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2 years ago

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A barbellbell is loaded with two 20 kg plates on its right side and two 20kg plates on its left side. The barbell is 2.2m long,

and its unloaded mass is 20kg. The two 20kg plates on the right side are locked in place at 35cm and 40cm from the right end of the bar. The two 20kg plates on the left side have slipped one iss 30 cm from the left end of the bar and the other is 20 cm from the left end of th bar. where is the center of gravity of this bar?

1 answer:

Aleks [24]2 years ago

7 0

Answer:

<h2>Center of gravity of the system of all masses is 105 cm from the left end of the rod</h2>

Explanation:

Let the position of Left end of the rod is our reference

So here we will have

$m_1 = 20 kg$

$x_1 = 20 cm$

$m_2 = 20 kg$

$x_2 = 30 cm$

$m_3 = 20 kg$

$x_3 = 110 cm$

$m_4 = 20 kg$

$x_4 = (220 - 40) = 180 cm$

$m_5 = 20 kg$

$x_5 = (220 - 35) = 185 cm$

so we will have

$x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4 + m_5x_5}{m_1 + m_2 + m_3 + m_4 + m_5}$

$x_{cm} = \frac{20(20 + 30 + 110 + 180 + 185)}{20 + 20 + 20 + 20 + 20}$

$x_{cm} = 105 cm$

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A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m Ã 3.00 m Ã 5.00 m. 1. what is the

cupoosta [38]

1) The volume of the raft is the product between the lenghts of its three dimensions:
$V = (2.00 m)(3.00m)(5.00m)=30 m^3$

2) The mass of the raft is the product between its density, d, and its volume, V:
$m=dV=(650 kg/m^3)(30 m^3)=19500 kg$

3) The weight of the raft is the product between its mass m and the gravitational acceleration, $g=9.81 m/s^2$ :
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( $650 kg/m^3$ ) is smaller than the water density ( $1000 kg/m^3$ ), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
$W_a = W-B=W-W=0$

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship $m=dV$ we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7) $19.5 m^3$ . This volume is now "filled" with part of the volume of the raft, therefore $19.5 m^3$ is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, $30 m^3$ :
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is $30 m^3-19.5 m^3 = 10.5 m^3$ . Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
$m_E = m = 19500 kg$

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: $24.2 m^3$ . Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
$30 m^3 - 24.2 m^3 = 5.8 m^3$
and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

8 0

2 years ago

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

lesya692 [45]

Answer:

H = 109.14 cm

Explanation:

given,

Assume ,

Total energy be equal to 1 unit

Balance of energy after first collision = 0.78 x 1 unit

= 0.78 unit

Balance after second collision = 0.78 ^2 unit

= 0.6084 unit

Balance after third collision = 0.78 ^3 unit

= 0.475 unit

height achieved by the third collision will be equal to energy remained

H be the height achieved after 3 collision

0.475 ( m g h) = m g H

H = 0.475 x h

H = 0.475 x 2.3 m

H = 1.0914 m

H = 109.14 cm

6 0

2 years ago

A small lead ball, attached to a 1.25-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

cluponka [151]

Explanation:

Radius of the circular path, r = 1.25 m

Angular velocity of the ball, $\omega=3\ rev/s=18.84\ rad/s$

It is placed 1.5 meters above the ground. Using the conservation of energy to find the height of the ball.

$mgh=\dfrac{1}{2}mv^2$

$h=\dfrac{v^2}{2g}$

Since, $v=r\times \omega$

$h=\dfrac{(r\omega)^2}{2g}$

$h=\dfrac{(1.25\times 18.84)^2}{2\times 9.8}$

h = 28.29 meter

So, the ball will rise to a maximum height of 1.5 m + 28.29 m = 29.796 meters. Hence, this is the required solution.

4 0

1 year ago

Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.

m_a_m_a [10]

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, $q_1=6\ \mu C=6\times 10^{-6}\ C$

Charge 2, $q_2=2\ \mu C=2\times 10^{-6}\ C$

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

$F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N$

So, the force applied between charges is 10.788 N.

3 0

2 years ago

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Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

Debora [2.8K]

Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

$H = \vec d$

work done is given as

$W = \vec F . \vec d$

Potential energy is given as

$PE_g = mgh$

force due to gravity is given as

$\vec F_g = mg$

now here if we plug in the value of distance and force in the formula of work done then we will have

$W = (mg)(h)$

so here we got

$W = PE_g$

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

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2 years ago

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