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WITCHER [35]
3 years ago
12

A 1.7m long barbell has a 20kg weight on its left and a 35kg weight on its right. (a) If you ignore the weight of the bar itself

, how far from the left end of the barbell is the center of gravity? (b) Where is the center of gravity if the 8.0 kg mass of the barbell is taken into account?
Physics
1 answer:
Levart [38]3 years ago
4 0

Answer:

(a) 1.08 m

(b) 1.06 m

Explanation:

<u>Step 1:</u> calculate the center of gravity from 20kg mass

Let the center of gravity from 20kg mass = X

Applying the principle of moment; clockwise moment = ant-clockwise moment

20*X = 35*(1.7-X)

20X = 59.5 - 35X

55X = 59.5

X = 59.5/55

X = 1.08 m

Ignoring the weight of the bar, the center of gravity is 1.08m from left end of the barbell.

<u>Step 2:</u> calculate the center of gravity from 20kg mass, if the 8.0kg mass of the barbell in considered.

Applying the principle of moment

(20*X)+\frac{X}{2}(\frac{8}{1.7}) = 35(1.7-X) + (\frac{1.7-X}{2})(\frac{8}{1.7})

20X + 2.353X = 59.5 -35X +2.353(1.7-X)

20X +2.353X = 59.5 -35X + 4 - 2.353X

59.706X = 63.5

X = 63.5/59.706

X = 1.06 m

considering the weight of the bar, the center of gravity is 1.06m from left end of the barbell.

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jolli1 [7]

Answer:

Explanation:

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the positively charged body loses some electrons and the negatively charged body gains some electrons.

According to the properties of charge when the plastic rod is rubbed with the fur, the plastic rod has a tendency to get the negative charge and the fur has a tendency to get positive charge.

So,

When we rub a plastic rod with a piece of fur, electrons--------------transfer from fur----------------to the -------plastic-----. In the result charging fur with -positive-----------charge and plastic rod with-----negative--------charge.

6 0
3 years ago
A car is travelling with a velocity of 10 m/s and has a mass of 550 kg. The<br> car has<br> energy.
Elenna [48]

Answer: The car has a kinetic energy (because it's in motion) of:  27500J

Explanation:

Formula: E=\frac{1}{2}mv^2

E=\frac{1}{2}mv^2

E=\frac{1}{2}(550kg)(10m/s)^2

E=\frac{1}{2}(550kg)(100m^2/s^2)

E=\frac{1}{2}(55000J)\\E=27500J

3 0
3 years ago
Is baking a cake conduction, convection, or radiation?
Stolb23 [73]

Answer:

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5 0
2 years ago
A small car with mass of 0.800 kg travels at a constant speed
Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

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7 0
2 years ago
In which electric circuit would the voltmeter read 10 volts ?
ki77a [65]

Given that,

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We need to calculate the equivalent resistance

Using formula of parallel

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Put the value into the formula

\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}

\dfrac{1}{R}=\dfrac{1}{4}

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Using ohm's law

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I=\dfrac{V}{R}

Where, V = voltage

R = resistance

Put the value into the formula

I=\dfrac{10}{4}

I=2.5\ A

Hence, The current in the circuit is 2.5 A

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