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AVprozaik [17]
3 years ago
8

Indicate all ways in which the graph used in question 4 would change if the Normal Force applied to the object was increased. Be

specific. (What happens to slopes? Y-values? X-values? Etc.)

Physics
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

Answer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNINGAnswer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNING

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A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance
yan [13]

Answer:

a)  h'= 5/7 h , b)   h ’= h

Explanation:

Let's use energy conservation for this exercise

Starting point. Upper left side

         Em₀ = mg h

Final point. Lower left side

        Emf = K = ½ m v² + ½ I w²

        Em₀ = emf

        mgh = ½ m v² + ½ I w²

Angular and linear velocity are related

           v = r w

           w = v / r

The moment of inertia of the marble that we take as a solid sphere is

           I = 2/5 m r²

We substitute

          m g h = ½ m v² + ½ 2/5 m r² (v / r)²

          g h = ½ v2 (1 + 2/5)

         v = √(g h 10/7)

This is the speed at the bottom of the bowl

Now let's apply energy conservation to the right side

a) right side if rubbing

             Em₀ = K

              Emf = U = mg h’

             ½ m v² = mg h’

              h’= ½ (g h 10/7) / g

              h'= 5/7 h

b) right side with rubbing

             Em₀ = K

             Emf = K + U = -½ I w² + m g h

             Emf = -½ 2/5 m r² v² / r² + m gh

            Em₀ = emf

            ½ v² = -1/5 v² + g h’

            h’= (1/2 +1/5) (gh 10/7) / g

            h ’= h

c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher

8 0
3 years ago
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity If the distance betwee
natima [27]

Answer:

The luminosity remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

The apparent brightness, F = L/4πr² where L = luminosity and r = distance between us and the star.

Since L is independent of the distance between us and the star, it is constant, then

F ∝ 1/r²

So, F₁/F₂ = r₂²/r₁² where F₁ = apparent brightness at r₁ and F₂ = apparent brightness at r₂

If the distance is doubled, that is r₂ = 2r₁, then

F₁/F₂ = r₂²/r₁²

F₁/F₂ = (2r₁)²/r₁²

F₁/F₂ = 4r₁²/r₁²

F₁/F₂ = 4

F₂ = F₁/4

So, since the luminosity is constant, <u>the luminosity remains the same, but the apparent brightness is decreased by a factor of four.</u>

5 0
3 years ago
PLEASE HELP!!! 50 points!!!!!! SEE ATTACHMENT!!! In which direction is there a net force of 200 N? left, right, up, down
Sladkaya [172]

The answer is right. The point and the 200 N sign maybe on the left side but the direction can be going right. I just took the quiz today, I wouldn't give the wrong answer cause i'm not a liar.  -_-

5 0
3 years ago
Read 2 more answers
A 1.25 x 10-4 C charge is moving5200 m/s at 37.0° to a magnetic fieldof 8.49 x 10-4 T. What is the magneticforce on the charge?
juin [17]

0.0003321 Newtons

Explanation

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation to find the force

\begin{gathered} F=qvBsinθ \\ where\text{ F is the magnetic force} \\ q\text{ is the charge} \\ v\text{ is the velocity of the charge} \\ Bis\text{ the magnetic field} \\ \theta\text{ is the angle} \end{gathered}

so

Step 1

Let

\begin{gathered} q=1.25*10^{-4\text{ }}C \\ v=5200\frac{m}{s} \\ \theta=37\text{ \degree} \\ B=8.49*10^{-4}T \end{gathered}

now, replace

\begin{gathered} F=qvBs\imaginaryI n\theta \\ F=1.25*10^{-4}\text{ C*5200 }\frac{m}{s}*8.49*10^{-4}Tsin(37) \\ F=0.0003321\text{ Newtons} \\  \end{gathered}

so, the answer is

0.0003321 Newtons

I hope this helps you

7 0
1 year ago
A blank is a statement the summerizes a pattern found in nature
MrMuchimi
What are you asking here?
4 0
3 years ago
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