Which of the following involves reflected light waves

Two tectonic plates moving toward one another are at a

nordsb [41]

Answer:

A. cause i just took the test

Explanation:

6 0

2 years ago

Read 2 more answers

Please I need with this

guajiro [1.7K]

I think its b im not sure im in 8th grade

8 0

3 years ago

Please help! the mass of an object is measured on a pan balance with a precision of 0.005 g and the recorded value of 128.01 g.

inysia [295]

Yes, these two objects have different masses.

<h3>How can we calculate that this statement is right ?</h3>

To calculate the precision of mass we are using the formula,

Precision(P) = $\frac{m_1}{m_1+\triangle m}$

Or, $\triangle$ m= $\frac{m_1}{P}$ -m₁

For the first case we are given,

m₁= The recorded value of mass

= 128.01 g

P= Precision of the mass

=0.005g

So, according to the formula, $\triangle$ m will be,

$\triangle$ m= $\frac{128.01}{0.005}$ -128.01

Or, $\triangle$ m=25,473.99 g

Or, $\triangle$ m=25.47 Kg

For the first case $\triangle$ m is 25.47 Kg..

For the second case we are given,

m₁= The recorded value of mass

= 0.13 Kg

P= Precision of the mass

=0.005 Kg

So, according to the formula, $\triangle$ m will be,

$\triangle$ m= $\frac{0.13}{0.005}$ -0.13

Or, $\triangle$ m= 25.87 kg

For the second case $\triangle$ m is 25.87 Kg.

For the two cases $\triangle$ m has different values, 25.47 Kg≠25.87 Kg.

Therefore we can conclude that, these two objects have different masses.

Learn more about Mass:

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7 0

1 year ago

Explains this: a light source can emit more than one type of light

Radda [10]

Answer:

The electromagnetic spectrum comprise a lot of waves length. Usually, different waves length are called as different lights, and a light source can emit in more than a different wave length, as the sun does, for example. The sun emit the visible light, UV light, infrared, etc.

6 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

3 0

3 years ago