Calculate the force that must be applied to an object weighing 25 pounds for it to be in equilibrium, if it is on a plane inclined at 45 °

Answer: In a way, yes and no. Chlorophyll's job in a plant is to absorb light. Light that's collected goes into a chemical equation to make glucose, the plant's food.

Hope it helps!

As the problem tells us, the field at point P is 0, thus, the field exerted by charges q1 and q2 have the same magnitude, and exactly the opposite direction, as it can be seen on the following drawing:

Thus, we know that charge q2 will have to be a source of field (as opposed to a sink), and thus, a positive charge. Now all we have to do is find out what charge could produce a field with the same magnitude of the one from q1. As the electric field can be written as:

We'll have:

Thus, our answer is q2=1.43775*10^(-8)C

Note: Your lesson requests the answer to be inserted as: +1.44

Answer:

s=0.204m

Explanation:

Assuming that the arrow is thrown horizontally and there is no air resistance, we can use the following formula to calculate the launch angle 0:

tan(0) = opposite/adjacent = height/distance

where opposite is the height that the arrow reaches and adjacent is the distance to the balloon.

Rearranging the formula, we get:

0 = arctan(height/distance)

0 = arctan(height/3)

Taking the tangent of both sides, we get:

tan(0) = tan(arctan(height/3))

tan(0) = height/3

Now, we need to find the height of the arrow. Using the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (0m/s, at maximum height), u is the initial velocity (2m/s), a is acceleration (-9.8m/s^2, due to gravity) and s is the distance travelled vertically until the arrow reaches maximum height.

At maximum height, the final velocity is 0m/s. Therefore, we have:

0 = (2m/s)^2 + 2(-9.8m/s^2)s

Solving for s, we get:

s = 0.204m

Therefore, the height of the arrow is approximately 0.204m.

As the charge is negative, the force, which has a magnitude of 33.05 N, is directed to the left, against the electric field.

E is a symbol for the magnitude of the electric field at a specific distance from a point charge. At twice the distance from the point charge, what is the electric field's strength? The field's strength is E/2 at twice the distance. The field's strength is still equal to E at a distance that is twice as great.

E = k*q/r²

r1 = 2.00 cm

r2 = 1.00 cm + 3.00 cm = 4.00 cm

r3 = 1.00 cm

Using these distances, we can calculate the electric field due to each charge:

E1 = kq1/r1² = (9.0 x 10⁹ Nm²/C²) * (1.50 x 10⁻⁶ C) / (0.02 m)² = 168.75 N/C (to the right)

E2 = kq2/r2² = (9.0 x 10⁹ Nm²/C²) * (-2.00 x 10^⁻⁶ C) / (0.04 m)² = -112.50 N/C (to the left)

E3 = kq3/r3² = (9.0 x 10⁹ Nm²/C²) * (6.00 x 10⁻⁶ C) / (0.01 m)² = 5.40 x 10⁶ N/C (to the right)

E = E1 + E2 + E3 = 168.75 N/C - 112.50 N/C + 5.40 x 10⁶ N/C = 5.39 x 10⁶ N/C (to the right)

F = q*E

F = (-6.13 x 10 C) * (5.39 x 10⁶ N/C) = -33.05 N

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Answer:34.6 m/s

Explanation: It is asking how long meaning the answer is in time

Explanation

Step 1

find the work done

All you need to do is take the force, F , and multiply it by the distance the object moved. The result will be the work performed on the object

hence,

Let

replace

Step 2

Work done has the same units as energy – joules. This is because energy is the ability to do work. So you must have energy to do work. ... Work done is equal to energy transferred.

so

the energy added to the cart is

I hope this helps you

Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.

the answer I need points for my math test

Calculate the horizontal component of the velocity. The horizontal component of the velocity is given by:

v_x = v * cos(theta)

where v is the original speed of the arrow and theta is the angle of projection.In this case, v = 2 m/s and theta is unknown. Solving for theta, we get:

theta = arccos(v_x / v)

theta = arccos(2 / 2) = 45 degrees

Calculate the vertical component of the velocity. The vertical component of the velocity is given by:

v_y = v * sin(theta)

In this case, v = 2 m/s and theta = 45 degrees. Solving for v_y, we get:

v_y = 2 * sin(45 degrees) = 1.414 m/s

Calculate the time of flight. The time of flight is given by:

t = 2 * v_y / g

In this case, v_y = 1.414 m/s and g = 10 m/s^2. Solving for t, we get:

t = 2 * 1.414 / 10 = 0.283 seconds

Calculate the height of the arrow. The height of the arrow is given by:

y = v_y * t - 0.5 * g * t^2

In this case, v_y = 1.414 m/s, t = 0.283 seconds, and g = 10 m/s^2. Solving for y, we get:

y = 1.414 * 0.283 - 0.5 * 10 * 0.283^2 = 0.303 meters

Therefore, the angle of projection is 45 degrees and the height of the arrow is 0.303 meters.

The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

Answer:

I think the answer is option 3

All the equations of motion are as follows, Displacement (s) equation, Final velocity (v) equation, Average velocity (v_avg) equation, Displacement (s) equation with average velocity, and Displacement (s) equation.

In terms of its motion as a function of time, equations of motion define how a physical system behaves. In more detail, the equations of motion define how a physical system behaves as a collection of mathematical functions expressed in terms of dynamic variables.

In conclusion, equations of motion define how a physical system behaves in terms of how its motion changes over time.

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The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.

The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.

In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.

By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.

By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.

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Given:

The voltage applied,

The resistance of the resistor, R=24.0 Ω

To find:

The reading in the ammeter.

Explanation:

An ac ammeter connected in series with the resistor reads the root mean square value of the current, i.e., I_rms.

Comparing the given equation of the voltage to the standard equation,

We get, V₀=80 V

The root mean square value of the voltage is given by the equation,

On substituting the known values,

Thus the value of I_rms is calculated by the equation,

On substituting the known values,

Final answer:

Thus the reading of the ac ammeter connected in series with the resistor is 2.4 A.

Answer:

isotopes

Hope this helps

Answer:

The speed of sound in air varies with temperature. At a temperature of -18°C, the speed of sound in air is approximately 306.3 m/s.

To find how far the sound wave has moved in 10 seconds, we can use the formula:

distance = speed x time

distance = 306.3 m/s x 10 s = 3063 m

Therefore, the sound wave has moved 3063 meters in 10 seconds.

Explanation:

Answer: In this scenario, we have two carts colliding with each other. One cart weighs 250 kg and is moving at a speed of 8 m/s, while the other cart weighs 100 kg and is initially at rest.

After the collision, the 250 kg cart continues moving forward, but at a slower speed of 3 m/s. We want to find out the speed at which the 100 kg cart moves after the collision.

To solve this, we use the principle that the total "push" or momentum before the collision should be the same as the total momentum after the collision.

Since the 100 kg cart is initially at rest, its momentum is zero. The momentum of the 250 kg cart before the collision is 250 kg * 8 m/s = 2000 kg·m/s.

After the collision, the momentum of the 250 kg cart becomes 250 kg * 3 m/s = 750 kg·m/s.

To find the momentum of the 100 kg cart after the collision, we subtract the momentum of the 250 kg cart after the collision from the total momentum before the collision: 2000 kg·m/s - 750 kg·m/s = 1250 kg·m/s.

Now, we divide this momentum by the mass of the 100 kg cart to find its velocity: 1250 kg·m/s / 100 kg = 12.5 m/s.

Therefore, the 100 kg cart moves at a velocity of 12.5 m/s after the collision, in the opposite direction of the 250 kg cart's motion.

Answer: Brainliest?

Explanation:

Isobars and isotherms are both types of contour lines used to represent data on weather maps, specifically for atmospheric pressure and temperature, respectively.

The similarities between isobars and isotherms are:

Both are contour lines that connect points of equal value on a map.

Both are used to depict weather patterns and conditions.

Both help to identify areas of high and low values.

The differences between isobars and isotherms are:

Isobars connect points of equal atmospheric pressure, whereas isotherms connect points of equal temperature.

Isobars are measured in units of pressure such as millibars, while isotherms are measured in units of temperature such as degrees Celsius or Fahrenheit.

Isobars are typically used to show pressure patterns associated with wind, while isotherms are used to show temperature patterns.

Isobars are often used to forecast weather conditions, including the movement and intensity of storm systems. Isotherms are used to identify areas of warm and cold air masses, which can affect local weather patterns.

In summary, both isobars and isotherms are useful tools for understanding weather patterns, but they represent different types of data and are used for different purposes.

Isobars and isotherms are both concepts used in meteorology and climatology to represent important variables that help to describe atmospheric conditions. While they share some similarities, they also have several key differences.

Isobars refer to lines of equal pressure, meaning they connect points on a map or graph where the atmospheric pressure is the same. Isobars are drawn on weather maps to indicate areas of high and low pressure, and to show the general movement of air masses. When isobars are closely spaced, it indicates a steep pressure gradient, which can result in strong winds.

On the other hand, isotherms refer to lines of equal temperature, meaning they connect points on a map or graph where the temperature is the same. Isotherms are often drawn on weather maps to show the boundaries between warmer and cooler air masses, and to indicate areas where temperature changes rapidly.

One similarity between isobars and isotherms is that they are both used to describe atmospheric conditions in terms of spatial variation. They are also both used to infer information about the movement of air masses and the development of weather patterns.

However, there are also some key differences between isobars and isotherms. The most obvious difference is that isobars represent pressure while isotherms represent temperature. Additionally, while isobars are generally oriented parallel to each other and indicate the direction of winds, isotherms are typically oriented perpendicular to isobars and indicate the location of temperature gradients. Finally, while isobars are more commonly used to describe weather conditions associated with areas of high and low pressure, isotherms are often used to identify the location of fronts and other weather boundaries.

In summary, isobars and isotherms are similar in that they both describe atmospheric conditions in terms of spatial variation, and can be used to infer information about the movement of air masses and the development of weather patterns. However, isobars represent pressure and are oriented parallel to each other, while isotherms represent temperature and are oriented perpendicular to isobars.

Answer:

What is Bohr’s Model of an Atom?

The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

Answer:

D) It does not change

Explanation:

Since there is no friction in the inclined plane. Therefore, there is no loss in the total mechanical energy of the system. So according to the law of conservation of energy we can write:

Total Mechanical Energy at Start = Total Mechanical Energy at End + Frictional Loss

Total Mechanical Energy at Start = Total Mechanical Energy at End + 0

Total Mechanical Energy at Start = Total Mechanical Energy at End

It means there is no change in the total mechanical energy of the system.

Therefore, the correct option is:

D) It does not change

Answer:

b)      v_y = 4.57 m / s

a)      vₓ = 4.43 m / s

Explanation:

This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity

          v = v₀ + a t

          v = 0 + a t

          v = 186  0.0342

          v = 6.36 m / s

let's use trigonometry to decompose this velocity

           sin 45.9 = v_y / v

           cos 45.9 = vₓ / v

           v_y = v sin 45.9

           vₓ = v cos 45.9

           v_y = 6.36 sin 45.9

           vₓ = 6.36 cos 45.9

           v_y = 4.57 m / s

           vₓ = 4.43 m / s

ₓ

The 10/90 principle can be a powerful tool for taking control of your situation and improving your life. By taking responsibility for what you can change and focusing on your reaction to the situation, you can make positive changes in your life and become the master of your own destiny.

The 10/90 principle refers to the idea that life is made up of 10% of what happens to you and 90% of how you respond to it. In other words, you may not be able to control what happens to you, but you can control your reaction to it. By taking responsibility for what you can change rather than being a victim of what you cannot change, you can take control of your situation and improve your life.One example of a situation where the 10/90 principle could be applied is losing a job. Losing a job can be a devastating experience, and it can be easy to feel like a victim in this situation. However, by applying the 10/90 principle, you can take control of your situation and make positive changes in your life.The first step in applying the 10/90 principle in this situation would be to take responsibility for what you can change. This could mean updating your resume, networking with others in your field, and applying for new jobs. By taking action and doing what you can to find a new job, you are taking control of your situation and improving your chances of finding a new job.
The second step would be to focus on your reaction to the situation. Instead of dwelling on the negative aspects of losing your job, try to focus on the positive aspects. This could mean using the extra time to pursue a new hobby or spend more time with family and friends. By focusing on the positive aspects of the situation, you are taking control of your reaction and improving your overall well-being.
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Answer:

La energía potencial elástica es la energía asociada con los materiales elásticos. Por ejemplo, un resorte al ser comprimido o elongado almacena energía potencial elástica y, al ser soltado, puede realizar trabajo sobre un objeto.

Para mantener el resorte comprimido o alargado una cierta longitud x, a partir de su largo natural, es necesario que, en este caso, la mano aplique una fuerza F_{M} sobre el resorte; esta fuerza es directamente proporcional a x.

Explanation:

ón conocida como ley de Hooke.

Para encontrar una expresión que describa la energía potencial asociada con la fuerza del resorte, se determina el trabajo que se requiere para comprimir el resorte desde su posición de equilibrio hasta cierta posición final arbitraria x. Debido a que la fuerza varía desde O hasta kx, se utiliza la fuerza promedio \frac{(F_{0}+F_{X})}{2}.

 \[ \bar{F}=\frac{0+K X}{2}=\frac{1}{2}kx \]

fuerza-sobre-un-resorte

Fuerza sobre un resorte. La fuerza para estirar un resorte aumenta linealmente con su elongación .

El trabajo realizado por la fuerza aplicada será: W=\bar{Fx}=\frac{1}{2}kx^{2}

El trabajo realizado se almacena en el resorte comprimido en forma de energía potencial elástica como:

 \[ \boxed{ Ep_{elas}=\frac{1}{2}kx^{2}} \]

Una vez que se ha comprimido o estirado el resorte respecto a su posición de equilibrio, la energía potencial elástica se puede considerar como la energía almacenada en el resorte deformado. Esta energía siempre es positiva en un objeto deformado al depender de x^{2}.

Por ejemplo, en la figura se observa que un resorte realiza trabajo sobre un bloque. El resorte que se encuentra sin deformar (a) cuando es empujado por un bloque de masa m, se comprime una distancia x (b). Cuando el bloque se suelta (c), partiendo del reposo, la energía potencial plástica almacenada en el sistema se transforma en energía cinética del bloque.

energia-potencial

Work done on the book is 10.595 Joules .

Work done on the book would be equal to the Potential Energy

Work Done = Potential Energy

W = m g h

where ,  ' m ' represents mass of the book

' g ' stands for  Acceleration due to gravity

' h ' represents the height to which the book is raised

So, according to the given question ,

m = 5.1 kg

g = 9.8 m / s ²

h = 21.2 cm

  =  0.212 m

W = m g h

    = 5.1 kg x 9.8 m / s ² x 0.212 m

    = 10 . 595 Joules

Hence, Work done on the book is 10.595 Joules .

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Answer:

See below

Explanation:

Average speed =  (0+ 10)/2 = 5 m/s

 then to cover 50 m     will take    50 m / 5 m/s  = 10 seconds

     change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2