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erik [133]
3 years ago
7

Caused by the tilting of the Earth s axis

Physics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

Seasons changing

Explanation:

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In a position vs. time graph depicting the motion of two different objects, the point at which the lines intersect is where the
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3 years ago
How long does it take oxygen to diffuse a distance of 1 mm in water at room temperature? How long does it take oxygen to diffuse
Viktor [21]

Answer:

option A

Explanation:

given,

travel distance = 1 mm

diffusion constant for the  molecule  of oxygen in water = 1 x 10⁻⁹ m²/s

we know that

 t = \dfrac{x^2}{2D}

t is the time taken for diffusion

x = 1 mm = 10⁻³ m

D is the diffusion

 t = \dfrac{(10^{-3})^2}{2\times 10^{-9}}

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so, the option to the answer is 170 s

the correct answer is option A

8 0
4 years ago
4. Calculate the kinetic energy of a 4.7 kg object moving at a speed of 7 m/s. SHOW YOUR WORK
SashulF [63]

Answer:

\boxed {\boxed {\sf 115.15 \ J}}

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated with the following formula.

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

The mass of the object is 4.7 kilograms. The velocity of the object is 7 meters per second.

  • m= 4.7 kg
  • v= 7 m/s

Substitute the values into the formula.

E_K= \frac{1}{2} (4.7 \ kg)(7 \ m/s)^2

Solve the exponent.

  • (7 m/s)²= 7 m/s * 7 m/s = 49 m²/s²

E_K= \frac{1}{2} (4.7 \ kg)(49 \ m^2/s^2)

Multiply the numbers together.

E_K = 2.35 \ kg * 49 \ m^2/s^2

E_K= 115.15 \ kg*m^2/s^2

Convert the units. 1 kilogram square meter per square second is equal to 1 Joule.

E_K= 115.15 \ J

The object has <u>115.15 Joules</u> of kinetic energy.

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2 years ago
Read 2 more answers
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

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