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3 years ago

Help with this plz!!!

Physics

1 answer:

Alex3 years ago

6 0

The answer is:

It is virtual and behind the mirror.

Mirrors produce images, either virtual or real.

Virtual images are produced when the image is behind the mirror. Real images are formed when they image is produced in front of the mirror.

So look at the image produced. It's small and found in behind the mirror. This means that it is a virtual image. How do I know that its behind the mirror?

When you look at the diagram, the front side of the mirror is the blue side. This is a convex mirror.

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2. A powerful experimental sewing machine is powered by a mass-spring system. This

Alexus [3.1K]

We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

For more information on Mass visit

brainly.com/question/15959704

7 0

2 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

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