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Mekhanik [1.2K]
3 years ago
14

Could someone help me with this question please?

Physics
2 answers:
Eduardwww [97]3 years ago
5 0
A.) Only "Amplitude" <span>can NOT be used to express the electromagnetic spectrum order among the option

Hope this helps!</span>
BabaBlast [244]3 years ago
3 0
You are correct...amplitude will be the answer
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Is cell using oxygen to break down sugar a physical or chemical change
Viefleur [7K]
Breaking down sugar (glucose) is a chemical change. Sugar is a compound that can be broken down.
5 0
3 years ago
Read 2 more answers
How does the increasing mass effect the force of an object in motion?​
irina [24]

Answer:

<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>

<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>

Explanation:

<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>

<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>

7 0
1 year ago
Justin notices a particular type of caterpillar feeds only on cottonwood trees in his neighborhood.In which way has Justin incre
Anna007 [38]
Justin notices a particular type of caterpillar feeds only on cottonwood trees in his neighborhood.In which way has Justin increased his scientific knowledge, Justin increased his powers of scientific observation. His observation is empirical evidence. Empirical relies on observation  and experimentation.

It will help you..........

7 0
3 years ago
What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

7 0
2 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
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