Which of the following is a practical example of gravity acting on an object.

A.) A leaf falls from a tree and floats to the ground.

B.) A diver jumps off a diving board and splashes into a pool below.

C.) A car speeds up as a traffic light turns green.

D.) Rain falls from the sky.

Answer:

Explanation:

See attached file.

Answer:

D

Explanation:

Answer:

Correct answer is polycarbonate not gold

Explanation:

Answer:

I think they are called balanced forces

Explanation:

Answer:

Star Y is showing radial motion

Star Y is moving toward the Earth

Explanation:

Just answered this question on a quiz and got it right.

Star Y is showing radial motion

Star Y is moving towards earth.

Spectrum is defined as the arrangement of radiations emitted by atoms or molecules based on their characteristic wavelength and frequency.

Here,

The spectrum of the star Y is given and is compared to a reference hydrogen spectrum. The range of wavelengths of the star is given. The spectrum of the star shows that the wavelengths are shifted such that from longer wavelengths to shorter wavelength. This phenomenon of shift in wavelengths from higher to lower is known as blue shift. The star shows blue shift that means shifted to shorter wavelength or it can be said as shifted from lower frequency to higher frequency. As a result, it can be concluded that the star Y is moving towards the earth which implies Star Y is showing radial motion.

Hence,

The star Y is showing radial motion.

Star Y is moving towards the earth.

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Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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Explanation:

so sorry

don't know but please mark me as brainliest please

Answer:

The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.

Explanation:

Let suppose that air bubble behaves as ideal gas, whose equation of state is:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (Eq. 1)

Where:

[tex]P[/tex] - Pressure of the bubble, measured in kilopascals.

[tex]V[/tex] - Volume of the bubble, measured in cubic meters.

[tex]n[/tex] - Molar amount of the bubble, measured in kilomoles.

[tex]T[/tex] - Temperature, measured in Kelvin.

[tex]R_{u}[/tex] - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

Then, we eliminate the molar amount and the ideal gas constant by constructing the following relationship:

[tex]\frac{P_{A}\cdot V_{A}}{T_{A}} = \frac{P_{B}\cdot V_{B}}{T_{B}}[/tex] (Eq. 2)

Where:

[tex]P_{A}[/tex], [tex]P_{B}[/tex] - Pressure of the bubble at bottom and surface, measured in kilopascals.

[tex]V_{A}[/tex], [tex]V_{B}[/tex] - Volume of the bubble at bottom and surface, measured in cubic meters.

[tex]T_{A}[/tex], [tex]T_{B}[/tex] - Temperature of the bubble at bottom and surface, measured in Kelvin.

The pressure experimented by the bubble at bottom and surface are, respectively:

[tex]P_{A} = 101.325\,kPa+\left(1027\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (25\,m)\cdot \left(\frac{1}{1000}\,\frac{kPa}{Pa} \right)[/tex]

[tex]P_{A} = 353.120\,kPa[/tex]

[tex]P_{B} = 101.325\,kPa[/tex]

If we know that [tex]P_{A} = 353.120\,kPa[/tex], [tex]P_{B} = 101.325\,kPa[/tex], [tex]V_{A} = 1.20\times 10^{-6}\,m^{3}[/tex], [tex]T_{A} = 290.15\,K[/tex] and [tex]T_{B} = 303.15\,K[/tex], then the volume of the bubble at surface is:

[tex]\frac{(353.120\,kPa)\cdot (1.20\times 10^{-6}\,m^{3})}{290.15\,K} = \frac{(101.325\,kPa)\cdot V_{B}}{303.15\,K}[/tex]

[tex]1.460\times 10^{-6} = 0.334\cdot V_{B}[/tex]

[tex]V_{B} = 4.372\times 10^{-6}\,m^{3}[/tex]

[tex]V_{B} = 4.372\,cm^{3}[/tex]

And the volume of the air bubble is determined by this formula:

[tex]V_{B} = \frac{4\pi\cdot R^{3}}{3}[/tex] (Eq. 3)

Where [tex]R[/tex] is the radius of the air bubble, measured in centimeters.

If we know that [tex]V_{B} = 4.372\,cm^{3}[/tex], then the radius of the air bubble is:

[tex]4.372 = \frac{4\pi\cdot R^{3}}{3}[/tex]

[tex]R^{3} = 1.044[/tex]

[tex]R \approx 1.015\,cm[/tex]

The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.

Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

Answer:

840 m/s

Explanation:

Given that,

In the first second the rocket ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s².

We need to find the speed of the exhaust gas relative to the rocket.

The thrust of rocket is given by :

[tex]T=v_{gas}\dfrac{dm}{dt}\\\\ma=v_{gas}\dfrac{dm}{dt}\\\\v_{gas}=\dfrac{ma}{\dfrac{dm}{dt}}\\\\v_{gas}=\dfrac{14m}{\dfrac{1}{60}m}\\\\v_{gas}=840\ m/s[/tex]

So, the speed of the exhaust gas relative to the rocket is 840 m/s.

The speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s

Given data :

In first round Rocket ejects  1/60 of mass as exhaust gas

Acceleration of rocket ( a ) = 14.0 m/s²

Determine the speed of the exhaust gas relative to rocket

We will apply the equation for Rocket thrust

T = Vgas * [tex]\frac{dm}{dt}[/tex]

where : T = ma

∴ Vgas = ma / [tex]\frac{dm}{dt}[/tex]

            = 14 m / [tex]\frac{1}{60}[/tex] m

therefore V gas = 840 m/s

Hence we can conclude that the  speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s

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Answer:

Explanation:

To find the mass of the box we use the formula

[tex]m = \frac{w}{d} \\ [/tex]

w is the workdone

d is the distance

From the question we have

[tex]m = \frac{7000}{1.2} \\ = 5833.3333...[/tex]

We have the final answer as

Hope this helps you

Answer:

[tex]F=3.61\times 10^{-47}\ N[/tex]

Explanation:

Mass of a proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]

Mass of an electron, [tex]m_e=9.11\times 10^{-31}\ kg[/tex]

The distance between the electron and the proton is, [tex]r=5.3\times 10^{-11}\ m[/tex]

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

[tex]F=G\dfrac{m_em_p}{r^2}[/tex]

Where G is the universal Gravitational constant

[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N[/tex]

So, the force between the electron and proton is [tex]3.61\times 10^{-47}\ N[/tex].

Answer:

c. arrow's push against the bowstring.

Explanation:

Newtons third law of motion describe the relationship between action and reaction forces. Thus when the action force is applied by the bowstring against the arrow, a reaction to this force is supplied by the arrow's push against the bowstring.

If these two forces are not equal and opposite, the arrow would not move as expected. The elastic property of the bowstring propels the arrow at a rate which is due to the potential energy it has gained due to the stretching force.

We are given:

extension of the spring (x) = 10 cm OR 0.1 m

Force applied to keep it in position (F) = 500 N

Solving for the spring constant:

We know that F = kx (when the spring is extended)

replacing the variables with the given values

500 = k * 0.1

k = 5000 N/m

This means that to extend the spring by 1 m, we have to apply a force of 5000 N

Answer:

Explanation:

a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;

a = mv²/r

m is the mass of proton = 1.67×10^−27kg

v is the velocity = 3.00×10^8m/s

r is the radius

Since C = 2πr

7000m = 2πr

r = 7000/2π

r = 1114.08m

Substitute

a = 1.67×10^−27 (3.00×10^8)²/1114.08

a =  1.67×10^−27 * 9×10^16/1114.08

a = 15.03*10^-11/1114.08

a = 0.001346*10^-11

a = 1.346*10^-14m/s²

b) Force on the proton = mass * acceleration

Force = 1.67×10^−27kg * 1.346*10^-14

Force = 2.246*10^-41N

hence the force on the proton is 2.246*10^-41N

Given :

Pressure, P = 0.4 lb/in².

Depth, h = 42 in.

To Find :

The density of a liquid.

Solution :

Pressure at height h of a liquid of density [tex]\rho[/tex] is given by :

[tex]P = \rho g h[/tex]   ....1)

Here, g = 386.04 in/s²( acceleration due to gravity )

Putting all values in equation 1, we get :

[tex]P = \rho g h\\\\0.4 = \rho\times 386.04 \times 42\\\\\rho=\dfrac{0.4}{42\times 386.04}\ lb/in^3\\\\\rho=\dfrac{0.4}{42\times 386.04}\times 12^3\ lb/ft^3\\\\\rho=0.0426\ lb/ft^3[/tex]

Hence, this is the required solution.

Answer:

so what is your question that's not a question

An asteroid is a minor planet in the inner solar system. The asteroid orbits the sun and millions of asteroids exit as remains of planetesimals.

Hence there is no contact as no gravitational force.

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Answer:

42

Explanation:

p=mv

momentum=mass*velocity

x=2.8*15

x=42

Answer:

Option (c) is correct.

Explanation:

Acceleration of an object is given by the formula as follows :

[tex]a=\dfrac{v-u}{t}[/tex]

Where

u and v are initial and final velocity

t is time

(v-u) is also called the change in velocity

So, the acceleration of an object is equal to the rate of change of velocity. Hence, the correct option is (c) " Change in its velocity divided by the change in time".

First of all, this isn't physics. But, here's the answer:

1 gram is a 1000 milligram and 250 milligram is 250 * 0.001 g which is 0.25 gram.

Since 1 kg is 1000 g, we divide 1000 by 0.25 and we get 1000 * 4 which is 4000.

There are [tex]4000[/tex] tablets inside the bottle.

Answer:

C

Explanation:

The most commonly accepted theory of muscle growth states that microscopic tears within a muscle activate an inflammatory response process that causes dormant cells to build and regenerate muscle tissue; this is option c.

It has negative consequences on the muscle, and it is also known as "microtrauma," which occurs as a result of muscle contraction and is a normal part of the muscle repair and growth process. Here, the fibers in the muscle are damaged, and in response, the body initiates a repair process. This leads to muscle growth and increased strength and is helpful for muscle repair.

Hence, the most commonly accepted theory of muscle growth states that microscopic tears within a muscle activate an inflammatory response process that causes dormant cells to build and regenerate muscle tissue; this is option c.

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Answer:

because of glass' higher refractive index

Explanation:

air and vacuum have a lower refractive index and are optically lighter, so light travels faster.

whereas glass has a higher optical density, so light travels slower

Answer:

20 K

Explanation:

It is given that,

The change in temperature is 20 C.

We need to find the change in thermodynamic temperature.

If teperauture T₁ = 0° C = 0+273 = 273 K

T₂ = 20° C = 20 + 273 = 293 K

The change in temperature,

[tex]\Delta T=T_2-T_1\\\\=293-273\\\\=20\ K[/tex]

So, the change in temperature of 20°C is equivalent to 20 K.

The change in thermodynamic temperature is equal to 20 Kelvin.

Given the following data:

To determine the change in thermodynamic temperature:

A thermodynamic temperature can be defined as an absolute measure of the average total internal energy possessed by a body or an object. Thus, thermodynamic temperature is typically measured in Kelvin (K).

Mathematically, the change in temperature of an object is given by the formula:

[tex]\theta = T_f - T_i[/tex]

Where:

Next, we would convert the values in degree Celsius to Kelvin:

Conversion:

0°C = 273 K

20°C = 273 + 20 = 293 K

For the change in thermodynamic temperature:

[tex]\theta = 293 - 273\\\\\theta =20\;K[/tex]

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The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

The specific heat capacity should be determined in joules per kilogram degree-celsius ( J k g − 1 ∘ C − 1 ).

It is to be supplied for the substance with respect to mass and it increased the temperature.

Hence, The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

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The unit used to measure energy when calculating specific heat capacity is the joule (J).

The joule (J), a unit used to quantify energy, is used to calculate specific heat capacity. In the International System of Units (SI), the joule serves as the default unit of energy.

It is described as the quantity of energy that is delivered when one newton of force is exerted across a one-meter distance.

Thus, the quantity of heat energy needed to increase the temperature of a particular substance by a specific amount is measured as specific heat capacity. J/kg°C, or joules per kilogramme per degree Celsius, is the unit of measurement.

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Answer:

True.

Explanation:

Exercises and yoga are important for the individual's health. The exercise makes the individual, fit, healthy and also reduces the chances of the deadly disease in human beings.

Different types of exercise can be performed by the individual. The aerobic exercises increase the amount of oxygen in the body. With the aerobic exercise, the exercise that can increase the strength, flexibility and total fitness must also be included.