A roller coaster accelerates from 0 m/s to 25 m/s in 5 seconds. Determine the acceleration of the roller coaster.

The height at which the diver must jump in other to hit the water with a speed of 24 m/s is 29.4 m.

The height of the player above the ground can be calculated using the formula below.

Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

⇒ Solve for s.

Hence, The height at which the diver must jump in other to hit the water with a speed of 24 m/s is 29.4 m.

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Answer:

melanie ran 8.3 meters per second

Explanation:

I think thats right, hope it helps.

Answer:

a

The point on the x-axis where the magnetic field will be zero is  \( d =  2.8 \ cm \)

b

The point on the x-axis where the magnetic field will be zero is \(Z =  5.333 \  cm \)Explanation:

From the question we are told that

   The first  current is \(I_1 = 5.5 \ A\)

    The point of x- axis  intersection is  \(x =  -2.0 \  cm\)

    The second current is  \(I_2  =  2.5  \  A\)

     The point of intersection of the x-axis is  \(x =  2 .0  \  cm\)

Generally given that the current of the two wires are same direction it means that the magnetic field in -between the wires will cancel out  giving zero

 So  

       \(B_1 -  B_2 =  0\)

=>     \(B_1  =  B_2 \)

=>      \(\frac{\mu_o *  I_1 }{2 *  \pi *  d }  = \frac{\mu_o *  I_2 }{2 *  \pi *  (4 -d) }\)

Here  d is the of one wire to the point where the magnetic field is 0

and  given that the total distance in-between the wire is  D  =  2 =  2 =  4

Hence the distance of the other wire to the point where magnetic field is zero is  (4 - d)

So

     \(\frac{5.5}{ d}  =  \frac{2.5}{4-d}\)

=>   \( 8d  =  22 \)

=>   \( d =  2.8 \ cm \)

So the point on the x-axis where the magnetic field will be zero is  \( d =  2.8  \)

 Generally given that the current of the two wires are opposition direction it means that the magnetic field at a position which is not in-between the wire will be zero  

Let that position be  k

Let  the distance from the middle of both wires to k  be Z  

So

      \(\frac{\mu_o *  I_1 }{2 *  \pi *  ( 2 +  Z) }  = \frac{\mu_o *  I_2 }{2 *  \pi *  (Z -2) }\)

=>  \(5.5 Z  - 11 =  5+ 2.5 Z\)

=>    \(3Z  =  16\)

=>    \(Z =  5.333\  cm\)

The speed with which a rat runs a maze is an example of a(n) dependent variable.

A dependent variable is the variable that is being measured or tested in an experiment. In this case, the speed of the rat running the maze is the dependent variable because it is the outcome that is being measured.

The independent variable, on the other hand, is the variable that is being manipulated or changed in the experiment. For example, the type of maze or the presence of food at the end of the maze could be independent variables that could potentially affect the speed of the rat running the maze. By measuring the dependent variable (speed of the rat) in response to changes in the independent variable(s), researchers can determine if there is a relationship between the two.

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The geothermal gradient in the tectonically stable region is approximately 21°C/km, indicating that the temperature increases by an average of 21 degrees Celsius per kilometer of depth.

To calculate the geothermal gradient, we need to find the rate at which the temperature increases with depth.

Temperature at the surface (T₁) = 14°C

Temperature at a depth of 5.0 km (T₂) = 119°C

Temperature difference = T₂ - T₁ = 119°C - 14°C = 105°C

Depth difference = 5.0 km - 0 km = 5.0 km

Geothermal gradient = Temperature difference / Depth difference

Geothermal gradient = 105°C / 5.0 km

Calculating this expression, we find:

Geothermal gradient ≈ 21°C/km

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The critical angle for light traveling from the liquid back into the air is 42.0°.

The critical angle is the angle of incidence in the first medium (air) at which all of the light is refracted into the second medium (liquid).

The formula for the critical angle is:

sin(c) = n₂/n₁

where:

c is the critical angle

n₂ is the index of refraction of the second medium

n₁ is the index of refraction of the first medium

In this case:

n₁ = 1.00

n₂ = 1.33

Substituting these values into the formula:

sin(c) = 1.33/1.00

c = sin⁻¹(1.33/1.00)

c = 42.0°

Therefore, the critical angle for light traveling from the liquid back into the air is 42.0°.

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Answer:

well no, but yes

Explanation:

depends on your religion tbh

Answer:

There is a 50-50 chance that we do live in a simulation.

Explanation:

When a sound source produces waves, the frequency of the waves determines the pitch of the sound. In this scenario, the sound source is producing 1.20-khz waves. However, the sound source is also moving towards a stationary listener at one-half the speed of sound. This means that the waves will be compressed as they travel towards the listener, resulting in a higher frequency.

To calculate the frequency that the listener will hear, we can use the formula:

f' = f (v +/- v_observer) / (v +/- v_source)

Where:
f = frequency of the sound source (1.20 kHz in this case)
v = speed of sound in air (approximately 343 m/s)
v_observer = speed of the observer (0 m/s since they are stationary)
v_source = speed of the source (one-half the speed of sound, or approximately 171.5 m/s)

Plugging in the values, we get:

f' = 1.20 kHz * (343 m/s + 0 m/s) / (343 m/s - 171.5 m/s)
f' = 2.14 kHz

Therefore, the listener will hear a frequency of 2.14 kHz instead of 1.20 kHz. This is because the sound waves are compressed as they travel towards the listener due to the motion of the sound source.

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Answer: c. The electric force increases

Explanation:

If the distance between two charged particles decreases, the electric force between them increases.

According to Coulomb's Law, the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, the equation can be represented as:

F = k * (q1 * q2) / r^2

Where:

F represents the electric force between the particles.

k is the electrostatic constant.

q1 and q2 are the charges of the particles.

r is the distance between the particles.

As the distance (r) between the particles decreases, the denominator of the equation (r^2) becomes smaller, causing the overall electric force (F) to increase. Conversely, if the distance between the charged particles increases, the electric force between them decreases. This inverse relationship between the distance and electric force is a fundamental characteristic of the electrostatic interaction between charged objects.

The distance and velocity of one of the galaxies, the universe "started its trip about 17.23 billion years.

A galaxy redshift = 0.228

velocity = redshift × speed of light

= 0.228 × 3 × 10⁸m/s

velocity = 6.84 × 10⁷ m/s

Distance = 1050 × 3.2 × 10⁶ light years

= 1050 × 3.2 × 10⁶× 9.46 × 10¹⁵ m

= 3.17856 × 10²⁵ m

So,

Age = D/v

= 5.436 × 10¹⁷ sec

1 sec = 3.17 × 10¹⁷ sec

Age = 5.436 × 10¹⁷ × 3.17 × 10¹⁷ sec

= 17.23 billion years

Thus, according to the relationship between distance and velocity, the universe started its trip about 17.23 billion years.

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Answer:

The speed with which a rock would have to be thrown to put it in 234 Ida's orbit, near its surface is approximately 12.917 m/s

Explanation:

The given parameters are;

The mass of Ida, M = 4 × 10¹⁶ kg

The radius of 234 Ida, r = 16 km = 16,000 m

The speed, v, required to put a rock in 234 Ida's orbit near its surface is given by the orbital velocity equation as follows;

\(v = \sqrt{{\dfrac{G \times M}{r} } }\)

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Substituting the known values gives;

\(v = \sqrt{{\dfrac{6.67408 \times 10^{-11} \times 4 \times 10^{16}}{16,000} } } \approx 12.917\)

Therefore, the speed required to put a rock in 234 Ida's orbit near its surface = v ≈ 12.917 m.

The magnitude of momentum for each neutron in the laboratory frame is 5.03 x 10^-19 kg m/s, expressed as a multiple of m0c to three significant figures.

The magnitude of momentum for each neutron in the laboratory frame can be calculated using the equation p = mv, where p is momentum, m is mass, and v is velocity.  Neutrons have a mass of approximately 1.675 x 10^-27 kg, expressed as m0, the rest mass of the neutron. Speed of light, c, is approximately 3 x 10^8 m/s.
We can express the magnitude of momentum for each neutron as a multiple of m0c to three significant figures. This can be done by multiplying the mass of the neutron by the speed of light, which gives us a value of approximately 5.025 x 10^-19 kg m/s. We can then round this value to three significant figures, giving us an answer of 5.03 x 10^-19 kg m/s.
Therefore, the magnitude of momentum for each neutron in the laboratory frame is 5.03 x 10^-19 kg m/s

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Answer: Friction that acts on a sliding object

Explanation:

Answer:

An antinode should be motionless by definition.

Vav = 0 m / n sec = 0 m/s

Answer:

A pure element.

Explanation:

Calcium, being whatever weight, is a pure element.

Hope this helped! Have a good day.

Answer:

the answer is

Explanation: Calcium is a chemical element with symbol Ca and atomic number 20. Classified as an alkaline earth metal, Calcium is a solid at room temperature.

Answer:

First tell what are the directions and what we've to do?? add subtract or multiply??

So, their net force on the object is 50 N in a forward direction from the first person.

Hi ! Here, I will help you with the net forces (results of forces) acting on a two-dimensional area and in opposite directions. Steps that can be taken are as follows :

The equation for calculating the net force from this two-dimensional straight line is as follows:

\( \boxed{\sf{\bold{\sum F = F_1 + F_2 + ... + F_n}}} \)

With the following condition :

We know that :

In my mind, I determined that the force will go to forward direction from the first person. So :

What was asked :

Step by step :

\( \sf{\sum F = F_1 + F_2} \)

\( \sf{\sum F = 200 + (-150)} \)

\( \sf{\sum F = 200 - 150} \)

\( \boxed{\sf{\sum F = 50 \: N} \)

The movement of the object is forward (from the first person) because the net force value that I calculated is not opposite (with negative sign) to the right direction. So, their net force on the object is 50 N to the forward direction from first person.

The cruiser must accelerate at a rate of 1.68 m/s²to catch the speeding car before the state line, 1.2 km away.

To determine the rate at which the cruiser must accelerate to catch the speeding car, we need to consider the relative positions and velocities of both vehicles. The speeding car is initially 100 m past the cruiser and has a constant velocity of 33.4 m/s. The cruiser starts from rest and needs to cover a distance of 1.2 km to catch the car before the state line.

We can use the equation of motion s = ut + (1/2)at², where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration. Since the car is moving at a constant velocity, its displacement is given by s_car = u_car * t_car. The cruiser needs to cover a distance of 1.2 km (1200 m) in order to catch the car. The displacement of the cruiser is given by s_cruiser = u_cruiser * t_cruiser + (1/2) * a_cruiser * t_cruiser².

We can set up a system of equations using the given information and solve for the acceleration of the cruiser. By equating the displacements of the car and the cruiser and solving for the time, we can substitute this time into the equation for the displacement of the cruiser. Finally, rearranging the equation for the displacement of the cruiser, we can solve for the acceleration.

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a. the final image formed by the second lens will be located approximately 34.4 cm (13.51 cm + 2.631 cm) from the second lens. b. the total magnification is approximately -1.21, following the sign conventions.

Part A) The final image formed by the second lens will be located 34.4 cm from the second lens.

To determine the image distance from the second lens, we can use the lens formula:

1/f = 1/v - 1/u

where f is the focal length, v is the image distance, and u is the object distance.

Given that the focal length of each lens is 22 cm and the object is placed 35.0 cm in front of the first lens, we can calculate the image distance formed by the first lens using the lens formula.

Using the lens formula for the first lens:

1/f1 = 1/v1 - 1/u1

Substituting the values:

1/22 = 1/v1 - 1/35

Simplifying the equation:

1/v1 = 1/22 + 1/35

1/v1 = (35 + 22) / (22 * 35)

1/v1 = 57 / 770

v1 = 770 / 57 ≈ 13.51 cm

Now, the image formed by the first lens acts as an object for the second lens. The distance between the two lenses is given as 16.5 cm. Therefore, the object distance for the second lens will be:

u2 = 16.5 cm - v1

u2 = 16.5 cm - 13.51 cm

u2 ≈ 2.99 cm

Applying the lens formula for the second lens:

1/f2 = 1/v2 - 1/u2

Substituting the focal length of the second lens (22 cm) and the object distance (u2):

1/22 = 1/v2 - 1/2.99

Simplifying the equation:

1/v2 = 1/22 + 1/2.99

1/v2 = (2.99 + 22) / (22 * 2.99)

1/v2 = 24.99 / 65.78

v2 = 65.78 / 24.99 ≈ 2.631 cm

Therefore, the final image formed by the second lens will be located approximately 34.4 cm (13.51 cm + 2.631 cm) from the second lens.

Part B) The total magnification is approximately -1.21.

To calculate the total magnification, we can multiply the individual magnifications of the two lenses. The magnification of a lens can be determined using the formula:

Magnification = -v/u

where v is the image distance and u is the object distance.

For the first lens, the object distance (u1) is 35.0 cm and the image distance (v1) is 13.51 cm. Therefore, the magnification of the first lens is:

Magnification1 = -13.51 cm / 35.0 cm ≈ -0.386

For the second lens, the object distance (u2) is 2.99 cm and the image distance (v2) is 2.631 cm. Therefore, the magnification of the second lens is:

Magnification2 = -2.631 cm / 2.99 cm ≈ -0.879

To calculate the total magnification, we multiply the individual magnifications:

Total Magnification = Magnification1 * Magnification2

Total Magnification ≈ -0.386 * -0.879 ≈ 0.339

Therefore, the total magnification is approximately -1.21, following the sign conventions.

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Explanation:

(Sorry but we can prosses your answer)

Answer:

20 m/s

Explanation:

Use the position/displacement formula to find out how long the ball is in the air:

(distance) = (initial velocity)*(time) + (1/2)*(acceleration)*(time)^2

20 = 0*t + 1/2*10*t^2

t^2 = 4

t = 2 seconds

Use the velocity formula:

(final velocity) = (initial velocity) + (acceleration)*(time)

Vf = 0 + 10*2

Vf = 20 m/s

Efficiency of a ramp is a ratio of output to input energy. It measures the amount of input energy that is converted to useful output energy.

When someone uses 10 J of work to push a box up a ramp, and lifting the box would only require 2 J of work, the efficiency of the ramp can be determined as follows:Efficiency of ramp = useful work output / total work inputTotal work input is the amount of work used to push the box up the ramp, which is 10 J.Useful work output is the amount of work required to lift the box, which is 2 J.Therefore, the efficiency of the ramp is:Efficiency of ramp = 2 J / 10 J = 0.2 or 20%Therefore, the efficiency of the ramp is 20%.

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In the experiment, the researcher investigates the relationship between effort distance and the applied effort on lifting a load, with effort distance as the independent variable, the applied effort as the dependent variable, and load weight and surface angle as controlled variables.

i. The variables involved in this experiment are:

Effort distance (E.d.): The distance over which the effort force is applied.

Load distance (L.d.): The distance over which the load is lifted.

Effort applied: The force exerted by the researcher to lift the load.

ii. The variables that need to be controlled in this experiment are:

Load weight: The weight of the load should be kept constant to ensure that only the effort distance is being tested.

Surface angle: The slanted surface angle should be consistent for each trial to isolate the effect of effort distance.

iii. The independent, dependent, and controlled variables can be identified as follows:

Independent variable: Effort distance (E.d.) is the independent variable as it is intentionally varied by the researcher to observe its effect on the effort applied.

Dependent variable: The dependent variable is the effort applied. It is measured or observed and expected to change based on the different effort distances.

Controlled variables: The load weight and surface angle are controlled variables. They are kept constant throughout the experiment to eliminate their influence on the results and isolate the effect of effort distance.

Therefore, Using effort distance as the independent variable, applied effort as the dependent variable, load weight and surface angle as the controlled variables, the researcher conducts an experiment to examine the link between these two factors when lifting a load.

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Answer:

As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease. ... Thus, entropy measurement is a way of distinguishing the past from the future.

Explanation:

Answer:

Ve respuesta abajo.

Explanation:

Para hacer esto, podemos asumir que la presión es constante, pues es un proceso adiabatico, por tanto se aplica la ley de Charles a presión constante:

V₁/T₁ = V₂/T₂    (1)

De ahí podemos despejar V₂, ya que conocemos las condiciones iniciales de temperatura y volúmen, y la temperatura final:

V₂ = V₁ T₂ / T₁   (2)

Las temperaturas deben estar en grados Kelvin, y solo es cuestión de sumarle 273 al valor de la temperatura dada en °C:

T₁ = 28 + 273 = 301 K

T₂ = -6 + 273 = 267 K

El volúmen podemos pasarlo a litros ahora o al final. En este caso, podemos dejarlo en m³ como está y luego pasarlo a Litros. Resolviendo tenemos:

V₂ = 0.4 * 267 / 301

Pasando este volumen a Litros, sabiendo que 1 m³ son 1000 L:

V₂ = 0.35482 * 1000

Finalmente para saber cuanto disminuyó el volumen en Litros, pasemos el volumen inicial a Litros y luego se hace la resta con el volumen final:

V₁ = 0.4 * 1000 = 400 L

V₁ - V₂ = 400 - 354.82

Esto es lo que disminuyó. Espero te ayude.

The time needed to take off is 0.8s

Acceleration can be defined as the rate of change of velocity. It is a vector quantity.

Given that a plane accelerates from rest at a constant rate of 5.0 m/s² along a runway that is 1800 mm long. Assume that the plane reaches the required takeoff velocity at the end of the runway.

The takeoff velocity V will be

V² = U² + 2aS

Where

Substitute all the parameters into the formula above

V² = 0 + 2 × 5 × 1.8

V² = 18

V = √18

V = 4.24 m/s

The time t needed to take off will be

V = U + at

4.24 = 0 + 5t

5t = 4.24

t = 4.24 / 5

t = 0.8 s

Therefore, the time needed to take off is 0.8s

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Answer:

D. 3 Ohms

Explanation: