An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in which frictional forces are NOT considered to be negligible. A graph of the net force exerted on the object as a function of the object’s distance traveled is shown. How could a student use the graph to determine the net work done on the object?

Answer:speed difference

Explanation:

Answer:they live in grass land also called forests hope the answer helps

Explanation:

Answer: 2

Explanation: Because if you add 1  to 8 you get 9 then subtract 9 by 7 then and your answer is

Answer:

The inductive time  constant is just T = L / R

T = 10E-3 henry/ 10 ohm = 10E-2 sec = .1 sec  

(.01 / 10 = .1)

Answer:

a 5 kilogram mass, at a height of 3 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.

Explanation:

Answer: The Rasalas's solar mass is 1.50 times that of our star, the Sun. The Sun's Mass is 1,989,100,000,000,000,000,000 billion kg.

Explanation:

Answer:

The Rasalas's solar mass is 1.50 times that of our star, the Sun. The Sun's Mass is 1,989,100,000,000,000,000,000 billion kg. which to calculate using this website is too large. To give idea of size, the Sun is 99.86% the mass of the solar system.

Explanation:

The amount of work done by the car engine if it is 100% efficient is 366.67 Joules.

Given the following data:

To determine the amount of work done by the car engine if it is 100% efficient:

Mathematically, the work done by an object with respect to power and time is given by the formula:

[tex]Work\;done = \frac{Power}{time}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Work\;done = \frac{22000}{60}[/tex]

Work done = 366.67 Joules.

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St. John’s

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The amount of current flowing in the wire is 1.69 × 10⁻¹⁶ A

From the question,

We are to calculate, the amount of current flowing in the wire.

From the formula,

[tex]I =\frac{nQ}{t}[/tex]

Where I is the current

n is the number of charge

Q is the quantity of charge

t is the time

From the given information

n = 8

Q = e = 1.69 × 10⁻¹⁹ C

t = 8 msecs = 8 × 10⁻³ sec

Putting the parameters into the formula, we get

[tex]I = \frac{8 \times 1.69 \times 10^{-19} }{8\times 10^{-3} }[/tex]

[tex]I = 1.69 \times 10^{-16} \ A[/tex]

Hence, the amount of current flowing in the wire is 1.69 × 10⁻¹⁶ A

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Answer:

I think it would be 0.1496 but I'm not really sure sorry

[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(40)(1)^2[/tex]

[tex]\\ \sf\Rrightarrow K.E=20J[/tex]

[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(80)(1)^2[/tex]

[tex]\\ \sf\Rrightarrow K.E=40J[/tex]

Answer:

try 2.4136 as an answer/ good luck

Answer:

Explanation:

The speed of the wave given only it's frequency and wavelength can be found by using the formula

[tex]c = f \times \lambda[/tex]

where

c is the velocity of the wave in m/s

[tex] \lambda[/tex] is the wavelength in m

f is the frequency in Hz

From the question

c = 2700 × 6 = 16,200

We have the final answer as

Hope this helps you

Answer:

"electric or magnetic"

F = q E    force on charge q due to electric field E

F = q v x B       force on charge q due to magnetic field B

Answer:

Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.

Explanation:

I got this from quizlet :)

Answer: To see objects surrounding more clearly

Explanation:

Explanation:

c. if the vector is oriented at 0° from the X -axis.

The main reason why wheelchair ramps are used to move people up a staircase is because it has a high mechanical advantage which leads to a greater efficiency in moving people up the staircase.

The mechanical advantage of the ramp is calculated as follows;

[tex]M.A = \frac{L}{h}[/tex]

As the length of the ramp increases, people will be move up to a greater distance up the staircase.

Thus, the main reason why wheelchair ramps are used to move people up a staircase is because it has a high mechanical advantage which leads to a greater efficiency in moving people up the staircase.

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1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ([tex]v[/tex]), en metros por segundo:

[tex]v = \sqrt{\frac{K}{\rho} }[/tex] (1)

Donde:

Si sabemos que [tex]\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}[/tex] y [tex]K = 2,16\times 10^{9}\,\frac{N}{m^{2}}[/tex], entonces la velocidad de las ondas sonoras es:

[tex]v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }[/tex]

[tex]v\approx 1469,694\,\frac{m}{s}[/tex]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ([tex]\lambda[/tex]), en metros, mediante la siguiente fórmula:

[tex]\lambda = \frac{v}{f}[/tex] (2)

Donde [tex]f[/tex] es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que [tex]v\approx 1469,694\,\frac{m}{s}[/tex] y [tex]f = 1000\,hz[/tex], entonces la longitud de onda de las ondas sonoras es:

[tex]\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}[/tex]

[tex]\lambda = 1,470\,m[/tex]

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: https://brainly.com/question/1070238

The final velocity of an object that starts from rest and travels for 5 seconds at an acceleration of 4.3 m/s² is 21.5m/s.

EQUATION OF MOTION:

The final velocity of a moving object can be calculated by using one of the equations of motion as follows:

V = u + at

Where;

According to this question,

v = 0 + 4.3(5)

v = 21.5m/s.

Therefore, the final velocity of an object that starts from rest and travels for 5 seconds at an acceleration of 4.3 m/s² is 21.5m/s.

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Answer:

Explanation:

The sun is much closer than the other stars

The distance traveled by the egg after 2 seconds is 19.6 m.

The given parameters:

The distance traveled by the egg after 2 seconds is calculated as follows;

[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\[/tex]

where;

[tex]h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m[/tex]

Thus, the distance traveled by the egg after 2 seconds is 19.6 m.

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Answer:

  4/3 m/s

Explanation:

Assuming momentum is conserved, the sum of products of mass and speed before the collision is the same as after:

  (2000 kg)(4 m/s) +(4000 kg)(0 m/s) = (2000 +4000 kg)(Vt)

  Vt = (8000 kg·m/s)/(6000 kg) = 4/3 m/s

The speed of the combined objects after the collision is 4/3 m/s.

Answer:

This is the answer.

Explanation:

is this right?

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

By Newton's second law, the net force acting on the block in the vertical direction is

∑ F [ver] = n - mg = 0

where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.

When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is

f = µ[s] n = µ[s] mg

The net horizontal force would be

∑ F [hor] = X - µ[s] mg = 0

so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.

Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is

∑ F [hor] = X - µ[k] mg = 0

so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.

The principles here are captured in the attached plot.

Answer: