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2 years ago

How could the location of tests affect the performance of a catapult ?

Engineering

2 answers:

Svetach [21]2 years ago

6 0

The location of tests affect the performance of a catapult depending on the length of the arm.

<u>Explanation:</u>

A ball from a catapult, will travel a longer distance by increasing the length of the arm. Using the catapult, if the arm is extended to different lengths the wooden ball how far it is thrown can be observed.

This shows that extending the arm length does increase the distance thrown. To shoot a far distance (example for sideways) you need to the pebble sideways and up. By doing this, it can travel a longer distance

Luba_88 [7]2 years ago

3 0

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

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In a series circuit, resistance is the simple _____ of all resistors in the circuit.

melisa1 [442]

In a series circuit, resistance is the simple _____ of all resistors in the circuit.
The Answer Is A (Sum)

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2 years ago

Determine the magnitude of the resultant force and the moment about the origin. Note: the symbol near the 140 N-m moment are not

arlik [135]

Answer:

R = 148.346 N

M₀ = - 237.2792 N-m

Explanation:

Point O is selected as a convenient reference point for the force-couple system which is to represent the given system

We can apply

∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N

∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N

Then

R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)

⇒ R = 148.346 N

Now, we obtain the moment about the origin as follows

M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)

We can see the pic shown in order to understand the question.

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2 years ago

What is the difference between a single-model production line and a mixed-model production line?

Nat2105 [25]

The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.

This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety

The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.

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2 years ago

Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un a

Nitella [24]

Answer:

can you translate

Explanation:

what Is that?

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2 years ago

For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

$x(t)=Asin(\omega t+\phi)$

where,

A is the amplitude of motion

$\omega$ is the angular frequency of the motion

$\phi$ is known as initial phase

part 1)

Now by definition of velocity we have

$v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )$

part 2)

Now by definition of acceleration we have

$a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )$

part 3)

The angular frequency is related to Time period 'T' as $T =\frac{2\pi }{\omega }$

where

$\omega$ is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is $A\omega ^{2}$ and $A\omega$ respectively.

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2 years ago