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2 years ago

8

Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require

d for the ball to make an entire loop

1 answer:

strojnjashka [21]2 years ago

3 0

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

Em₀ = U = m g h

Final point. In the highest part of the loop

Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

I = m R²

v = w R

we substitute

Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

em_f = m v² + 2 m g R

Energy is conserved

Emo = Em_f

mgh = m v² + 2m g R

h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

h> 2R

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a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

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θ= 45°

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F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

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3 years ago

Which of the following best represents potential energy being converted to kinetic energy? A thrown baseball lands on the ground

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3 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

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Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

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$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

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On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

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If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

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3 years ago

A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if

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Answer:

Image distance is -52.5 cm

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Explanation:

u = Object distance

v = Image distance

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$v=-2.5\times 21=-52.5\ cm$

Image distance is -52.5 cm

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3 years ago