To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary, in what instance does a person become unhealthy

Answer:

I feel exited and happy I enjoy it with my friend

Answer:

thx for the points

Explanation:

no need brainliest

Answer:

8.01e-22

Explanation:

The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

The extension produced by one mole of lead atom is calculated by applying Hooke's law;

F = kx

mg = kx

x = mg/k

x = (0.207 x 9.8) / (20)

x = 0.101 m

The Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;

E = ¹/₂kx²

E = ¹/₂ (20)(0.101)²

E = 0.102 J

Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

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The actual weight of the gas = apparent weight + weight.

The actual weight = [tex]W_{A}[/tex] + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/[tex]m^{2}[/tex]

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight =  [tex]W_{A}[/tex] + W

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The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.

A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.

Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.

According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.

Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.

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The motion of the rope which is perpendicular to the direction of the

propagation of the wave is a transverse wave motion.

Reasons:

The given function for the wave speed is presented as follows;

[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]

Where;

[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]

Taking the mass of the rope as, m = 2.00 kg

The length of the rope, L = 80.0 m

The mass hanging on the rope, M = 20.0 kg

We have;

T = 20.0 kg × 9.81 m/s² = 196.2 N

[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]

Therefore;

Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;

v = f × λ

Therefore;

v = 7.9 Hz × 7.9 m = 62.41 m/s

Which gives;

[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]

T = 62.41² × 0.025 = 97.3752025

[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]

Where;

g = The acceleration due to gravity which is approximately 9.81 m/s²

[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]

Therefore;

The mass of the box, m ≈ 9.93 kg

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The parameters obtained from a similar question online are;

[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]

Length of the rope, L = 80.0 m

Mass of the rope, m = 2.0 kg

Frequency of a point on the rope, f = 20 Hz

Answer:

its b it must be charge

Explanation:

Answer:

Con el cielo del agua

Explanation:

espero que te ayude

Answer:

i think the correct answer is B. cancel out

Answer:

i dont know need points

Explanation:

Answer:

 |F| = 35 kN

Explanation:

a = F/m

100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s

v² = u² + 2as

   a = (v² - u²) / 2s

F/m = (v² - u²) / 2s

   F = m(v² - u²) / 2s

   F = 5000(0² - 27.8²) / 2(55)

   F = - 35,072.9517...

The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]

The distance between the glass is calculated as follows;

[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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Answer:

I don't see anything so you should repost

This question involves the concepts of tension in a string and the wavelength of the wave in a string.

The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.

The speed of a wave on a string is given by the following formula:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

where,

v = speed of wave = fλ

f = frequency of the wave

λ = wavelength of the wave

T = tension force

μ = linear mass density of the string

Therefore,

[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]

It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,

[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]

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This question involves the concepts of density, volume, and mass.

(a) The volume occupied by this amount of water is "90 m³".

(b) The length of an edge of each small cube is "83 μm".

(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.

(a)

The volume can be found using the following formula:

[tex]V = \frac{nM}{\rho}[/tex]

where,

V = volume occoupied = ?

n = no. of moles = 5 mol

M = molar mass = 18 g/mol

[tex]\rho[/tex] = density of water = 1 g/m³

Therefore,

[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]

V = 90 m³

(b)

First, we will find the volume of an individual molecule:

[tex]V_i =\frac{V}{nN_A}[/tex]

where,

[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol

Therefore,

[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]

Vi = 3 x 10⁻²³ m³

This volume can be given as a volume of the sphere:

[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]

r = 4.15 x 10⁻⁷ m

Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)

d = 8.3 x 10⁻⁷ m = 83 μm

since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.

Edge Length = L = d

L = 8.3 x 10⁻⁷ m = 83 μm

(c)

The edge length is equal to the diameter of the molecule.

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The internal energy after heating in terms of U is 100U.

The given parameters;

Assuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.

[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]

Thus, we can conclude that the internal energy after heating in terms of U is 100U.

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The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.

The relationship between the shutter speed and the focal ratio is an inverse relationship.

Ratio of Areas = 5.6²/4²

Since; T1A1 = T2A2

In essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.

Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.

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How to find g (acceleration due to gravity)

We know,

Acceleration due to gravity (g)

[tex] = \frac{GM}{ {R}^{2} } [/tex]

where, G = Gravitational constant

[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]

M = Mass of the earth

[tex] = 6 \times {10}^{24} \: kg[/tex]

R = Radius of the earth

[tex] = 6.4 \times {10}^{6} m[/tex]

Putting these values of G, M and R in the above formula, we get

[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]

So, the value of acceleration due to gravity is

[tex]9.8m/s ^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Answer:

Explanation:

I'm not 10% sure but i think that if you were to do:

603 + 277 = 880

1000 - 880 = 120

So to keep it from moving, i think you would need to pull with a force of 120N

The force with which it should be pulled to keep the knot from moving is 663.58 N.

Newton's third law states that, for every action, there is an equal and opposite reaction.

Here,

Force exerted by Roberta, F₁ = 277 N

Force exerted by Michael, F₂ = 603 N

Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.

The force needed to keep the knot from moving is the resultant force of these two forces.

Therefore, the resultant force,

Fₙ = √F₁² + F₂²

Fₙ = √(277)² + (603)²

Fₙ = 663.58 N

Hence,

The force with which it should be pulled to keep the knot from moving is 663.58 N.

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Answer:

No - It is connected in parallel instead of series

Explanation:

The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.

An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.

The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).

The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.

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The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.

The given parameters;

The normal force on the wooden box is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]

The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;

[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]

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Answer:

600

Explanation:

GPE=mgh (mass*gravitational force*height)

m=10 kg

g=10

h=6m

10*10*6= 600 Joules

Answer:

  1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

  5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

  10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

  (5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

  (10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

  v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

Answer:

F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]

Explanation:

In pic

_________________

(hopet his helps can I pls have brainlist (crown)☺️)

The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is

∑ F = -f = ma = (4.00 kg) a

Assuming constant acceleration a , the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)

Solve for the acceleration :

a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²

Then the frictional force exerted a magnitude of

-f = (4.00 kg) (-14.8 m/s²)

f ≈ 59.4 N

and was directed opposite the block's motion.

Answer:

Our atmosphere has five different layers. They are:

1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.

2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.

3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.

4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.

5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.

Answer:

Explanation:

I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was

Conservation of momentum

14000(8.1) + m(0) = (14000 + m)(3.6)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

The mass of the second car is 17500 kg.

The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.

Newton's third law of motion has a direct impact on the idea of momentum conservation.

mass of the first car = 14000 kg.

Initial speed of first car = 8.1 m/s.

Final speed of the two cars = 3.6 m/s.

From law of conservation of momentum; we can write:

Total initial momentum = total final momentum

14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

Hence, the mass of the second car is 17500 kg.

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The acceleration due to gravity on the given planet is 8.71 m/s².

The given parameters;

The acceleration due to gravity on the planet is calculated by applying following formula as follows;

[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2[/tex]

Thus, the acceleration due to gravity on the given planet is 8.71 m/s².

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Answer:

Explanation:

350 cm / 25 cm = 14 radians

14 rad / 2π rad/rev = 2.23 revolutions

Answer:

option B is the correct answer

Explanation:

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