True/False: A light's color is determined by the light wave's frequency.
True
False

Answer:7

Explanation:none I'm a computer

The electron must be placed at a distance of 3.734×10^-5 m from the point charge q in the opposite direction of the gravitational force.

To find the position where the electric force on the electron is exactly opposite to its weight, we need to set up the equation for the electric force and weight and solve for the position.

The electric force (Fe) between the point charge q and the electron is given by Coulomb's law:

Fe = (1/4πε₀) * (q * e) / r²

where  

ε₀ is the electric constant,

r is the distance between the two charges, and

e is the charge of the electron.

The weight (Fg) of the electron is given by:

Fg = m * g

where

m is the mass of the electron and

g is the acceleration due to gravity.

We want to find the distance r where

Fe = -Fg

So we have:

|Fe| = |Fg|

(1/4πε₀) * (q * e) / r² = m * g

Solving for r, we get:

r = sqrt[(1/4πε₀) * (q * e) / (m * g)]

Plugging in the values for the constants and the given charge and mass, we get:

r = sqrt[(1/4π * 8.99×10^9 N⋅m²/C²) * (0.55×10^-9 C * 1.60×10^-19 C) / (9.11×10^-31 kg * 9.81 m/s²)]

r = 3.734×10^-5 m

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The standard form equation for the circle with center (3, 10) and radius 2 is: [tex]\rm 2.x^{2} + y^{2} - 6x - 20y + 105 = 0.[/tex]

The radius, diameter, circumference, arc, chord, secant, tangent, sector, and segment are all parts of a circle.

A circle is a round figure with no corners or edges. A circle is a closed shape, a two-dimensional shape, and a curved shape in geometry.

A circle with center (h, k) and radius r has the following standard form equation:

[tex]\rm (x-h)^{2} + (y-k)^{2} = r^{2}[/tex]

Substituting the values of center and radius:

[tex]\rm (x-3)^{2} + (y-10)^{2} = 2^{2}[/tex]

Simplifying the equation:

[tex]\rm x^{2} - 6x + 9 + y^{2} - 20y + 100 = 4[/tex]

[tex]\rm x^{2} + y^{2} - 6x - 20y + 105 = 0[/tex]

[tex]\rm x^{2} + y^{2} - 6x - 20y + 105 = 0[/tex]

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The joule is a basic unit of work or energy​

Answer:

Joule

Explanation:

If you ever feel like you are not doing enough work or energy in your life, just remember that you are constantly producing joules. Every time you move a muscle, breathe, blink, or even think, you are using up some joules. And every time you plug in your phone, laptop, or toaster, you are transferring some joules. Joules are everywhere, and they are awesome.

But how awesome are they? Well, let's put them in perspective. One joule is enough to lift an apple one meter off the ground. That may not sound impressive, but what if you had a million joules? Then you could lift a car one meter off the ground. Or what if you had a billion joules? Then you could launch a rocket into space. Or what if you had a trillion joules? Then you could... well, I don't know what you could do with that much energy, but it would be something amazing.

So, the next time you feel low on energy, just think of all the joules you have at your disposal. You are a powerhouse of work and energy, and you can do anything you set your mind to. Just don't forget to measure it in joules.

Explanation:

Sure, here are some examples of drag, elastic force, and tension force:

1. Drag force: A swimmer experiences drag force while swimming through water - this force tends to slow the swimmer down. Another example is a flag flapping in the wind - the wind pushes on the flag, creating airflow around it that creates drag.

2. Elastic force: A stretched elastic band has stored elastic energy, which can be released when the band is allowed to snap back into its original shape. Similarly, a compressed spring has stored elastic energy, which can be released when the spring is allowed to expand back to its original length.

3. Tension force: When a rock climber climbs a cliff face, they use ropes to help them ascend. The ropes are under tension, which helps support the climber's weight and prevent them from falling. Another example is lifting weights - the muscles exert tension on the weight in order to lift it.

4. Drag force: When a car drives on a highway, it experiences aerodynamic drag force from the air resistance. This force slows the car down and makes it harder to drive at high speeds.

5. Elastic force: A rubber ball bounces when it hits a hard surface because of the energy stored in the ball due to its elasticity - as the ball compresses upon impact, it stores elastic energy that then causes it to bounce back up into the air.

6. Tension force: When a crane lifts a heavy load, the cables and ropes are under tension to support the weight of the load. This allows the crane to lift and move the load safely.

These are just a few examples - drag, elastic, and tension forces can be found in many different physical systems and everyday situations.

The net force on the fully submerged scuba tank at the beginning of a dive is 18.484 N upward.

What is the net force on a fully submerged scuba tank at the beginning of a dive?

The net force acting on the fully submerged tank is equal to the difference between its weight and the buoyant force acting on it.

The weight of the tank can be calculated as the product of its mass and the acceleration due to gravity (9.81 m/s²):

weight = mass x gravity

weight = 12.6 kg x 9.81 m/s²

weight = 123.606 N

The buoyant force on the tank is equal to the weight of the seawater displaced by the tank. The displacement volume of the tank is given as 14.4 L. We can convert this to cubic meters, as follows:

1 L = 0.001 m³

14.4 L = 14.4 x 0.001 m³ = 0.0144 m³

The density of seawater is approximately 1025 kg/m³, so the weight of the displaced seawater can be calculated as follows:

weight of displaced seawater =

density x volume x gravity

weight of displaced seawater = 1025 kg/m³ x 0.0144 m³ x 9.81 m/s²

weight of displaced seawater = 142.09 N

Therefore, the buoyant force on the tank is equal to 142.09 N.

The net force on the fully submerged tank can now be calculated as follows:

net force = weight - buoyant force

net force = 123.606 N - 142.09 N

net force = -18.484 N

The negative sign indicates that the buoyant force is greater than the weight of the tank, so there is a net upward force on the tank.

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Answer: the magnitude of the total electric force exerted on the moved charge is proportional to the square of the charge q and inversely proportional to the square of the radius R.

Explanation:

The total electric force exerted on a charge moved to the center of a circle with 12 equally spaced charges can be found by calculating the electric field at the center due to the remaining 11 charges and multiplying by the charge of the moved particle. Coulomb's law gives the electric field at the center due to a single charge located at a distance R from the center as

E = kq/R^2

The total electric field due to the 11 other charges is

E_total = 11E

Thus, the force on the moved charge is

F = qE_total = 11kq^2/R^2.

Therefore, the force is proportional to the square of the charge q and inversely proportional to the square of the radius R.

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The correct answer for the first question is b. compliance. Milgram's research found that people were more likely to comply with authority figures, even if it meant delivering potentially lethal shocks to another person.

However, follow-up studies have suggested that a teacher's willingness to deliver shocks may be more a product of compliance than obedience because people often comply with social norms and expectations without necessarily believing in them or being obedient to authority figures.

The public service messages that encourage parents to talk to their children about drugs are promoting b. direct instruction as a method of attitude formation. Direct instruction involves providing explicit information or guidance to shape a person's attitudes or beliefs. In this case, the messages are encouraging parents to directly instruct their children about the dangers of drugs and how to avoid them. Direct contact, on the other hand, involves personal experiences or interactions that shape attitudes, such as meeting someone who has struggled with drug addiction.

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Answer:

To solve this problem, we can use the equations of motion for projectile motion. We know that the horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to gravity. Let's consider the horizontal and vertical components separately.

Horizontal Component:

The horizontal component of velocity (Vx) remains constant throughout the motion. Therefore, we can write:

Vx = b/t

where t is the time taken by the golf ball to hit the tree.

Vertical Component:

The vertical component of velocity (Vy) changes due to gravity. We can use the following equation to find Vy at any time t:

Vy = u*sin(0) - gt

where u is the initial velocity, 0 is the angle of projection, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken by the golf ball to hit the tree.

At the maximum height (h) reached by the golf ball, Vy becomes zero. Therefore, we can write:

h = u^2*sin^2(0)/2g

Solving for u, we get:

u = sqrt(2gh)/sin(0)

Now, we need to find the time taken by the golf ball to hit the tree. We can use the following equation:

b = Vx*t

Solving for t, we get:

t = b/Vx

Substituting this value of t in the equation for Vy, we get:

h = u*sin(0)*t - gt^2/2

Substituting the values of u and t, we get:

h = b*tan(0) - (g*b^2)/(2*Vx^2)

Finally, substituting the value of Vx, we get:

h = b*tan(0) - (g*b^2)/(2*b^2/t^2)

Simplifying, we get:

h = b*tan(0) - (g*t^2)/2

Solving for u, we get:

u = sqrt(2*(h + g*t^2/2)/sin(0))

Therefore, the initial velocity in terms of b and h is given by:

u = sqrt(2*(h + g*(b/Vx)^2/2)/sin(0))

Answer:

Mass = 56.83 Kg

Explanation:

[tex]PE = mgh\\9094 J = 16m*10 m/s^2*mass\\mass = \frac{PE}{g*h} \\mass = \frac{9094 J}{16*10} = 56.83 kg[/tex]

The maximum size of the logo the company wants to use for the wheelbarrow is 162000 mm².

5.1 To calculate the approximate amount of metal sheeting needed to build the wheelbarrow, we could divide it up into 5 shapes as follows:

Shape 1: A rectangle for the wheelbarrow tray with dimensions 1000 mm x 600 mm

Shape 2: A rectangle for the front of the tray with dimensions 324 mm x 50 cm

Shape 3: Two identical rectangles for the sides of the tray, each with dimensions 600 mm x 50 cm

Shape 4: A rectangle for the back of the tray with dimensions 600 mm x 324 mm

Shape 5: A trapezoid for the front wheel support with dimensions 50 cm (top), 324 mm (bottom), and 600 mm (height)

Here is a net of the 5 shapes:

  +-----------------------------------+

  |              Shape 1                |

  |                                           |

  |                                           |

  |                                           |

  |                                           |

  |                                           |

  +-----------------------------------+

      +-----------------------+

      |         Shape 2      |

      |                             |

      |                             |

      |                             |

      +-----------------------+

+-----------------------------------------+

|                 Shape 3                   |

|                                                  |

|                                                  |

+-----------------------------------------+

      +-----------------------+

      |         Shape 3      |

      |                             |

      |                             |

      |                             |

      +-----------------------+

  +-----------------------------------+

  |              Shape 4               |

  |                                           |

  |                                           |

  |                                           |

  |                                           |

  |                                           |

  +-----------------------------------+

  +-----------------------------------+

  |              Shape 5               |

  |                 /|\                       |

  |                  |                        |

  |                  |                        |

  |                 \|/                       |

  |                                           |

  +-----------------------------------+

5.2 To calculate the maximum size their logo could be in mm², find the area of the front of the wheelbarrow where the logo will be printed. This is Shape 2, which has dimensions 324 mm x 50 cm. Converting 50 cm to mm, we get 500 mm. Therefore, the area of Shape 2 is:

Area = 324 mm x 500 mm = 162000 mm²

So, the maximum size the logo could be is 162000 mm².

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Answer:

The answer is D

26KJ is adsorbed

The temperature of the boiler is 150.9°C, the thermal energy absorbed each hour is 1.62 x 10^9 joules, and the amount of thermal energy lost per hour is 1.61 x 10^9 joules.

The Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is a highly efficient cycle that provides the maximum possible efficiency for a heat engine operating between two given temperatures.

18. The efficiency of a Carnot engine is given by the equation:

efficiency = 1 - (T_C / T_H)

where T_C is the temperature of the cold reservoir and T_H is the temperature of the hot reservoir. We can rearrange this equation to solve for T_H:

T_H = T_C / (1 - efficiency)

Substituting the given efficiency and waste heat temperature values, we get:

T_H = 36.3°C / (1 - 0.761) = 150.9°C

Therefore, the temperature of the boiler is 150.9°C.

19. The efficiency of a Carnot engine is given by the equation:

efficiency = 1 - (T_C / T_H)

where T_C is the temperature of the cold reservoir and T_H is the temperature of the hot reservoir. We can rearrange this equation to solve for the thermal energy absorbed each hour:

thermal energy absorbed per hour = power output / efficiency

Substituting the given values, we get:

thermal energy absorbed per hour = 115 kW / (1 - (13°C / 569°C)) = 1.62 x 10^9 J/h

Therefore, the thermal energy absorbed each hour is 1.62 x 10^9 joules.

20. In a Carnot engine, all the input energy must be converted either into work output or waste heat. Therefore, the amount of thermal energy lost per hour is equal to the difference between the thermal energy absorbed and the power output:

thermal energy lost per hour = thermal energy absorbed per hour - power output

Substituting the given values, we get:

thermal energy lost per hour = 1.62 x 10^9 J/h - 115 kW x 3600 s/h = 1.61 x 10^9 J/h

So, the amount of thermal energy lost per hour is 1.61 x 10^9 joules.

Hence, The boiler has a temperature of 150.9°C, absorbs 1.62 × 109 joules of thermal energy every hour, and loses 1.61 x 109 joules of thermal energy every hour.

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Let us get to solutions real quick:

Using the formula for the resonant frequency of a tube with one closed end:

f = (2n - 1) v / 4L

where:

f = frequency of the sound wave

n = harmonic number (8th harmonic in this case, so n = 8)

v = speed of sound in air (approximately 343 m/s at room temperature)

L = length of the tube from the closed end to the water level = 63.75 cm = 0.6375 m

f = [tex]\frac{(2*8 - 1) * 343}{4*0.6375}[/tex] = [tex]\frac{5145}{2.55}[/tex]

Therefore f = 2017.64 Hz

Now we can easily get our wavelength

λ = v/f,

λ = wavelength

λ = 343/2017.64

  = 0.17m = 17 cm (conversion from meter to centimeter)

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(a)  Average speed is 463.04 m/s

(b) The centripetal force is  5,361 N

(c) The centripetal force acts towards the center of the circle.

In this scenario, the actual force providing the centripetal force required to keep a person in uniform circular motion on Earth's surface is the force of gravity.

(d) The weight of a person is 686.7 N

(e) The normal force exerted on a person by the Earth's surface is 686.7 N

A centripetal force is described as a force that makes a body follow a curved path. The direction of the centripetal force is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path.

The weight of a person can be found using the formula:

Weight = mass * gravitational acceleration

Assuming the individual has a  mass 70 kg, we have:

Weight = 70 kg * 9.81 m/s^2

weight = 686.7 N

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