A disk 8.08 cm in radius rotates at a constant rate of 1 210 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 2.94 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction ---Select--- (d) Determine the total distance a point on the rim moves in 2.02 s. m

The force applied in one direction is higher than the force applied in the opposing direction, resulting in imbalanced forces. Unbalanced force is that force shows the condition when an object moves, changes speed or changes direction.

The force applied in one direction is higher than the force applied in the opposing direction, resulting in imbalanced forces.

There is a shift in speed and/or direction when unbalanced forces occur on an item. You can affect the motion of an object by exerting an imbalanced force.

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Answer:

The Recovery Principle applies both to immediate rest needed between bouts of exercise, as well as to longer time intervals of several hours to about two days. When you sprint as fast as possible or lift heavy weights, you will notice that your heart still pumps hard and you breathe heavy for a while after you stop.

Explanation:

Hope this helped!

If the car b travels a distance d before stopping, the car a travelled a distance of 4D.

The third equation of motion can be used to calculate the distance how far the cars drove before coming to a stop:

Vf2 - Vi2 = 2as

(Vf2 - Vi2)/2a = s

where,

s is the distance covered.

Final Speed = 0 m/s for Vf

Initial Speed = Vi

an is the deceleration rate.

We start by thinking about Car B and giving it a subscript 2:

Vf₂ = 0 m/s (As, automobile finally stops) (As, car finally stops)

s₂ = D

a₂ = - a (due to slowdown) (due to deceleration)

D = (0² - Vi₂²) /(-2a) (-2a)

the equation D = Vi22/2a (1)

We now take Car A into consideration and give it a subscript 1:

Vf₁ = 0 m/s (As, automobile finally stops) (As, car finally stops)

s₁ = ?

a₁ = - a (due to slowdown) (due to deceleration)

Vi₁ = 2 Vi₂ (Since, car A was previously moving at twice speed of car B) (Since, car A was initially traveling at twice speed of car B)

s₁ = (0² - Vi₁²) /(-2a) (-2a)

s₁ = (2Vi₂)²/2a

s₁ = 4 (Vi₂²/2a)

Using formula (1), we arrive at:

s₁ = 4D.

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The image of the truck formed by the rear-view mirror is smaller and virtual, located in front of the mirror.

To calculate the size of the image relative to the size of the truck formed by the rear-view mirror, we can use the mirror equation:1/f = 1/u + 1/v,

where f is the focal length (radius of curvature) of the mirror, u is the object distance, and v is the image distance.

Given that the radius of curvature of the mirror is 4m and the truck is located 5m from the mirror, we can determine the object distance:

u = -5m (since the object is behind the mirror and its distance is negative)

Substituting these values into the mirror equation, we get:

1/4 = 1/(-5) + 1/v

Simplifying the equation:

1/v = 1/4 + 1/5

1/v = (5 + 4) / (4 * 5)

1/v = 9/20

v = 20/9

Now, we can calculate the size of the image relative to the size of the truck using the magnification formula:

magnification = -v/u

Substituting the values:

magnification = -(20/9) / (-5)

magnification = 4/9

Therefore, the size of the image relative to the size of the truck is 4/9. This means the image formed by the rear-view mirror is smaller than the actual truck.

As for the position and nature of the image, since the image distance (v) is positive, the image is formed on the same side of the mirror as the object. In this case, it means the image is formed in front of the rear-view mirror. The positive image distance also indicates that the image is virtual.

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Answer:

The constriction in the thermometer is to prevent the mercury from dropping back to the bulb when the reading is being taken. It is easier for us to take reading. It prevents the thermometric substance (Mercury or Alcohol) from running back into the bulb.

Explanation:

Hope it is helpful....

Answer:

constriction is made in a clinical thermometer because it prevents the thermometric liquid from running back into the bulb.

ANSWER

B) the opposite sign

EXPLANATION

The force between charged particles is inversely proportional to the square of the distance between particles, from Coulomb's law:

If the force increases it can mean two things: either one or both particles are gaining charge - which is not the case of this problem, or the distance between them is decreasing. In this case, the distance must be decreasing for the force to increase. This means that the particles are moving towards each other, they are being attracted by each other. Since only particles with opposite signs attract, the answer is option B, they have opposite sign.

A heavy rock will encounters the greatest amount of air resistance than the light rock due to its mass. Thus, the correct option is B.

Air resistance is a force which is caused by the air. The force which acts in the opposite direction to an object that is moving through the air. A car with a flat front will experience high air resistance while a sports car with a streamlined shape will experience lower air resistance, allowing the car with low air resistance to go faster.

Air resistance happens when an object moves through the direction of air. Depending on the velocity, shape, and area of the object, the resistance differs. The faster an object moves in the direction and the greater its area, the higher will be the air resistance gets.

Therefore, the correct option is B.

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Answer:

it is increasing

Explanation:

gpe is highest at highest point of travel

If a 1200 N box were lifted off the ground for a distance of 5 m, the amount of work done would be 6000 J.

Work (W) is defined as the product of the force (F) applied on an object and the distance (d) over which the force is applied. Mathematically, W = Fd.

In this case, the force applied is the weight of the box, which is 1200 N. The distance through which the box is lifted is 5 m. Therefore, the work done in lifting the box can be calculated as W = Fd = 1200 N x 5 m = 6000 J.

This means that lifting the box requires an input of 6000 Joules of energy.

It is important to note that as work is a scalar quantity and the direction of the force and displacement are parallel, therefore, work done is simply the product of force and displacement.

Additionally, the work done on the box is equal to the potential energy gained by the box, which is at a height above the ground.

Therefore, if the box is allowed to fall back to the ground, it will release the same amount of potential energy as it gained, which will be converted into kinetic energy.

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Answer:

79.38 J

Explanation:

W = mgz = 4.5*9.8*1.8 (with g = 9.8 m/s^2)

= 79.38 K

The point in the picture in which the cart have the greatest potential energy is point D (option D)

To know which point have the greatest potential energy, we shall obtain the potential energy at each point. This is shown below:

For point A

PE = mgh

= m × 9.8 × 10

= 98m J

For point B

PE = mgh

= m × 9.8 × 0

= 0 J

For point C

PE = mgh

= m × 9.8 × 3

= 29.4m J

For point D

PE = mgh

= m × 9.8 × max h

= 9.8m × max h J

Since the height at D is maximum, we can conclude that the point D has the greatest potential energy.

Hence, Option D is the correct answer to the question

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Answer:

20 km

Explanation:

You answered the question already. 60 min = an hour. Reread your question :)

Answer:

Answer: oxygen

Explanation: Helium burns into carbon and carbon combines with helium to make oxygen.

Answer:

The internal core collapses, and heats up, until it is hot enough to fuse helium into larger atoms, for instance, by combining three helium atoms into carbon. At this same time, some helium will fuse with that carbon to produce oxygen. ... it becomes essentially a free-for-all for creating heavier and heavier atoms.

Explanation:

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As the intensity increases the energy increases.

Answer:

C. the energy of the sound wave increases

Answer:

Work done = 838470 Joules.

Explanation:

Given the following data;

Power = 19 hp

Time = 1 minute to seconds = 60 seconds.

Next, we would convert the unit of power in "hp" to "Watt."

1 Watt = 0.00135962 horsepower

x Watt = 19 horsepower

Cross-multiplying, we have;

19 = 0.00135962x

x = 19/0.00135962

x = 13974.5 Watts.

Now, to find the work done in moving the mower;

Work done = power * time

Substituting into the formula, we have;

Work done = 13974.5 * 60

Work done = 838470 Joules.

Answer:In primary succession, newly exposed or newly formed rock is colonized by living things for the first time. In secondary succession, an area previously occupied by living things is disturbed—disrupted—then recolonized following the disturbance.The first organisms to appear in areas of primary succession are often mosses or lichens. These organisms are known as pioneer species because they are the first species present; pioneer species must be hardy and strong, just like human pioneers.A heterotroph is an organism that eats other plants or animals for energy and nutrients. The term stems from the Greek words hetero for “other” and trophe for “nourishment.” Organisms are characterized into two broad categories based upon how they obtain their energy and nutrients: autotrophs and heterotrophs.

Explanation:I forgot Extinct

Given the resistance of the copper wire, length of the wire, and area of cross-section, we can calculate the resistivity of copper by the following formula.ρ = RA/L, .

Where: R is the resistance of the copper wireA is the area of cross-section of the copper wireL is the length of the copper wireSubstituting the given values, we haveρ = (2×10-2 ohms)(1.7×10-6 m2)/(2 m)ρ = 1.7×10-8 ohm-m.

Therefore, the resistivity of copper is 1.7×10-8 ohm-m. Given the resistance of the copper wire, length of the wire, and area of cross-section, we can calculate the resistivity of copper by the following formula.ρ = RA/L, .Where: R is the resistance of the copper wireA is the area of cross section of the copper wireL is the length of the copper wire .

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The acceleration of the bicyclist is 5.56 m/s2, and it takes 2.27 seconds to reach a speed of 20 km/h. Given,Initial velocity of the bicyclist, u = 0 Distance traveled, s = 25 m Final velocity of the bicyclist, v = 20 km/h = 20 × (5/18) m/s = 100/9 m/s.

We know that acceleration, a = (v - u) / t Where,a is acceleration v is the final velocityu is the initial velocityt is the time taken. Initially the bicyclist was at rest so, u = 0 Now, we can calculate the time taken to reach the speed of 20 km/h (100/9 m/s) using the formula, v = u + atv = u + at => t = (v - u) / at = (100/9 - 0) / aa = (v - u) / t = (100/9) / tRearranging this equation for a, we geta = (v - u) / t = (100/9) / t.

To calculate t, we can use the formula for distance, s = ut + 1/2 at2 Substituting the given values,25 = 0 × t + 1/2 a t2 => t2 = 50/a => t = √(50/a)Substituting the value of t in the expression for acceleration, we geta = (100/9) / √(50/a)On solving, we get a = 5.56 m/s2, which is the acceleration of the bicyclist, and the time taken to reach the speed of 20 km/h is t = 2.27 seconds (approx).

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Balloon A is positively charged, and balloon C is negatively charged. If balloon A approaches balloon C, there will be an electrostatic force of attraction between them.

When two objects carry opposite charges, they exert an attractive force on each other. This force is known as the electrostatic force and follows Coulomb's law. According to Coulomb's law, the magnitude of the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this scenario, since balloon A is positively charged and balloon C is negatively charged, they have opposite charges. Therefore, the electrostatic force between them will be attractive. The magnitude of the force depends on the charges of the balloons and the distance between them. It is important to note that without specific information about the charges of the balloons and their distance, it is not possible to determine the exact magnitude of the force. To calculate the force, you would need the values of the charges and the distance between the balloons.

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If we consider the cantilever to be a perfect, massless spring, then K = 5.952N/m is its effective force constant.

The meaning of the term "force" is obvious. At about this point, it's perfectly permissible to describe a force as shoving or pulling. An object doesn't "have a force in it" or "contain a force." A force is applied to one thing by another.

A force can be a push or a pull, and it affects how we go about our daily lives because we can't open or close items, raise our legs or arms or accomplish a lot more without it.

Numerous E coli bacteria, m = 7.10 × 10⁻¹⁶

frequency =17.1 MHz

frequency of vibration, F = 1/2π√K/M

17.1×10⁶= 1/2π√K/ 7.10 × 10⁻¹⁶

K/ 7.10 × 10⁻¹⁶ = 7.10 ×10¹⁵

K = 5.952N/m

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For each planet/comet in Kepler's Second Law experiment: The areas swept out by the planet/comet in equal time intervals are equal, and The perihelion distance (Rp) is the closest distance of the planet/comet to the Sun and the aphelion distance (Ra) is the farthest distance from the Sun. The difference between Rp and Ra is known as the eccentricity of the planet/comet's orbit. The closer the orbit is to be circular, the smaller the eccentricity and the more similar Rp and Ra will be.

Kepler's Second Law experiment demonstrates that a planet moves faster when it is closer to the Sun and slower when it is farther away. It involves tracking the position of a planet as it orbits the Sun and measuring the area swept out by the planet in a given time interval.

The Neptune:

1. Neptune has an elliptical orbit around the sun, with the Sun located at one of the foci of the ellipse.

2. No, the Sun is not at the center of Neptune's orbit. The Sun is located at one of the foci of the elliptical orbit.

3. Neptune moves fastest when it is closest to the Sun (at perihelion) and slowest when it is farthest from the Sun (at aphelion), in accordance with Kepler's Second Law. Neptune's speed is not constant throughout its orbit because it experiences varying gravitational forces due to its elliptical orbit.

4. Without a specific diagram or graph to reference, it is unclear what is meant by "each area." Please provide more information or context.

5. The perihelion distance (Rp) is the distance between Neptune and the Sun when it is closest to the Sun in its orbit, while the aphelion distance (Ra) is the distance when it is farthest from the Sun. Since Neptune has an elliptical orbit, Rp and Ra are different values. Specifically, Neptune's perihelion distance is about 4.45 billion km, while its aphelion distance is about 4.55 billion km.

Hence, Kepler's Second Law experiment shows that planets/comets sweep out equal areas at equal times and that the perihelion and aphelion distances are related to the eccentricity of the orbit, with more circular orbits having smaller differences between the two.

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Answer:

Approximately \(1.3\; {\rm m}\) (assuming that \(g = 9.81\; {\rm m\cdot s^{-2}}\), the table is level, and that air resistance on the coin is negligible.)

Explanation:

Assume that the air resistance on the coin is negligible. As long as the coin is in the air, its horizontal velocity of the coin will be constantly equal to the original value, \(v_{x} = 0.40\; {\rm m\cdot s^{-1}}\).

It is given that the horizontal displacement of this coin is \(x_{x} = 0.20\; {\rm m}\) during the flight. Divide this horizontal displacement by horizontal velocity to find the duration \(t\) of the flight:

\(\begin{aligned} t &= \frac{x_{x}}{v_{x}} \\ &= \frac{0.20\; {\rm m}}{0.40\; {\rm m\cdot s^{-1}}} \\ &= 0.50\; {\rm s}\end{aligned}\).

Also because air resistance on the coin is negligible, the vertical acceleration of the coin will be constantly \(a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}\). If the table is level, the initial vertical velocity of the coin \(u_{y}\) will be \(0\; {\rm m\cdot s^{-1}}\). Apply the SUVAT equation \(x = (1/2)\, a\, t^{2} + u\, t\) to find the vertical displacement of the coin during the \(t = 0.5\; {\rm s}\) that it is in the air:

\(\begin{aligned}x_{y} &= \frac{1}{2}\, a_{y}\, t^{2} + u_{y}\, t \\ &= \frac{1}{2} \, (-9.81\; {\rm m\cdot s^{-2}})\, (0.5\; {\rm s})^{2} + (0\; {\rm m\cdot s^{-1}})\, (0.50\; {\rm s}) \\ &\approx (-1.3)\; {\rm m}\end{aligned}\).

In other words, the coin lands approximately \(1.3\; {\rm m}\) below the table edge. Hence, the height of the table would be approximately \(1.3\; {\rm m}\!\).

To solve the given problem, we can use the right-hand rule that applies to a positive charge, with the magnetic field going into the page. The given problem can be represented as shown below:

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Newton's second law allows to find the result for the friction coefficient of the street is:

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the body.

           ∑ F = m a

Where bold indicates vectors, m is to mass and acceleration.

In the attached we see a free body diagram, it is a diagram of the forces without the details of the body, the x-axis is parallel to plane also shown with the positive in the direction of movement, going down the plane and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight.

        Sin θ = \(\frac{W_x}{W}\)  

        cos θ = \(\frac{W_y}{W}\) / W

        Wₓ = W sin θ

         \(W_y\)  = W cos θ

We write Newton's second law for each axis.

y-axis

         N- \(W_y\)  = 0

         N = mg cos θ

x-axis

        Wₓ - fr = ma

Since they indicate that the body goes down at a constant speed, the acceleration is zero.

         W sin θ = fr

The friction force is the macroscopic representation of the interactions between the two surfaces and the formula.

         fr = μ N

we substitute.

          fr = μ mg cos θ

         mg sin θ = μ cos θ  

        μ = tan θ

Let's calculate.

         μ = tan 38.0

         μ = 0.78

In conclusion using Newton's second law we can find the results for the friction coefficient of the street is:

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Answer:

31.6

Explanation:

velocity÷ time

212 ÷ 6.7 = 31.6m/s

We can calculate the refractive index by substituting the values into the formula: n = √(1 - (2.774 × 10^6 / f)^2). The refractive index of the ionospheric layer can be determined using the formula n = √(1 - (f_ce / f)^2)

(a) The refractive index of the ionospheric layer can be determined using the formula n = √(1 - (f_ce / f)^2), where n represents the refractive index, f_ce is the electron gyrofrequency, and f is the frequency of the incident wave.

The electron gyrofrequency (f_ce) can be calculated using the formula f_ce = 8.978 × √(N), where N is the electron density. Substituting the given electron density value, we have f_ce = 8.978 × √(24.536 × 10^11) ≈ 2.774 × 10^6 Hz.

Now, we can calculate the refractive index by substituting the values into the formula: n = √(1 - (2.774 × 10^6 / f)^2).

(b) The critical frequency for the communication link can be determined using the formula f_c = f_ce / sin(θ), where f_c represents the critical frequency and θ is the angle of incidence. Substituting the given angle of 60°, we have f_c = 2.774 × 10^6 Hz / sin(60°).

(c) The maximum usable frequency (MUF) can be calculated using the formula MUF = f_c / sin(θ). Substituting the critical frequency and angle of incidence given in parts (b) and (a), respectively, we can find the MUF.

(d) Transmissions would fail at the frequencies of 10 MHz and 30 MHz because they are below the critical frequency. The critical frequency represents the maximum frequency that can be reflected back to Earth by the ionospheric layer. If the frequency of the transmission is below the critical frequency, the wave would penetrate through the ionosphere and not be reflected back, leading to a failed transmission.

(e) The ionosphere bends high-frequency radio waves towards Earth due to the phenomenon of refraction. When a radio wave encounters the ionosphere, which is composed of charged particles, it experiences a change in speed and direction. This change in speed and direction is due to the varying density and composition of the ionosphere at different altitudes.

As the radio wave passes through the ionosphere, its path is curved downward towards the Earth's surface. This bending phenomenon occurs because the refractive index of the ionosphere is greater than that of the surrounding vacuum or atmosphere. The higher the frequency of the radio wave, the greater the bending effect due to the higher electron density in the ionosphere at higher altitudes.

This bending of high-frequency radio waves allows for long-distance communication by enabling the waves to travel beyond the line-of-sight. It plays a crucial role in long-distance radio communication, especially in areas where direct line-of-sight transmission is obstructed by the Earth's curvature or other obstacles.

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The turtle's position and direction appropriately after each instruction to ensure accurate drawing. You can also customize the colors, sizes, and shapes to enhance the visual appearance of the AND gate.

To draw an AND gate using turtle graphics, you can use the following set of instructions:

Set up the turtle:

a. Set the turtle's initial position.

b. Set the turtle's pen color and size.

Draw the first input line:

a. Move the turtle forward to the starting point of the line.

b. Draw a straight line segment to represent the first input.

Draw the second input line:

a. Move the turtle to the starting point of the second line.

b. Draw a straight line segment to represent the second input.

Draw the output line:

a. Move the turtle to the starting point of the output line.

b. Draw a straight line segment to represent the output.

Draw the logic gate shape:

a. Move the turtle to the starting point of the gate.

b. Draw a rectangle to represent the gate.

c. Add any necessary labels or symbols to indicate it as an AND gate.

Add connections between lines and gate:

a. Move the turtle to the intersection point of the first input line and the gate.

b. Draw a small line segment to connect the input line to the gate.

c. Repeat the above step for the second input line and the gate.

d. Draw a small line segment to connect the output line to the gate.

Repeat the above steps as necessary to draw multiple AND gates or any additional components.

Remember to adjust the turtle's position and direction appropriately after each instruction to ensure accurate drawing. You can also customize the colors, sizes, and shapes to enhance the visual appearance of the AND gate.

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Answer:

The car's initial velocity is 195.96 m/s

Explanation:

Given;

the length of the mark, x = 9.6 km = 9600 m

acceleration of the car, a = -2 m/s²

Apply the kinematic equation below

v² = u² + 2ax

where;

v is the final velocity = 0

u is the initial velocity

simplify the equation and you will obtain;

0 = u² + 2ax

0 = u² + 2(-2)(9600)

0 = u² - 38400

u² = 38400

u = √38400

u = 195.96 m/s

Therefore, the car's initial velocity is 195.96 m/s

From around 6500 K at the bottom to 4000 K at the top, the photosphere's temperature changes (11,000 and 6700 degrees F, 6200 and 3700 degrees C). Granulation covers a large portion of the photosphere.

The results of these many techniques indicate that the Sun's surface actually has a temperature of about 5,800 kelvin (9,980 degrees Fahrenheit [5,520 degrees Celsius]). The photosphere contains areas of darkness known as sunspots, where the temperature is comparatively lower—about 3800K. The photosphere is therefore the layer that is the coldest. It is 15 million K inside the sun (27 million degrees Fahrenheit). It is 5780 K at the surface (9900 degrees Fahrenheit).

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