All categories

3 years ago

What is the difference between a general and a specific tolerance, and how can you tell the difference on a drawing?

1 answer:

6 0

<span>During the reading of a blueprint, a general tolerance pertains to all dimensions that are not independently identified. All tolerances are +\- .030. General tolerances are mostly found in the blueprint’s set of information. Let’s say for example, a compact tolerance is required, then a specific tolerance is considered for specific areas of the blueprint. These kinds of tolerances are commonly found along the affected area.</span>

You might be interested in

A. How far does a 100-newton force have to move to do 1,000 joules

Aloiza [94]

Work done by a force is given as the product of force and the distance moved by the force.

Work done is the product of force and the distance moved by the the force.

Work done = Force × distance

Thus, distance required by the 100 N force is given as:

Distance = work done/force

Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

Force applied = work done/ distance

Force applied by the hoist = 500/2

Force applied by the hoist = 250 N

Distance moved in one push-up = 25 cm = 0.25 m

Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

Distance moved by the force = 0 m

Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

5 0

2 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$