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>

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Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a

Crank

A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.

6 0

3 years ago

Read 2 more answers

s=1/2(u+v)t check the dimensional consistency for this? where s is displacement, t is time, u is initial velocity, v is final ve

Anettt [7]

Answer:

See the explanation below.

Explanation:

The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.

s = displacement [m]

v and u = velocity [m/s]

t = time [s]

Now using these units in the given equation.

$s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]$

So the expression is good, and dimensional has consistency.

8 0

3 years ago

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$