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3 years ago

When and where is the bar graph used?

Physics

2 answers:

ira [324]3 years ago

4 0

bar graphs are used to compare things between different groups or to track changes over time

Artyom0805 [142]3 years ago

3 0

Bar graphs are used to compare things between different groups or to track changes over time.

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Water evaporators is this a physical change or a chemical change?

harina [27]

It will be a physical change

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3 years ago

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How did treating workers well benefit henry ford ??

Diano4ka-milaya [45]

He treated then with care he would get more help or what he needed

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3 years ago

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

Anastasy [175]

Answer:

Part A:

$\rm 2.8\times 10^3\ Volts.$

Part B:

$\rm 5.6\times 10^{-5}\ J.$

Explanation:

Part A:

Potential energy of charge at the given point, $\rm U=42\ \mu J=42\times 10^{-6}\ J.$

Charge, $\rm q=15\ nC = 15\times 10^{-9}\ C.$

The potential energy at a point due to a charge is defined as

$\rm U=qV$ .

where,

V = electric potential at that point.

Therefore,

$\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.$

Part B:

Now, if the charge at that point is replaced with $\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.$ , then the electric potential energy at that point is given by

$\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.$

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3 years ago

A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potenti

Korolek [52]

Call the capacitance C.
Note the energy in a capacitor with voltage V is E =½CV². 

Initial energy = ½C(12)² = 72C 

40% of energy is delivered, so 60% remains.in the capacitor. 
Remaining energy = (60/100) x 72C =43.2C 

If the final potential difference is X, the energy stored is ½CX² 
½CX² = 43.2C 
X² = 2 x 43.2 = 86.4 
X = 9.3V

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3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

$I = \frac{P}{4\pi r^2},$