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Arada [10]
3 years ago
9

A bowling ball weighing 71.7 N is attached to the ceiling by a rope of length 3.73 m . The ball is pulled to one side and releas

ed; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.60 m/s. At this instant, what are:
a. the acceleration of the bowling ball, in magnitude and direction
b. the tension in the rope?
Physics
1 answer:
vekshin13 years ago
4 0

Answer:

A) a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) T = 113.2 N

Explanation:

A) We are given;

Weight of bowling ball; W = 71.7 N

Speed; v = 4.6 m/s

Rope length; r = 3.73 m

Now, formula for the centripetal acceleration is;

a = v²/r

Thus; a = 4.6²/3.73

a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) since weight is 71.7 N, it means that;

Mass = weight/acceleration = 71.7/9.8

Mass(m) = 7.316 kg

Thus,

Centripetal force is;

F_cent = 7.316 × 5.673

F_cent = 41.5 N

Thus, Tension in the rope is;

T = W + F_cent

T = 71.7 + 41.5

T = 113.2 N

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Remark
When you are asked a question like this, the first thing to do is search out a formula and put some limits on it.

Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
E/R = I*R/R
E/R = I

Discussion
This is an inverse relationship. That means that as one goes up the other one will go down. 

So in this case you keep E constant and you manipulate R and look at your results for I

Case 1

Let us say that E = 10 volts
Let us also say the R = 10 ohms

I = E/R
I = 10/10
I = 1 ohm

Case Two
Let's raise the Resistance to 100 ohms
E = 10
R = 100
I = 10/100 = 0.1 

Conclusion
As the Resistance goes up, the current goes down. Answer: A
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3 years ago
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A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength ou
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The cylinder's electric field magnitude, at a distance <em>r</em> from the axis of the cylinder (greater than the cylinder's radius), is equal to E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}

<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a <em>charge</em>, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density \lambda, this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

\Phi = \frac{\lambda \cdot L}{\epsilon_0}

Where \Phi is called the Electric flux. Since the electric field is constant for a given distance <em>r</em> from the axis of the cylinder we can write that:

\Phi = E \cdot 2\pi r \cdot L

Joining both our expressions we can get that:

E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}

<h3 /><h3>Learn more</h3>
  • Description on Electric fields: brainly.com/question/8971780
  • Relation between electric fields and magnetism: brainly.com/question/2838625
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<h3>Keywords</h3>

Electrons, protons, electric field, cylinder, electric flux

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3 years ago
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Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on
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Answer:

Xcm = 1.95 cm  and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

     R cm = 1/M  ∑ m_{i}  r_{i}

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

     M = m₁ + m₂ + m₃

     M = 140 + 45 + 85

     M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

      R₁ = d / 2 i ^ + d j ^

      R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

      R₂ = 0

Point 3. Right vertex

      R₃ = d i ^

      R₃ = 3.4 i ^ cm

a) The x component of the massage center

      Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

      Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

      Xcm = d / M (m₁ / 2 + m₃)

b)   Let's write the mass center component x

      Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

      Xcm = 238/270

      Xcm = 1.95 cm

c) let's find the component and center of mass

     Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

    Ycm = 1 / M (m₁ d + 0 + 0)

    Ycm = m₁ / M d

d) let's calculate

    Y cm = 1/270 (140 3.4 + 0 + 0)

    Ycm = 1.76 cm

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4 years ago
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Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

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Acceleration,

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For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

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Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

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