ANSWER SO U DONT HAVE TO( I GOT A 100 TRUST ME)

Would you personally get better results and be more motivated by participating in physical activities that are competitive or recreational? Explain your choice and choose at least five physical activities that you would be interested in participating in that fall under the category you have chosen.

ANSWER:

i would be more motivated by participating in physical activities that are recreational. Competitive activities may be a big responsibility because you have to aim for a win. If you will be exercising regularly, it will make you happy without any aim of winning in a contest. Just like playing sports such as badminton, basketball, volleyball, baseball, and soccer which need a team. These recreational activities will be advisable because you are with others who love the sport also but not treating it as a competition.

Answers

Answer 1

Yes i would personally get better results and be more motivated by participating in physical activities that are competitive or recreational.

My motivation would increase if I engaged in leisure physical activity. Because you must aspire for victory, competitive hobbies may come with a lot of responsibilities. Regular exercise will make you happy regardless of whether you're doing it to compete or just for the sake of it. similar to team sports like baseball, soccer, basketball, badminton, and volleyball. Because you'll be with people who share your passion for the sport but aren't competing with you, these leisure activities are a good idea.

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Related Questions

How would life on earth be different if its axis was not tilted with respect to its orbit?.

Answers

Answer:

oone side at a time would be hot and the other oone

Explanation:

One side will be hot and the other cold no in-between

If a 1,000-kg car is traveling at 20.0 m/s when it reaches the bottom of a circular hill of radius 8.00 m, what normal force does the road exert on the car at the bottom of the hill

Answers

Answer:

i don't know this one.

Explanation:

A particle moves in a circle with radius 10 cm and with a uniform speed 1. 3 m/s. What is the centripetal acceleration of this particle?.

Answers

Centripetal acceleration is = 10π^2    cm/s.

Centripetal acceleration is = r×(W)^2.

The radius of the circular path is 0.1 m or 10 cm.

Centripetal acceleration is = r×(W)square.

Area=0.1×(π)square                 cm/s square.

Area=10π^2                              cm/s square.

Acceleration of an object in a circular orbit. Since velocity is a vector quantity that is it has both magnitudes velocity and direction as an object moves along a circular path, its direction and velocity constantly change, resulting in acceleration. The centripetal force causes centripetal acceleration.

In the special case of the earth orbiting the sun or satellites orbiting celestial bodies the centripetal force causing motion is the result of the gravitational pull between them. Centripetal acceleration is defined as the property of motion of a body through a circular orbit. An object that moves in a circle and has an acceleration vector pointing to the center of the circle is called centripetal acceleration.

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Two planes take off at the same time from an airport. The first plane is flying at 272 miles per hour on a course of 155. 0°. The second plane is flying in the direction 175. 0° at 336 miles per hour. Assuming there are no wind currents blowing, how far apart are they after 2 hours

Answers

The planes are 246.06 miles apart after 2 hours.

According to the question, Two planes take off at the same time from an airport. The first plane is flying at 272 miles per hour on a course of 155.0° and the second plane is flying at 336 miles per hour on a course of 175.0°.

The distance covered by the first plane in 2 hours (a) = 272 * 2

                                          a = 544 miles

The distance covered by the second plane in 2 hours (b) = 336 * 2

                                           b = 672 miles

The angle of the first plane from the ground (x) = 180 - 155

                                           x = 25°

The angle of the second plane from the ground (y) = 180 - 175

                                            y =

So, The angle between the two planes (C) ⇒ x - y = 25 - 5

                                            C = 20°

Using the law of cosines, we can calculate the distance between the two planes.

               Law of cosines ⇒  \(c^{2} = a^{2} + b^{2} - 2ab.cosC\)    → 1

Substitute a = 544, b = 672 and C = 20° in 1, then we get

                           \(c^{2} = 544^{2} + 672^{2} - 2(544)(672).cos(20)\)

                               =  295936 + 451584 - 2(365568).(0.9396)

                               =  747520 - (731136).(0.9396)

                               =  747520 - 686975.3856

                           \(c^{2}\)  =  60544.6144

                           c = \(\sqrt{60544.6144}\)

                           c = 246.058 ≅ 246.06

Therefore, the distance between the two planes after two hours is 246.06 miles.

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What are the factors that change the pattern observed on a screen during Young’s double-slit experiment?

Answers

The factors that can change the pattern observed on a screen during Young's double-slit experiment are given below:1. Width of the slit. 2. Distance between slits. 3. Distance between slits and screen. 4. Wavelength of the incident light. 5. Refractive index of the medium.

The factors that can change the pattern observed on a screen during Young's double-slit experiment are given below:

1. Width of the slit. The width of the slit can influence the diffraction pattern that is observed on a screen. When the width of the slit decreases, the central maximum of the diffraction pattern becomes broader, and the intensity of the secondary maxima reduces.

2. Distance between slits. The distance between the slits in the double-slit experiment also affects the pattern on the screen. The distance between the slits is equal to the spacing between the maxima. If the spacing between the slits decreases, the distance between the maxima decreases, and vice versa.

3. Distance between slits and screen. The distance between the slits and the screen is also a factor that can affect the diffraction pattern. When the distance increases, the spacing between the maxima becomes wider, and the intensity of the maxima decreases.

4. Wavelength of the incident light. The wavelength of the incident light is another factor that affects the diffraction pattern on the screen. When the wavelength increases, the spacing between the maxima increases, and vice versa.

5. Refractive index of the medium. The refractive index of the medium in which the light travels can also influence the diffraction pattern observed on a screen.

When the refractive index of the medium changes, the position of the maxima changes as well. These are the factors that can change the pattern observed on a screen during Young's double-slit experiment.

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Two small, identical conducting spheres A and B are a distance R apart. Each carries the same charge Q.
(a) What is the force sphere B exerts on sphere A?
(b) An identical sphere with zero charge, sphere C, makes contact with sphere B and is then moved very far away.
What is the net force now acting on sphere A?
(c) Sphere C is brought back and now makes contact with sphere A and is then moved far away.
What is the force on sphere A in this third case?

Answers

(a) The force between the two spheres can be calculated using Coulomb's Law:

F = k * Q^2 / R^2

where k is Coulomb's constant, Q is the charge on each sphere, and R is the distance between the centers of the spheres.

The force on sphere A due to sphere B is the same as the force on sphere B due to sphere A, and it is given by:

F_AB = F_BA = k * Q^2 / R^2

(b) When sphere C is brought into contact with sphere B, the charge on sphere B will distribute evenly between sphere B and C, since they are in contact. The total charge on sphere B plus C remains equal to Q, since sphere C started with zero charge.

When sphere C is moved far away, the net force on sphere A due to sphere B and C will be the same as the force due to sphere B alone, since the charge on sphere C has been moved far away and has no effect on the force between the spheres. So the net force on sphere A is:

F_AB = F_BA = k * Q^2 / R^2

(c) When sphere C is brought into contact with sphere A, the charge on sphere A will distribute evenly between sphere A and C, since they are in contact. The total charge on sphere A plus C remains equal to Q, since sphere C started with zero charge.

When sphere C is moved far away, the net force on sphere A due to sphere A and C will be zero, since they now have equal and opposite charges and are separated by a large distance. Therefore, the net force on sphere A is the force due to sphere B alone:

F_AB = F_BA = k * Q^2 / R^2

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A two-slit experiment with blue light produces a set of bright fringes. Part A Will the spacing between the fringes increase, decrease, or stay the same if the separation of the slits is decreased? • Increase
• decrease • stay the same
Part B Will the spacing between the fringes increase, decrease, or stay the same if the experiment is immersed in water? • increase • decrease • stay the same

Answers

Part A: If the separation of the slits in a two-slit experiment is decreased, the spacing between the fringes will increase.

Part B: If the two-slit experiment is immersed in water, the spacing between the fringes will decrease.

Part A: If the separation of the slits in a two-slit experiment is decreased, the spacing between the fringes will increase.

The spacing between the fringes in a two-slit experiment is determined by the wavelength of the light used and the separation of the slits. According to the formula for fringe spacing, given by

dλ = mλ / D

Where d is the fringe spacing, λ is the wavelength of light, m is the order of the fringe, and D is the distance from the slits to the screen.

If the separation of the slits is decreased, the value of d in the equation increases. Since the wavelength of light remains constant, an increase in d leads to an increase in the spacing between the fringes.

Therefore, the spacing between the fringes will increase if the separation of the slits is decreased.

Part B: If the two-slit experiment is immersed in water, the spacing between the fringes will decrease.

The refractive index of water is greater than that of air. When light passes from air to water, its speed decreases. Since the speed of light in a medium affects the wavelength, the wavelength of light in water is shorter compared to its wavelength in air.

In the formula for fringe spacing mentioned earlier, the wavelength of light is involved. If the wavelength decreases due to the change in medium (from air to water), the spacing between the fringes will also decrease.

Therefore, the spacing between the fringes will decrease if the two-slit experiment is immersed in water.

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solve the problem. select the choice that indicates the correct answer and the correct number of significant figures for each measurement. 91/2.8

Answers

Some examples of significant figures that can help you better understand them are:

104.1097 contains seven significant digits. This is because all zeros that are on the right of a decimal point and also to the left of a non-zero digit is never significant.  0.00798 contains three significant digits

Furthermore, some extra tips are:

All non-zero numbers are always significant.All zeroes before a non-zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant.

What is a Significant Figure?

This refers to the digits that carry meaning contributing to its measurement resolution and each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit.

Hence, we can see that:

All non-zero numbers are always significant.All zeroes before a non-zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant.

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Why is the use of carbon- 14 dating limited?​

Answers

Answer:

because carbon 14 has only a short half life, rather than other elements with longer half lives.

Explanation:

✨science✨

the projection lens in a certain slide projector is a single thin lens. a slide 23.9 mm high is to be projected so that its image fills a screen 1.84 m high. the slide-to-screen distance is 3.05 m. (a) determine the focal length of the projection lens.

Answers

The focal length of projection lens is  0.039m.

The opposite of a concave lens is a convex lens. Contrary to concave lenses, light rays converge in convex lenses. In contrast to the concave lens, the convex lens is bigger in the middle and thinner at the edges, therefore it converges the incident rays into its central axis.

Instead of being bent inward, the convex lens' edges are curved outward. The image seems smaller and inverted when the light is extremely focused beyond the focal length of the lens.

A convex lens's opposite side's center of curvature, , creates the image of an object placed in the center of curvature, , of the lens. As seen in the given illustration, the picture that forms is inverted and the same size as the object.

The image is inverted

h/H

= -1.8/0.024

the distance from slide to screen d=p+q=76p

M =75

d=p+q=76p

p= 0.0395m

p= 39.5cm and q = 75p = 2.96

and

a)1/0.395+1/2.96=1/f

f = 0.039m

b) from above p = 39.5 mm

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What protects astronauts from extreme temperatures.

Answers

Astronauts are protected from extreme temperatures in space by a combination of different materials and technologies.

The first line of defense is the astronaut's spacesuit, which is specially designed to protect against temperature extremes. The suit includes layers of insulation to help regulate body temperature and keep astronauts warm or cool as needed. Additionally, the suit's outer layer is made of special materials that reflect sunlight and prevent the absorption of harmful radiation.

In addition to the spacesuit, spacecraft are also equipped with various heating and cooling systems to help maintain a comfortable temperature inside the vehicle. Finally, astronauts may also use thermal blankets or other materials to shield sensitive equipment from extreme temperatures.

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Is work being done on a barbell when a weight lifter is holding the barbell
over his head?

Answers

Answer:

Because in order for work to be done on an object, the object must be moving. Why isn't work being done on a barbell when a weight lifter is holding the barbell over his head? Work is maximized when force is applied in the same direction that the object is moving. ... In order to do work faster, more_is required.

which of the following statements about gravitational waves are true? select all that apply. which of the following statements about gravitational waves are true?select all that apply. the emission of gravitational waves from merging black holes is predicted by newton's universal law of gravitation. two orbiting neutron stars or black holes will gradually spiral toward each other as a result of energy being carried away by gravitational waves. the emission of gravitational waves from merging black holes is predicted by einstein's general theory of relativity. scientists seek to detect gravitational waves by using powerful gamma-ray telescopes. although gravitational waves are an important theoretical prediction, we do not yet have any observational evidence that they exist. the first direct detection of gravitational waves, announced in 2016, came from the ligo observatory. submit

Answers

The correct statements regarding gravitational waves are: The emission of gravitational waves from merging black holes is predicted by Einstein's General Theory of Relativity, Two orbiting neutron stars or black holes will gradually spiral toward each other as a result of energy being carried away by gravitational waves, The first direct detection of gravitational waves, announced in 2016, came from the LIGO observatory.

The correct options are (B), (C) and (F).

Einstein's General Theory of Relativity predicts the production of gravitational waves from merging black holes. According to the theory, any two heavy objects that circle one other will cause ripples in spacetime that propagate away as gravitational waves.

As a result of gravitational waves carrying away energy, two circling neutron stars or black holes will progressively spiral towards one other. The gravitational waves increase stronger as they go closer, driving the objects to spiral faster and faster until they ultimately join.

The Laser Interferometer Gravitational-Wave Observatory (LIGO) claimed the first direct detection of gravitational waves in 2016. The discovery validated Einstein's theory and opened the door to a new technique of investigating the cosmos.

Therefore, options B, C and F are correct.

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A child whirls a ball around in circles on the end of a 48 cm long string at a frequency of 2.5 Hz. What is the ball’s centripetal acceleration?

Answers

The ball's centripetal acceleration is 119.2 m/s^2.

The centripetal acceleration (a) of an object moving in a circle can be calculated using the formula:

a = (v²) / r

where v is the velocity of the object and r is the radius of the circle.

In this problem, we are given the frequency (f) of the ball's motion, which is related to its velocity (v) and the radius (r) of the circle by the equation:

v = 2πrf

We are also given the length (L) of the string, which is equal to the radius (r) of the circle.

Substituting the given values into the equations, we get:

r = L = 48 cm = 0.48 m

f = 2.5 Hz

v = 2πrf = 2π(0.48 m)(2.5 Hz) = 7.54 m/s

Using the formula for centripetal acceleration, we get:

a = (v²) / r = (7.54 m/s)² / 0.48 m = 119.2 m/s²

Therefore, the ball's centripetal acceleration is 119.2 m/s².

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10-kg box is sliding across an ice rink at 10 m/s . A skater exerts a constant force of 10 N against it. How long will it take for the box to come to a complete stop?

Answers

Answer:

Time, t = 10 seconds

Explanation:

Given the following data;

Mass = 10kg

Force = 10N

Final velocity = 10m/s

Initial velocity = 0m/s

To find the time;

First of all, we would find the acceleration of the box.

Force = mass * acceleration

10 = 10 * acceleration

Acceleration = 10/10 = 1m/s²

Now, we can find the time by using the first equation of motion;

V = U + at

10 = 0 + 1t

10 = t

Time, t = 10 seconds

Therefore, it will take 10 seconds for the box to come to a complete stop.

The reason that most seti programs choose to listen at microwave radio frequencies is that.

Answers

The reason that most seti programs choose to listen at microwave radio frequencies is that They have an ability of transmitting along a vast range of frequencies without any overlap .

Microwaves are the most energy efficient way to send information In electromagnetic spectrum, microwaves comes  between radio waves and infrared waves. They are high frequency waves and their wavelengths are short . Due to which they are used in radio and television broadcasting too . They have an ability of transmitting along a vast range of frequencies without any overlap . Hence , most seti programs choose to listen at microwave radio frequencies .

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.Which equation represents the most accurate model of cellular respiration? Note: C6H12O6 is sugar.


6CO2 + 6H2O + sunlight→ 6O2 + C6H12O6

carbon dioxide + water + sunlight → oxygen + sugar

6CO 2 + 6H 2 O + sunlight→ 6O 2 + C 6 H 12 O 6 , , carbon dioxide + water + sunlight → oxygen + sugar,

CO2 + H2O + energy → O2 + C6H12O6 +CO2

carbon dioxide + water + energy→ oxygen + sugar + carbon dioxide

CO 2 + H 2 O + energy → O 2 + C 6 H 12 O 6 +CO 2, , carbon dioxide + water + energy→ oxygen + sugar + carbon dioxide,

C6H12O6 → 6CO2 + 6H2O + 6O2 + energy

sugar→ carbon dioxide + water + oxygen + energy

C 6 H 12 O 6 → 6CO 2 + 6H 2 O + 6O 2 + energy, , sugar→ carbon dioxide + water + oxygen + energy,

C6H12O6 + 6O2→ 6CO2 + 6H2O + energy

sugar + oxygen → carbon dioxide + water + energy

Answers

The equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.

CELLULAR RESPIRATION:

Cellular respiration is the process whereby living organisms obtain energy by breaking down food molecules in their cells.

The process of cellular respiration breaks down sugar molecules (glucose) in the presence of oxygen to produce carbon dioxide and water as products, as well as energy in form of ATP.

Therefore, the equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.

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when an outlet supplied through an underfloor raceway is removed, the sections of the circuit conductors supplying the outlet shall be removed from the raceway.

Answers

When an outlet supplied through an underfloor raceway is removed, the sections of the circuit conductors supplying the outlet shall be removed from the raceway is True at 390.8.

Where can conductors be spliced in a raceway system?

All conductors of the same circuit and, where used, the grounded conductor and all equipment grounding conductors and bonding conductors shall be contained within the same raceway, cable tray, cable bus assembly, trench, or cord, unless otherwise permitted in accordance with 300.3

When an outlet supplied by an underfloor raceway is removed, the circuit conductor sections supplying the outlet must also be removed from the raceway. Raceway conductors must be continuous between all points of the system [300.13]. This means that splices in raceways are not permitted unless specifically authorized by 376.56, 378.56, 384.56, 386.56, 388.56 or 390.80.

The complete question is:

When an outlet, supplied through an underfloor raceway, is removed, the sections of the circuit conductors supplying the outlet shall be removed from the raceway. True/False.

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a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​

Answers

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula  d = \(\frac{at^{2} }{2} ,\)

where d is the distance, t is the time and "a" is the acceleration.
\(d=\frac{9*54*6*5^{2} }{2} = 201.53 m\)

A box of mass m is initially at rest at the top of a ramp that is at an angle with the horizontal. The block is at a height h and length L

from the bottom of the ramp. The block is released and slides down the ramp. The coefficient of kinetic friction between the block and

the ramp is u. What is the kinetic energy of the box at the bottom of the ramp?

Answers

Hi there!

We can use the work-energy theorem to solve.

Recall that:
\(E_i = E_f\)

The initial energy equals the final energy (Conservation of Energy). However, we must take into account energy dissipated due to friction in this instance.

The energy lost due to friction is equivalent to the work done by friction. Recall the following:

Normal force on an incline: \(N = Mgcos\theta\)Force due to friction: \(F_f = \mu N = \mu mgcos\theta\)

The work due to a force is:
\(W = F \cdot d \\\)

Since the displacement is in the same direction as the force, the dot-product becomes Fd.

The work due to friction then becomes:

\(W_f = \mu mgdcos\theta\)
The work due to friction is SUBTRACTED from the initial potential energy.

Initial energy = GPE = mgh

Final energy = KE

Therefore:

\(\boxed{mgh - \mu mgdcos\theta = KE}\)


what would be the noontime altitude of the sun at the time of the summer solstice?

Answers

At the time of the summer solstice, the noontime altitude of the sun is at its highest point, around 90°.

What is altitude?

Altitude is the height above sea level. It is typically measured in either metres or feet. In aviation, altitude can also refer to the vertical distance between an aircraft and a certain reference point on the ground. Altitude can be used to determine the air pressure, temperature, and density of the air. Altitude can also be used to calculate the distance a plane can travel without refueling. Altitude can play an important role in the weather of an area, as air pressure and temperature tend to decrease with altitude. Altitude can also affect the type of vegetation found in an area. In mountain regions, the altitude can have a dramatic effect on the climate, creating distinct areas of vegetation and wildlife. Altitude can also affect the types of crops that can be grown in an area, depending on the air pressure, temperature, and precipitation.

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Calculate the total charge Q of a thin plate with the charge density distribution p(x,y) = x²y mC/m². The plate shape is restricted by the lines: y=2-x² and y=2x-1, where x and y are measured in metres. a) Sketch the plate shape. [10 marks] [4 marks] b) Present the total charge through the double integral. c) Reduce the double integral to the repeated integrals and show limits of integration. [6 marks] d) Calculate the integral and present your answer with five significant figures. [20 marks]

Answers

a) Sketch the plate shape: we get a shape that resembles a trapezoid.

The plate shape is determined by the lines y = 2 - x² and y = 2x - 1. To sketch the plate shape, we can plot these two lines and shade the region in between them. The intersection points of the lines are found by solving the equations simultaneously:

2 - x² = 2x - 1

Simplifying, we get:

x² + 2x - 3 = 0

Factoring, we have:

(x - 1)(x + 3) = 0

So, x = 1 and x = -3. Plugging these values into the equations of the lines, we find the corresponding y-values:

For x = 1:

y = 2 - (1)² = 1

For x = -3:

y = 2(-3) - 1 = -7

Plotting these points and connecting them with the lines, we get a shape that resembles a trapezoid.

b) Total charge through the double integral:

To find the total charge Q, we need to integrate the charge density p(x, y) over the entire plate. We can express this as a double integral:

Q = ∬ p(x, y) dA

c) Reducing the double integral to repeated integrals: The limits of integration for x are the values of x that define the boundaries of the plate shape, which are -3 to 1.

Since the plate shape is described by the lines y = 2 - x² and y = 2x - 1, we can rewrite the double integral as a repeated integral by integrating with respect to x and y separately:

Q = ∫∫ p(x, y) dy dx

The limits of integration for y are from the lower curve y = 2 - x² to the upper curve y = 2x - 1. The limits of integration for x are the values of x that define the boundaries of the plate shape, which are -3 to 1.

d) Calculating the integral: The total charge Q of the thin plate is approximately 12.4 mC.

Now, we can evaluate the double integral to find the total charge Q:

Q = ∫(-3 to 1) ∫(2 - x² to 2x - 1) x²y dy dx

Performing the inner integral with respect to y first, we get:

Q = ∫(-3 to 1) [x²(y²/2 - y)] from 2 - x² to 2x - 1 dx

Simplifying the inner integral, we have:

Q = ∫(-3 to 1) [(x²/2)(2 - x²) - x²(2x - 1)] dx

Expanding and simplifying further, we get:

Q = ∫(-3 to 1) (x² - x⁴/2 - 4x³ + 2x²) dx

Integrating term by term, we have:

Q = [x³/3 - x⁵/10 - x⁴ + 2x³/3] from -3 to 1

Evaluating the integral at the limits, we get:

Q ≈ 12.4 mC (rounded to five significant figures)

Therefore, the total charge Q of the thin plate is approximately 12.4 mC.

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A textbook that weighs 2 Newtons is lifted 4 meters. How much work is done on the book?

Answers

Answer:

The answer is 8 J

Explanation:

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

workdone = weight × distance

From the question

weight = 2 N

distance = 4 m

We have

workdone = 2 × 4

We have the final answer as

8 J

Hope this helps you

ILL MARK BRAINLEST
Question 13: A food web of the Yellowstone National Park ecosystem is
provided below. If a sudden outbreak of disease severely reduced the
number of wolves in Yellowstone National Park very rapidly, what are two
changes that might happen to the local food web? Fully explain using
claim, evidence, and reasoning.

ILL MARK BRAINLEST Question 13: A food web of the Yellowstone National Park ecosystem isprovided below.

Answers

Answer:

they all will die and so will we

Explanation:

Answer:

3 animals will thrive better

Explanation:

If the number of wolves is reduced, 3 animal species usually eaten by the wolves will have a better chance of survival.

If it takes 500 seconds for the light from the sun to reach earth what is the distance to the sun in maters the speed of light is 300,000,000 meters second someone please awnser and explain in detail

Answers

Answer:

150000000000

Explanation:

speed= distance/time

HELP STUCK IN PHYSICS 100PTS!!!

A small piece of putty of mass 27 g and
negligible size has a speed of 2.2 m/s. It makes
a collision with a rod of length 5 cm and mass
78 g (initially at rest) such that the putty hits
the very end of the rod. The putty sticks to
the end of the rod and spins around after the
collision; the putty-rod center of mass CM has
a linear velocity v and an angular velocity ω
about the CM.
(a)What is the angular momentum of the system relative to the CM after the collision?
The rod has a moment of inertia equal to (1/12)MR^2
about its center of mass; consider
the piece of putty as a point mass. Assume
the collision takes place in outer space where
there is no gravitational field and no friction.
Answer in units of kg · m2
/s.
(b)What is the system’s angular speed about the
CM after the collision?
Answer in units of rad/s.
I'm so confused :(

HELP STUCK IN PHYSICS 100PTS!!! A small piece of putty of mass 27 g andnegligible size has a speed of

Answers

Answer:

a) 20.65kg

b)4.7rad

Explanation:

An object weighs 59N on earth where g is 9.8N/kg.Find the acceleration due to gravity on a planet where the same object weighs 40.5N​

Answers

Use Newton's second law.

The object's mass is the same everywhere, so first compute the mass using the object's weight on Earth:

F = ma   ⇒   59 N = m (9.8 N/kg)   ⇒   m ≈ 6.02041 kg

Now compute the acceleration on the other planet:

F = ma   ⇒   40.5 N ≈ (6.02041 kg) a

⇒   a ≈ 6.72712 m/s² ≈ 6.7 m/s² ≈ 1.5g

where g = 9.8 m/s², just to express this acceleration relative to the gravitational acceleration at Earth's surface.

Explain at least 3 different ways that you can motivate yourself past any distractions to reach your goals. (The subject is Physical Education)
Help!
Due today

Answers

Answer:

Explanation:

Set specific, achievable goals: Break down your larger goal into smaller, more manageable targets that you can work towards each day. When you accomplish these smaller goals, it will give you a sense of accomplishment and motivation to continue working towards your larger goal.

Visualize your success: Imagine yourself successfully achieving your goal and all of the positive feelings that come along with it. This visualization can help keep you focused and motivated when distractions arise.

Find an accountability partner: Find someone who shares your goals and who can help you stay focused and motivated. This could be a friend, family member, or a coach. Having someone to share your progress with and to hold you accountable can be a powerful motivator.

Reward yourself: Set up a system of rewards for yourself when you reach certain milestones on the way to your goal. These rewards could be something as simple as a treat or a movie, or something more substantial like a new piece of workout equipment.

Get organized: Plan out your workouts and stick to a routine. Having a clear plan and structure can help you stay focused and motivated, as you can see how each workout is helping you to reach your goal.

Surround yourself with positive influences: Surround yourself with people who support and encourage your goals. Having positive influences in your life can help you stay motivated and inspired to reach your goals.

A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on the table. What conclusion can be made about how they each will feel the vibrations?

Answers

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

Alika finds a certain kind of butterfly near her house. She reads that ultraviolet light is absorbed by the butterfly’s wings, red light is transmitted through the butterfly’s wings, and green light is reflected off the butterfly’s wings. Alika’s friend Claire tells her that if you shine light on the butterfly, its wings will glow.

Can light cause the butterfly’s wings to glow? Why or why not? Does it matter what type of light Alika shines on the butterfly?

Answers

Answer:

No the butterflies wings won't glow.

Explanation:

Butterflies have wings that have different color of attractiveness. Butterflies are however attracted to light but can't absorb the light.

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