Which of the organisms in the food web above is the top level carnivore

Answer:

(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.

Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.

(b). 27001.2 seconds(s)..

(c). 141 metre(m).

Explanation:

So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;

=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.

=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."

=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"

Therefore, the solution is given below;

(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.

Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.

Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.

(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.

Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;

[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.

Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.

(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.

The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m

Answer:

Explanation:

If the ray were not deviated, it would travel straight through the mirror. Due to the mirror, the incident ray is reflected at 30°. The ray travels 30° + 30° = 60°. The angle of deviation is 180° - 60° = 120°.

Answer:

5

Explanation:

Bextra in bf x vi d sj by

Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.

Answer:

His weight would be zero on the scale i.e he is weightless at that instance.

Explanation:

weight = mg

where m is the mass of the object, and g is the acceleration of gravity.

⇒ 810 = mg

During free fall, the weight of an object can be determined by:

W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)

where a is the acceleration of the object.

But since James fall at the acceleration of gravity, then:

g = a

mg = ma = 810 N

So that;

W = 810 - 810

    = 0 N

Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex]   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

[tex]\beta[/tex] = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex]   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]

therefore : V = 1.624* 10^-5 m/s

Answer:

The change in internal energy of the system is 2,054 J

Explanation:

The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.

Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:

ΔU= Q + W

where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.

By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.

In this case:

Replacing:

ΔU= 0.523 kJ + 1.531 kJ

Solving:

ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)

The change in internal energy of the system is 2,054 J

Answer:

I don't do physics but I think the answer would be -3674

Explanation:

Newton's 2 law of motion

Answer:

300000 Joules

Explanation:

Recall that the unit Watts is Joules per second, derived from the quotient of energy per unite of time. Therefore, to calculate the energy used on the given time, we need to multiply the power used (1000 W) times the time used. Then we need to express the 5 minutes in seconds to get our answer in Joules:

5 minutes = 5 * 60 seconds = 300 seconds

Final energy calculation gives:

E = 1000 W * 300 sec= 300000 Joules

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

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

Power lost in the first wire;

[tex]P_1 = \frac{V_1^2}{R}[/tex]

[tex]R = \frac{V_1^2}{P_1}[/tex]

Power lost in the second wire

[tex]P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}[/tex]

Keeping the resistance constant, we will have the following equation;

[tex]\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\[/tex]

[tex]P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1[/tex]

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

Answer:

(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the bowling ball is 113.272 joules.

Explanation:

The statement is incomplete. The complete question is:

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.

(a) What is the kinetic energy of the ball just before it hits the mattress?  

(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?  

(c) How much work does the gravitational force do on the ball while it is compressing the mattress?

(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）

(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:

[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.

Now we expand the expression by definition of gravitational potential energy:

[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]

[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)

Where:

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

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

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:

[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]

[tex]K_{2} = 102.974\,J[/tex]

The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]

[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]

[tex]\Delta W = 102.974\,J[/tex]

The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]

Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.

[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]

Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]

[tex]\Delta W = 10.298\,J[/tex]

Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:

[tex]\Delta W' = K_{2}+\Delta W[/tex]

([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])

[tex]\Delta W' = 113.272\,J[/tex]

The work done by the mattress on the bowling ball is 113.272 joules.

Answer:

Covalent Bond

Explanation:

i took the test , mark me brainliest.

Answer: Electrical

Explanation: Atoms are tied together by electrical bonding forces.

v² - u² = 2 a ∆x

where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.

So

v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)

v² = 4645 m²/s²

v ≈ 68.15 m/s

Explanation:

the link enjoy

Answer:

Explanation:

Have y’all seen steeleflag19 at all on here?

Answer:

The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

Explanation:

The value of g can be found using the following equation:

[tex] F = \frac{GmM}{r^{2}} [/tex]

[tex] ma = \frac{GmM}{r^{2}} [/tex]

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

Where:

a is the acceleration of gravity = g

G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)

M: is the Earth's mass = 5.97x10²⁴ kg

r: is the Earth's radius = 6371 km      

Since we need to find g at 6700 m, the total distance is:

[tex] r_{T} = 6371000 m + 6700 m = 6377700 m [/tex]

Now, the value of g is:

[tex] a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2} [/tex]

Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

I hope it helps you!

Answer:

D: gas and plasma

Explanation:

Answer:

 Em₀ = U = m g h ,  Em_{f} = K = ½ m v²

Explanation:

When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by

          Em₀ = U = m g h

As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy

            Em = K + U = ½ m v² + mg y

        y <h

when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy

          Em_{f} = K = ½ m v²

This energy transformation is in the case that the friction force is zero.

If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor

         [tex]W_{nc}[/tex] = - fr L

therefore the numeraire values ​​of the velocity are lower, due to the energy lost by friction.

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

Answer: It will be about 44.1m

Explanation:

Answer:

A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B)  T = 4.7 10⁴ K, C) n₂ = 42

Explanation:

A)  For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.

Let's reduce the units to the SI system

        E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J

The kinetic energy of the electron is

        K = ½ m v²

         E₀ = K

         v = √ 2E₀ / m

         v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)

         v = √ (2.14857 10¹²)

         v = 1.47 10⁶ m / s

now the speed of a calcium ion is asked, let's find sum

        m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg

        v = √ (2E₀ / M)

         v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)

        v = √ (0.2994457 10⁸)

        v = 0.5426 10⁴ m / s

B) the terminal energy of an ideal gas is

             E = 3/2 kT

              T = ⅔ E / k

              T = ⅔ (9,776 10-19 / 1,381 10-23)

              T = 4.7 10⁴ K

C) To calculate the energy of these lines we use the Planck expression

              E = h f

where wavelength and frequency are related

              c =λ f

              f = c /λ

let's substitute

              E = h c /λ

let's look for the energies

λ = 396.8 nm

  E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹

           E₁ = 5.0126 10⁻¹⁹ J

λ = 393.3 nm

           E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹

           E₂ = 5.0572 10⁻¹⁹ J

The difference in energy between these two states is

          ΔE = E₂ -E₁

          ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J

          ΔE = 0.0446 10⁻¹⁹ J

let's reduce eV

         ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         ΔE = 2.787 10⁻² eV

Now let's use Bohr's atomic model for atoms with one electron,

               E = -13.606 Z² / n²

where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium

               n = √ (13.606 Z² / E)

λ = 396.8 nm

E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV

               n₁ = √ (13.606 20² / 3.132875)

               n₁ = 41.7

since n must be an integer we take

               n₁ = 42

λ = 393.3 nm

E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV

              n₂ = √ (13.606 20² / 3.16075)

              n₂ = 41.5

Again we take n as an integer

               n₂ = 42

We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin

Answer:

Explanation:

the distance have the following relation:

d = (1/2)gt2

D=32.0 m

t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s

it take 2.56s from the glasses to hit the ground

when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s

x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m

the pen only travel 1.54m

so the pen is above the ground 32.0m - 1.54m = 30.46m

The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.

Height is a numerical representation of the distance between two objects or locations on the vertical axis.

The height can refer to a physical length or an estimate based on other factors in physics or common use. |

The given data in the problem is;

h is the height from the top of a stadium = 32.0 m

t is the time period when the pen is dropped later =  2.00 s

x is the height above the ground

a is the air resistance. a = -g = -9.81 m/s²

From the second equation of motion;

[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]

When the glasses fall to the ground, the pen only travels a short distance;

[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]

So the pen travel the distance;

[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]

The pen above the ground is found as;

[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]

Hence the pen is 30.46m above the ground. when the glasses hit the ground.

To learn more about the height refer to the link;

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Answer:

46-D

47-C

48-F

49-A

50-B

I am not very sure I am right about those answers though.

Answer: D 160kg object moving 126 m/s

Explanation:

An object having a mass of 162 kg and moving with a velocity of 124 m/sec will require the greatest amount of force to change its motion. The correct option is D.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

If the object has to stop, the final velocity must be zero. If the time is constant, the amount of force only depends on the mass and the velocity at which the body is moving.

The amount of force on the object depends on the momentum of the body.

The momentum of the body is;

P = mv

Object D will require the greatest amount of force to change its motion. Because the momentum of the body for option D is the greatest.

Hence, the correct option is D.

Learn more about the Force, here;

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x = 1/2 at²

where x = length of runway, a = acceleration, and t = time.

600 m = 1/2 (12 m/s²) t²

t² = (1200 m) / (12 m/s²)

t² = 100 s²

t = 10 s

Answer:

Explanation:

The work done by an object can be found by using the formula

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

Hope this helps you

The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.

Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.

Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.

The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.

Learn more about Density here:

https://brainly.com/question/29775886

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