what belongs in the center section
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Answer:

The mass of the other ball is 397.775 kg.

Explanation:

Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.

The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.

The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.

In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:

[tex]F=G\frac{M1*M2}{d^{2} }[/tex]

where:

In this case:

Replacing:

[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]

Solving for M2:

[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]

M2= 397.775 kg

The mass of the other ball is 397.775 kg.

Answer:

The correct option is  c

Explanation:

From the question we are told that

    The ratio of the noise to human hearing threshold is  [tex]I = 150 000[/tex]

Generally from the equation given we have that

       dB =  10 log I

So

      dB =  10 log 150000  

The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct

Given the equation for calculating the intensity of sound which is measured in decibel expressed as:

dB=10 log I

where;

I is the ratio of the sound to the human hearing threshold

Given that noise is 150,000 times greater than the human hearing threshold.

Substitute I = 150,000 into the expression above;

dB =  10 log 150,000

Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000

Learn more here: https://brainly.com/question/14468501

Answer: Areobic Resporation

Explanation; -Aerobic respiration is the process in living things that evolved as oxygen levels increased in earth's atmosphere.

Answer:

5.096*10^-8

Explanation:

Given that

The average value of the electromagnetic wave is 310 mW/m²

To find the maximum value of the magnetic field the wave is closest to, we say

Emax = √Erms

Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]

Emax = √233.7648

Emax = 15.289

Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer

15.289 / (3*10^8) = 5.096*10^-8 T

Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8

Answer:

127.53 J

Explanation:

Given that

Height of the floor, h = 1.3 m

Acceleration due to gravity, g = 9.81 m/s²

Mass of the box, m = 10 kg

To find the work done on the box by the janitor, we use the formula for potential energy. Remember that potential energy is actually the energy stored in an object as a result of its position.

P.E = mgh, substituting each parameters for its value, we then have

P.E = 10 * 9.81 * 1.3

P.E = 127.53 J

Therefore, we can conclude that the amount of work done on the box by the janitor is 127.53 J

Answer:

The law zero of thermodynamics.

Explanation:

The law zero of thermodynamics, which tells us that heat flows from a body at a higher temperature to another body with lower temperature, when the heat transfer is zero, it is said that the two bodies are in thermal equilibrium, their temperatures are equal

Answer:

d₂ = 6.18 mm

Explanation:

We can work on this interesting exercise using the concepts of energy conservation

alpha (1) particle, with have a charge q = 2e

Starting point

            Em₀ = U = q V₁

final point

            [tex]Em_{f}[/tex] = ½ m v²

             Em₀ = Em_{f}

            2eV₁ = ½ m₁ v²

           v² =[tex]\frac{4 \ e \ V_{1}}{m_{1} }[/tex]

proton (2) particle, with have a charge q = e

Starting point

          Em₀ = qV₂

Final point

          Em_{f} = ½ m₂ v²

          Em₀ = Em_{f}

          eV₂ = ½ m₂ v²

          v² = [tex]\frac{2 \ e \ V_{2} }{m_{2} }[/tex]

in the exercise, it should be noted that the two particles have the same velocity when reaching the plate, therefore let us solve the velocity in each equation and equal

        \frac{4 \ e \ V_{1}}{m_{1} }= \frac{2 \ e \ V_{2} }{m_{2} }

         [tex]V_{2} = \frac{2 m_{2} }{m_{1} } \ V_{1}[/tex]

the alpha particle is composed of two protons and two neutrons, therefore in first approximation

          m₁ = 4 m₂

subtitute

         V₂ =[tex]\frac{1}{2}[/tex] V₁

let's calculate

         V₂ = [tex]\frac{1}{2}[/tex] 600

         V₂ = 300 V

To find the distance we use the relationship between the electric field and the potential difference

         V = -d E

The electric field between the plates is constant, so

          E = - V / d = - V₂ / d₂

          d₂ = [tex]\frac{V_{2} }{V} \ d[/tex]

let's calculate

          d₂ = [tex]\frac{300}{680} \ 15[/tex]  

          d₂ = 6.18 mm

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

Answer:

Explanation: about 2 Mph when it falls and sorry if I’m wrong

Explanation:

Force, F = 60 N

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find horizontal and vertical component of the force.

Horizontal component,

[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]

Vertical component,

[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]

So, the horizontal and vertical component are 51.96 N and 30 N respectively.

Answer: the amount of mass is oscillating is 34.8 kg

Explanation:

Given that;

amplitude A = 20.0 cm

time t = 10 s

amplitude decreases x = 15.0 cm

damping coefficient b = 2.00 N.s/m

amount of mass is oscillating = ?

we know that; amplitude can be expressed as;

x = Ae^-(∝t)

we substitute

15 = 20e^-∝(10)

∝ = 0.02877 s⁻¹

Hence mass m will be;

m = b/2∝

we substitute

m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)

m = 34.8 kg

Therefore the amount of mass is oscillating is 34.8 kg

Answer:

A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.

Explanation:

Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).

When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:

?F(c) = ma(c) = n - mg

n = mg + ma(c)

= m[g + a(c)]

Since centripetal acceleration equals v² / r, the equalization enhances:

n = m[g + (v² / r)]

Answer:

(A)

Explanation:

Answer:

275 x 10"N

Explanation:

Answer:

1.25s

1) 90.9m

Explanation:

From the question, we are given the following;

Speed v = 30m/s

Maximum height H = 45m

Required;

Time

Using the equation of motion S = ut + 1/2gt²

Substitute;

45= 30t + 1/2(9.8)t²

45 = 30t + 4.9t²

4.9t² + 30t - 45 = 0

Factorize;

t = -30 ±√30²-4(-45)(4.9)/2(4.9)

t = -30 ±√900+882/9.8

t = -30 ±42.21/9.8

t = -30 + 42.21/9.8

t = 12.21/9.8

t = 1.25secs

Hence the ball spent 1.25secs in air.

i) To get the horizontal distance, we will use the formula;

R = U√2H/g

R = 30√2(45)/9.8

R = 30√90/9.8

R = 30√9.18

R = 30(3.03)

R = 90.9m

Hence the horizontal distance is 90.9m

Answer:

  (b)  Friction increases as the normal force (weight) increases, and also depends on the surface

Explanation:

The coefficient of friction relates the normal force between surfaces to the friction force opposing motion. The coefficient depends on the surface.

The appropriate choice is ...

  Friction increases as the normal force (weight) increases, and also depends on the surface

Answer:

yes

Explanation:

i hope this helps not sure im right

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]

[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]

The second distance, r₂, can be determined from sound intensity formula given as;

[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]

Therefore, the second distance of the sound from the source is 431.78 m.

Answer:gamma>beta>alpha

Explanation: Just took the test

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

Answer:C

Explanation:

Answer:

B

Explanation:

Because the earth shifts axes points so it would make it cllser hotter etc

Answer: the answer is B. 2.0

Explanation:

Answer:

Earth-crossing asteroid, asteroid whose path around the Sun crosses Earth's orbit. Two groups of such asteroids—Aten and Apollo asteroids—are distinguished by the size of their orbits and how closely they approach the Sun.

Explanation:

Answer:

0 because j×j =0

Explanation:

so a×b is 0

Answer:

38.40

Explanation:

This is the answer for this question

Answer:

B_total = 1694 N ,  T = - 937.7 N

Explanation:

This is an exercise of the Archimedes principle that establishes that the thrust of a liquid is equal to the weight of the dislodged volume

          B = ρ g V_body

For this case we will assume that the weight of the body is in equilibrium with the thrust of the liquids and the tension of the rope

          B₁ + B₂ - W + T = 0

suppose liquid 1 is water and liquid 2 is oil

          ρ₁ g V₁ + ρ₂ g V₂ = W

for body weight let's use the definition of density

         ρ = m / V

         m = ρ V

we substitute

         T = ρ g V - (ρ₁ g V₁ + ρ₂ g V₂)

In the problem we are told that the volume in the water is 75% and the volume in the oil is 25% of the body's volume.

         T = ρ g V - g (ρ₁ 0.75 V + ρ₂ 0.25 V)

let's calculate the volumes

Body

         V = 4/3 π r³

liquid 1 (water)

         V₁ = 0.75 V = ¾ V

liquid 2 (oil)

         V₂ = 0.25 V = ¼ V

we substitute

        T = ρ g [tex]\frac{4}{3}[/tex] π r³ - g (ρ₁ [tex]\frac{3}{4}[/tex]  [tex]\frac{4}{3}[/tex] π r³ + ρ₂  [tex]\frac{1}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³)

        T = [tex]\frac{4}{3}[/tex] ρ g r³ - g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)

The term of the floating force is

        B_total = g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)

let's calculate its value

        B_total = 9.8 π 0.35³ (1000 + ⅓ 850)

        B_total = 1694 N

therefore the tension in the string is

         T = [tex]\frac{4}{3}[/tex] 1350 9.8 0.35³ - 1694

         T = 756.3 - 1694

         T = - 937.7 N

The negative sign indicates that the system tends to come out of the liquid, for which a downward force must be applied to keep it in position.

Answer:

5× 3 over 10 = 1.5

Explanation:

F = 1m/s × kg over n

The net work done on the block in 1 minutes is most nearly 1800 Joule.

A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.

Work has no direction and only magnitude. Work is a scalar quantity as a result.

Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.

As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.

So, The net work done on the block in 1 minutes is =

force× displacement

= force × velocity × time

= 10 × 3 × 60 Joule

= 1800 Joule.

Learn more about work here:

https://brainly.com/question/18094932

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Answer:

Explanation:

The the mass of the car can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{400}{0.85} \\ = 470.58823...[/tex]

We have the final answer as

Hope this helps you