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Natali [406]
3 years ago
8

Daltons theory was identified using

Physics
1 answer:
MariettaO [177]3 years ago
8 0
Dalton's theory was identified using light microscope images of atoms.
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How long is a football field?
Nana76 [90]

Answer:

91.44 meters or 300 feet

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3 years ago
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The velocity v of a particle moving in the xy plane is given by v = (6.0t -4.0t2 )i + 8.5j, in m/s. Here v is in m/s and t (for
atroni [7]

Answer:

Explanation:

Given

Velocity of the particle in vector form is given by

v=\left ( 6t-4t^2\right )\hat{i}+8.5\hat{j}

acceleration is rate of change of velocity thus acceleration is

a=\frac{\mathrm{d} v}{\mathrm{d} t}

a=(6-8t)\hat{i}+0\hat{j}

at t=3\ s

a=6-8\times 3

a=-18\ m/s^2

7 0
4 years ago
The North American and European plates of the
faust18 [17]
Distance = speed/time

width/distance given is 2.9*10^3 miles
1 mile=1.6*10^6mm
total distance now is 2.9*1.6*10^9
speed is 25 mm/yr
plug the values,
therefore the rift takes approximately 185600000 years for a width <span>of 2.9 x 10^3 mi.</span>
5 0
3 years ago
When a negatively charged object moves in the opposite direction of an electric force field, the potential energy of the object
Simora [160]
The best and most correct answer among the choices provided by the question is decreases <span>.


</span>The potential energy of the object <span>decreases.</span>

Hope my answer would be a great help for you.    
If you have more questions feel free to ask here at Brainly.
3 0
3 years ago
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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
4 years ago
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