(27.7 m/s)(0.543 s)=15.0 m

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

d_b=48.6 m

So the total distance your car covers is

S'=d_t+d_b=15.0 m +48.6 m=63.6 m

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

d=S'-S=63.6 m-48.6 m=15.0 m


6 0
3 years ago
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