The eight plants of the Solar System orbit the Sun in a chaotic random way.

True
False

Answer:

55.897905

Explanation:

1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth

9.80665 times 5.7=55.897905

Brainliest?

When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.

Answer:

11.5 m/s²

Explanation:

The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s

So, a = rω²

a = r(2πN)²

a = 4π²rN²

substituting the values of the variables into the equation, we have

a = 4π²rN²

a = 4π²(0.107 m)(1.65 rev/s)²

a = 4π²(0.107 m)(2.7225 rev²/s)²

a = 4π² × 0.2913075 mrev²/s)²

a = 11.5 m/s²

Answer:

1) It seems that he would need the central gravitational force

   (the sun)

2) Also the planets would need to be included (orbits around the sun)

    Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.

3. Then, many of the planets have significant objects (moons) rotating about them.

Those would seem to be objects to be included in a model of the solar system.

1) He would need the central gravitational force (the sun)

2) The planets would need to be included: Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.

3) Many of the planets have specific moons rotating about them.

Larry should put the Earth between the planets Venus, and Mars.

Answer:

The mass of the mercury remaining in the bottle is 497.57 grams.

Explanation:

The mass of the mercury remaining in the bottle is found by subtracting the mass expeled due to heating from initial mass inside the bottle. That is:

[tex]m_{f} = m_{o}-\Delta m[/tex] (1)

Where:

[tex]m_{o}[/tex] - Initial mass, in grams.

[tex]\Delta m[/tex] - Mass expelled due to heating, in grams.

[tex]m_{f}[/tex] - Final mass, in grams.

If we know that [tex]m_{o} = 500\,g[/tex] and [tex]\Delta m = 2.43\,g[/tex], then the mass of the mercury remaining in the bottle is:

[tex]m_{f} = m_{o}-\Delta m[/tex]

[tex]m_{f} = 497.57\,g[/tex]

The mass of the mercury remaining in the bottle is 497.57 grams.

Answer:The answer is because of the gravity and the mass formed in the sun, the magnetic field reacts to it and leaves a fault on earth. wind goes by the earth and procides to be ok

Explanation:that is it

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

a = ∆v / ∆t = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force F is proportional to the acceleration a according to

F = m a

where m is the object's mass. Solving for m gives

m = F / a = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

Answer:

xplanation:

Angle of reflection is measured between the incident ray and the angle which it makes with the normal at the point where incident ray strikes the mirror surface.

Further on reflection, it makes the same angle i.e. angle of reflection is equal to angle of reflection.

Hence, as angle of incidence is 55∘ angle of reflection too is 55∘ and the angle between the incident ray and the reflected ray is 55∘+55∘=110∘

Answer:

 u = 14 m / s

Explanation:

For this exercise let's use conservation of energy

starting point. On the floor just when u speeding out

         Em₀ = K = ½ m v²

final point. When on top of the building, no speed

         Em_f = U = m g h

energy is conserved

          Em₀ = Em_f

          ½ m v² = m g h

          v = u

          u = [tex]\sqrt{2 g h}[/tex]

          u=  [tex]\sqrt{2 \ 9.8 \ 10}[/tex]

          u = 14 m / s

Answer:

The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.

Explanation:

mark me brainlest

Answer:18 watts

Explanation:i just got this question trust me

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Answer:

1.72 m

Explanation:

Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height

76 = (3.5)(9.8)h

76=44.1h

h=1.72335600907 ≈1.72 m

Answer:

:r

Explanation:r

Tobnbv346468this Ishmael

Answer:

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  [tex]\tau = T \alpha[/tex]

then,

⇒  [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]

⇒  [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]

⇒  [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]

⇒      [tex]=8.34 \ m/s[/tex]

hence,

⇒  [tex]T=mg+M(g-a_t)[/tex]

On substituting the values, we get

⇒      [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]

⇒      [tex]=6.86+2(1.46)[/tex]

⇒      [tex]=6.86+2.92[/tex]

⇒      [tex]=9.8 \ N[/tex]

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )

given that

I = 2.6 A

[tex]B^>[/tex] = 0.17

[tex]L^>[/tex] = 0.52

so we substitute

|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )

|F[tex]_{mg}^>[/tex] | = 0.2298 N

Therefore, the magnitude of the magnetic force on the wire is 0.2298 N

Answer:

air pressure

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Answer:

The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality

Explanation:

Answer:

The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality

Answer:

e. Combustion

Explanation:

In Combustion reaction, a substance reacts with oxygen from the air and resultant product is that it releases carbon dioxide and water.

Here,

2C2H6 is the substance that reacted with 7O2 (Oxygen) to release 4CO2 (Carbon Dioxide) and 6H2O (Water).

Answer:

3. (a) - is the answer most likely

4: (a)

5 (d or c)

6 (b)

Answer:

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

Explanation:

a. Who reaches the bottom first

The kinetic energy of the objects is given by

K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²

K = 1/2mv² + 1/4mv²

K = 3mv²/4

For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop

So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²

K' = 1/2m'v'² + 1/2m'v'²

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

3mv²/4 = m'v'²

v²/v'² = 4m/3m'

v²/v'² = 4/3(m/m')

v/v' = √[4/3(m/m')]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

b. Why

Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder

K - 1/2Iω² = K₀

3/4mv² - 1/2(mr²/2)(v/r)² = K₀

3/4mv² - 1/4mv² = K₀

K₀ = 1/2mv²

For the steel hoop,

K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop

K' - 1/2I'ω'² = K₁

m'v'² - 1/2(m'r'²)(v'/r')² = K₁

m'v'² - 1/2m'v'² = K₁

K₁ = 1/2m'v'²

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called  Kinetic energy.

The kinetic energy of the objects is given by

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]

where

m = mass of object,

v = velocity of object,

I = moment of inertia and

ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy,

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]

[tex]K = \dfrac{3mv^2}{4}[/tex]

For the steel hoop,

I' = mr'²

where

m' = mass of steel hoop and

r' = radius of steel hoop and

v' = velocity of steel hoop

So, its kinetic energy,

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]

[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]

[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]

[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

(b) Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]

where

K₀ = translational kinetic energy of wooden cylinder

[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]

[tex]K_o = \dfrac{1}{2}mv^2[/tex]

For the steel hoop,

[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]

where

K₁ = translational kinetic energy of steel hoop

[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]

[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

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Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Answer:

Potential and kinetic

Explanation:

Answer:

the net work done on the object is 80 J.

Explanation:

Given;

force applied on the object, F = 100 N

mass of the object, m = 20 kg

speed of the object, v = 0.2 m/s

distance moved by the object, d = 0.8 m

The net work done on the object is calculated as follows;

W = F x d

W = 100 N x 0.8 m

W = 80 J

Therefore, the net work done on the object is 80 J.

The force 1 is tension force.

To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.

Thus, we can conclude that the correct option is (B).

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1 mole of nitrogen atoms moving at 35.0 m/s possesses approximately 27.8 joules of energy.

To calculate the energy possessed by 1 mole of nitrogen atoms moving at 35.0 m/s, we need to consider both the kinetic energy and the molecular mass of nitrogen.

The kinetic energy (KE) of an object is given by the equation KE = 1/2 * m * v^2, where m is the mass and v is the velocity.

The molar mass of nitrogen (N₂) is approximately 28.0134 g/mol, which can be converted to kilograms by dividing by Avogadro's number (6.022 × 10^23). This gives us a mass of approximately 4.65 × 10^(-26) kg for one nitrogen atom.

Plugging in the values, we have KE = 1/2 * (4.65 × 10^(-26) kg) * (35.0 m/s)^2.

Evaluating the equation, we find that the kinetic energy possessed by one nitrogen atom is approximately 4.62 × 10^(-23) joules.

Since we are considering 1 mole of nitrogen atoms, we need to multiply this value by Avogadro's number to get the energy possessed by 1 mole. Avogadro's number is 6.022 × 10^23, so the total energy is approximately 2.78 × 10^1 joules, or 27.8 J.

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