calculate sound intensity

Answer and Explanation:

TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.

[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]

The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).

Numbers:

Sun’s mass: 2 x 10^30 kg

Neptune’s mass: 1 x 10^26 kg

Distance of Sun to Uranus: 3 x 10^9 km

Closest approach of Uranus and Neptune: 1.5 x 10^9 km

Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.

The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.

The formula for calculating the Force of Gravity between two masses is:

F = G*m₁*m₂/r²

Where;

F = force of gravity

G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²

m₁ = mass of the larger object

m₂ = mass of the smaller object

r = the distance between the centers of the two masses

Now, from online values, we have the following;

mass of Neptune; m₁ =  102.413 × 10²⁴ kg

mass of Uranus; m₂ = 86.813 × 10²⁴ kg

average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m

Thus, force exerted by Neptune on Uranus is;

F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²

F = 2.240 × 10¹⁷ N

We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.

That is greater than the force Neptune exerts on Uranus.

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Answer:

C. Some social patterns are helpful, while others are harmful.

Explanation:

Hope this was helpful, Have an amazing,spooky Halloween!!

Answer:

100 N

Explanation:

The following data were obtained from the question:

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 50 cm

Force (F) =.?

Next, we shall convert 50 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

50 cm = 50 cm × 1 m / 100 cm

50 cm = 0.5 m

Therefore, 50 cm is equivalent to 0.5 m.

Finally, we shall determine the magnitude of the net force on the ball by using the following formula:

F = mv²/r

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 0.5 m

Force (F) =.?

F = mv²/r

F = 2 × 5²/ 0.5

F = 2 × 25/ 0.5

F = 50 / 0.5

F = 100 N

Therefore, the magnitude of the net force on the ball is 100 N.

The magnitude of the net force on the ball will be "100 N".

According to the question,

Mass, m = 2 kg

Velocity, v = 5 m/s

Radius, r = 50 cm or,

               = 50 × [tex]\frac{1}{100}[/tex]

               = 0.5 m

We know the relation,

Force, F = [tex]\frac{mv^2}{r}[/tex]

By substituting the values, we get

              = [tex]\frac{2\times (25)^2}{0.5}[/tex]

              = [tex]\frac{2\times 25}{0.5}[/tex]

              = [tex]\frac{50}{0.5}[/tex]

              = 100 N

Thus the above response is appropriate.

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Answer:

b. The bicyclist is exerting 49 N of force

Explanation:

Given;

mass of bicyclist start, m = 50 kg

acceleration, a = 1 m/s²

density of air, ρ = 1.2 kg/m³

velocity, v = 2 m/s

Area of the bicyclist body, A = 0.5 m²

The drag force on the bicyclist is given by ;

Fd = 0.5CρAv²

where;

C is drag coefficient = 0.9 for bicycle

Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²

Fd = 1.1 N

The force of the bicyclist is given by;

F = ma

F = 50 x 1

F = 50 N

The effective force exerted by the bicyclist is given by;

Fe = F - Fd = 50 N - 1.1 N

Fe = 49 N

Therefore, the force exerted by the bicyclist is 49 N

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

Answer:

(a)  24.56 N

(b) 142.28 N

Explanation:

(a)

The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.

The centripetal function is generated from either scenario by Equation:

⇒  [tex]Fc = \frac{mv^2}{r}[/tex]

On putting the values, we get

⇒       [tex]=\frac{12\times 1.33^2}{0.864}[/tex]

⇒       [tex]=24.56 \ N[/tex]

(b)

Use T to denote whatever arm stress we can get at the bottom including its circle:

⇒  [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]

⇒  [tex]T = mg + Fc[/tex]

⇒      [tex]=12\times 9.81+24.56[/tex]

⇒      [tex]=142.28 \ N[/tex]

Answer:

Hyperbole

Explanation:

this is an extreme exaggeration or overstatement/ magnification

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G[tex]\frac{m1m2}{r}[/tex]

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

Answer:

luster

Explanation:

Answer:

μk = 0.58

Explanation:

       [tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]

       ⇒  [tex]F_{n} = - F_{apph} = -53.3 N[/tex]

       [tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]

       [tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]

      [tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]

        μk = 0.58

Answer:

The observed wavelength is  [tex] \lambda = 700nm[/tex] (color -  Red)

Explanation:

From the question we are told that

   The wavelength of the emitter is  [tex]\lambda_ e  = 500 nm  =  500 *10^{-9} \  m[/tex]

  The redshift is  R  =  0.4  

     Generally red shift is mathematically represented as

    [tex]R =  \frac{ \lambda -  \lambda_e }{\lambda_e}[/tex]

=>  [tex]0.4  =  \frac{ \lambda -   500 *10^{-9} }{500 *10^{-9} }[/tex]

=>  [tex] \lambda - 500*10^{-9} = 200*10^{-9}  [/tex]

=>   [tex] \lambda = 700 *10^{-9}[/tex]

=>   [tex] \lambda = 700nm[/tex]

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

Answer:

Answer and steps in the pic

Answer:

The angle is 3.95 rad.

Explanation:

The angle can be calculated as follows:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]

Where:

[tex]\omega_{f}[/tex]: is the final angular speed

ω₀: is the initial angular speed = 0 (it starts from rest)

α: is the angular acceleration

θ: is the angle=?

The centripetal acceleration is:

[tex]a_{c} = \omega_{f}^{2}*r[/tex]

And the tangential acceleration is:

[tex] a_{T} = \alpha*r [/tex]

Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:

[tex]a_{c} = 7.9a_{T}[/tex]

[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]

Now, the angle is:

[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]

[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]

Therefore, the angle is 3.95 rad.

I hope it helps you!          

The angular distance traveled by the electric drill is 3.95 radians.

The given parameters;

The angular distance traveled by the electric drill is calculated as follows;

[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]

The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;

[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]

Substitute the value of angular acceleration into the first equation;

[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]

Thus, the angular distance traveled by the electric drill is 3.95 radians.

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Answer:

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Explanation:

From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:

[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)

[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]m[/tex] - Mass of the motocyclist, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.

[tex]R[/tex] - Radius of the circular turn, measured in meters.

The static coefficient of friction is cleared in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]

From (Eq. 2) we get that normai force is:

[tex]N = m\cdot g[/tex]

And we expand the resulting expression in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]

[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)

If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:

[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

Answer: The correct answer is A) The track was not level and was tilted slightly downward.

Explanation: This is because of the two values: 0.4 kg and 0.5 kg. I won't go into much detail but due to this difference of mass, we know that the track was not level.

"The track was not level and was tilted slightly downward" could explain the difference in the mass.

Thus Option A is appropriate.

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Answer:

12.5 m

Explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m

Answer:

A mixture is a combination of two or more substances in any proportion. This is different from a compound, which consists of substances in fixed proportions. ... The lemonade pictured above is a mixture because it doesn't have fixed proportions of ingredients.

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

   The number  of alphabet is  n= 4  (a , b , c , d )

Generally the total  number of  string of length 10 over the 4 alphabets is  

     [tex]N  =  4^{10}[/tex]

Gnerally the number of string of length 10 that does not include b is  

     [tex]T =  3^{10}[/tex]    

Generally the number of strings of length 10 over the 4  alphabets that have at least one alphabet b  somewhere in the string is  

        [tex]G  =  N - T[/tex]

=>    [tex]G  =  4^{10} -  3^{10}[/tex]

Answer:

6000 joules

Explanation:

I jus learned dis

Answer:6000j

Explanation:

Hope that helps

Answer:

In two significant figure 360K

Explanation:

The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.

Hence,

ΔT = 212 - 32

ΔT = 180°F

To convert to °F to kelvin, we use the formula below

= (°F - 32) × 5/9 + 273.15

= (180°F - 32) × 5/9 + 273.15

= 355.37K ⇔ 360K

Answer:

a- More heat is required for the constant-pressure process than for the constant-volume

Explanation:

we have to solve using the thermodynamic first law. this is the heat applied to the system

dQ = dU + dW

definition of terms:

dU = change in internal energy

dW = work done

we have it that

change in internal energy dU is directly proportional to work done dW

but when we are in constant volume process, work done of the gas is zero

therefore

dQ of constant pressure is > than that of constant volume

so constant pressure process requires more heat

The process that requires more heat is the constant-pressure process than the constant-volume process.

According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.

In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.

Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.

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Answer:

4 bobux

Explanation:

one bobux

two bobux

three bobux

four bobux

Answer:

True.

Explanation:

        [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)

        [tex]\Delta x} = v_{avg}* {\Delta t}[/tex]

Complete question:

use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr

Answer:

The distance to the quasar is 1.02 x 10¹⁰ Lyr

Explanation:

Given;

speed of light, v = 300, 000 km/sec

Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr

percentage of the quasar recession = 75% of speed of light

Hubble's Law is given by;

[tex]v =H_od\\\\d = \frac{v}{H_o}\\\\d= \frac{(0.75*300,000)}{2.2*10^{-5}}.Lyr\\\\d = 1.02*10^{10} \ Lyr[/tex]

Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr

Answer:

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

[tex]x=x_{2}-x_{1}[/tex]

Put the value into the formula

[tex]x=4-(0.75+0.75)[/tex]

[tex]x=2.5\ m[/tex]

We need to calculate the displacement of the canoe

Using conservation of momentum

[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]

[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]

[tex]86.5\times2.5=(61+86.5)\times x'[/tex]

[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]

[tex]x'=1.46\ m[/tex]

Hence, The displacement of the canoe is 1.46 m