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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

the top vertex. The second bead of mass m2=45g is placed on the left vertex. The third bead of mass m3=85g is placed on the right vertex.
(a) Write a symbolic equation for the horizontal component of the center of mass relative to the left vertex of the triangle.

(b) Find the horizontal component of the center of mass relative to the left vertex, in centimeters.

(c) Write a symbolic equation for the vertical component of the center of mass relative to the base of the triangle.

(d) Find the vertical component of the center of mass relative to the base of the triangle, in centimeters.

1 answer:

Vlad [161]3 years ago

8 0

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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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed <em>v </em>of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where <em>q</em> is the water flow (1 cubic foot per minute).

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3 years ago

A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Î© resistor, and an open switch.A 8.2-V battery is connected

tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$

Put the value into the formula

$I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})$

$I(t)=0.01925\ A$

$I(t) = 19.25\ mA$

(B). We need to calculate the store energy in the inductor

Using formula of energy

$E=\dfrac{1}{2}LI^2$

Put the value into the formula

$E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2$

$E=7.04\times10^{-6}\ J$

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

8 0

2 years ago

an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi

algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

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2 years ago

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artcher [175]

Answer:

I would say the answer is B

Explanation:

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3 years ago