Convert 4 radians to degrees A. 242.6° B. 9.6° C. 229.2° D. 6.8°

At which labeled points would the skateboarder have the most kinetic energy?

a. W and X

c. Y and Z

d. W and Z

Here's the picture for everyone :0

The skateboarder will have the most kinetic energy at point X and Y.

Kinetic energy is the energy possessed by an object due to its motion. Kinetic decreases with decrease in speed of the object.

K.E = ¹/₂mv²

In the given diagram, the skateboard will have have the highest speed at point X and Y. Kinetic energy is directly proportional to speed  

Thus, we can conclude that the skateboarder will have the most kinetic energy at point X and Y.

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stress is the correct answer to this problem

In comparing a kilogram of feathers and a kilogram of bricks, they have the same mass but do not have the same volume or density. Bricks are more dense than feathers.

The comparison between a kilogram of feathers and a kilogram of bricks deals with two primary concepts: mass and density. When we are comparing a kilogram of feathers and a kilogram of bricks, they have the same mass because a kilogram is a measure of mass. Therefore, option (a) is correct.

However, they do not have the same volume and the density of each is different. Density is defined as mass per unit volume. Bricks are denser than feathers. This means that for a given mass, bricks take up less space or volume than feathers. Therefore, neither option (b) nor options (c) and (d) are correct.

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Answer: The given statement is true.

Explanation:

A change that is observable is known as a stimuli. Every individual has their own way of thinking and interpreting things.

For example, Lisa got a chocolate as a reward for completing her homework whereas his elder brother gets upset with his mom because of this.

So, in this situation Lisa will be happy as she has got a reward whereas his brother is taking this thing in a negative way by getting jealous or upset about it.

Hence, we can conclude that the statement one's perception depends on the way the brain interprets the stimuli, is true.

Only about 11% of the volume of the iceberg is exposed.

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The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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Given:

Density of Ice = ρ_ice = 0.92 g/cm³

Densiy of Seawater = ρ_seawater = 1.03 g/cm³

Asked:

Volume of the exposed iceberg = ΔV = ?

Solution:

We will use Archimedes' principle to solve the problem as follows:

[tex]w = F[/tex]

[tex]mg = \rho_{seawater} g V'[/tex]

[tex]\rho_{ice} \ g V = \rho_{seawater} \ g V'[/tex]

[tex]V' = ( \rho_{ice} \div \rho_{seawater} ) V[/tex]

[tex]V' = (0.92 \div 1.03) V[/tex]

[tex]V' = \frac{92}{103} V[/tex]

[tex]\texttt{ }[/tex]

[tex]\Delta V = V - V'[/tex]

[tex]\Delta V = V - \frac{92}{103} V[/tex]

[tex]\Delta V = \frac{11}{103}V[/tex]

[tex]\Delta V \approx 11 \% V[/tex]

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Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

The change in thermal energy of a wooden block with a mass of 20.0 kg and a specific heat of 1700 J/kg°C warming from 15.0°C to 25.0°C is 340,000 Joules (or 340 kJ).

To find the change in thermal energy of the block as it warms from 15.0°C to 25.0°C, we can use the formula for specific heat capacity, which is Q = mcΔT. Here, Q represents the thermal energy transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the given wooden block, we know:

Now we can calculate the change in thermal energy:

Q = mcΔT = (20.0 kg)(1700 J/kg°C)(10.0°C) = 340,000 J

Therefore, the change in thermal energy of the wooden block is 340,000 Joules (or 340 kJ).

In the darkest aquatic ecosystem, light cannot reach in the The open ocean .

Aquatic ecosystems, which include wetlands, rivers, lakes, and coastal estuaries, are vital to human economy and health as well as crucial components of Earth's dynamic processes.

Wetlands are the link between land and water, operating as natural filters to filter out pollutants, control flooding, and serve as cnidiaries for a variety of aquatic animals.

Estuaries, lakes, and rivers are crucial hubs for transportation, recreation, and biodiversity. Wetlands, rivers, lakes, and coastal estuaries are all aquatic ecosystems—critical elements of Earth’s dynamic processes and essential to human economies and health.

Therefore, In the darkest aquatic ecosystem, light cannot reach in the The open ocean .

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Final answer:

In a compound microscope, the ocular lens is located near the eye for viewing and the objective lens is closer to the specimen for initial magnification.

Explanation:

The two lenses located in a compound microscope used in most classrooms today are the ocular lens and the objective lens. The ocular lens, or eyepiece, is found at the top of the microscope and is the lens that you look through with your eye. It often contains multiple lenses within a cylindrical barrel and contributes to magnifying the image. The second lens, known as the objective lens, is located closer to the specimen. It is a convex lens with a short focal length that provides the initial magnification, typically ranging from 5× to 100×. The objectives are commonly mounted on a revolving nosepiece and are designed to be parfocal, allowing the sample to remain in focus when switching between different objective lenses. Both the ocular and objective lenses work together to produce a highly magnified, enlarged image that is far enough from the observer to be easily viewed.

The relevant equations we can use in this problem are:

d = v0 t + 0.5 a t^2

v^2 = v0^2 + 2 a d

a. how high does it go?

We are to find for the distance d. The formula is:

v^2 = v0^2 + 2 a d

where v is final velocity = 0 at the peak, v0 = 25 m/s, a is negative gravity since opposite to direction = - 9.8 m/s^2, d is height achieved = ?

0 = (25)^2 + 2 (-9.8) d

d = 31.89 m

Therefore the maximum height is about 31.89 meters

b. how long is it in the air

We use the formula:

d = v0 t + 0.5 a t^2

in going up, a = -g = -9.8 m/s^2, find for time t:

31.89 = 25 t + 0.5 (- 9.8) t^2

-4.9 t^2 + 25 t = 31.89

t^2 – 5.1 t = 6.51

Completing the square:

(t – 2.55)^2 = 6.51 + 6.5025

t – 2.55 = ± 3.61

t = -1.06, 6.16

Since time cannot be negative, therefore:

t = 6.16 seconds

However this is not the total time yet, because this only accounts for the time going up.

The time going up and going down are equal therefore:

t(total) = 2 t

t(total) = 12.32 seconds

Therefore the foul ball spent 12.32 seconds in the air.

Final answer:

The planet which is approximately 20 times farther from the Sun than Earth is Uranus, orbiting the Sun at an average distance of about 19.8 AU.

Explanation:

The question at hand is: Which planet is approximately 20 times farther from the sun than Earth is? To answer this, we can refer to the concept that the light a planet receives from the Sun decreases with the square of the distance from the Sun. Considering the average distance from Earth to the Sun is about 150 million kilometers (1 Astronomical Unit, AU), a planet that is 20 times farther from the Sun would be at 20 AU.

In our solar system, the planet that falls into this category is Uranus, which orbits the Sun at an average distance of approximately 19.8 AU. This makes Uranus about 20 times farther from the Sun than Earth is, hence receiving much less sunlight compared to Earth.

Answer:88 days

Explanation: The sidereal.orbital period is calculated from observations. Mercury is observed to come into

superior conjunction with the Sun every 116 days. This is called the "Mercury's synodic year". The sum of the inverse of earth's sidereal year and merciry's synod year equals the inverse of mercury's sidereal year. That is:

1/365.24 + 1/116 = 1/P where P is mercurry's sidereal year.

Therefore, P equals 88 days.

The earth's velocity as it orbits the sun is about 2.99 * 10⁴ m/s. The angular velocity is approximately 7.27 *10⁻⁷ rad/s, and the centripetal acceleration is around 5.93 m/s².

The earth's velocity (V), angular velocity (ω), and centripetal acceleration (A) can be calculated using the radius of the earth's orbit (r) around the sun and the period of orbit (T).

1. Earth's velocity:
As the earth orbits the sun, it covers a distance of 2πr (the circumference of a circle) in one complete orbit (T). Using the formula V = Δd/Δt, where Δd is the displacement in distance and Δt is the change in time, we can find earth's velocity to be approximately 2π* (1.5 *10¹¹ m)/(1 year) = 2.99 * 10⁴ m/s.

2. Angular velocity:
The earth's angular velocity can be determined by angular displacement over time. Since the earth completes 2π radians in one period of orbit (1 year), ω = Δθ/Δt = 2π/T , which when converted to SI units is approximately 7.27 *10⁻⁷ rad/s.

3. Centripetal acceleration:
Centripetal acceleration is determined using the formula A = V²/r. Using V = 2.99 * 10⁴ m/s and r = 1.5 * 10¹¹ m, we find A is approximately 5.93 m/s².

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The resulting angular speed of the student and the stool system after catching the tossed mass is 0.354 rad/s, calculated using the conservation of angular momentum.

To find the resulting angular speed of the student and the stool after catching the mass, we need to apply conservation of angular momentum since no external torques act on the system. Initially, the system (student and stool) is at rest, so the initial angular momentum is 0. Upon catching the mass, the student-stool system gains angular momentum from the tossed mass.

The angular momentum of the tossed mass just before it is caught is given by L = mvr, where m is the mass of the object, v is its tangential velocity, and r is the distance from the axis of rotation to where the mass is caught. Plugging in the given values, L = (1.5 kg)(2.3 m/s)(0.42 m) = 1.449 kg·m²/s. This is the angular momentum that the system will have after the mass is caught because of the conservation of angular momentum.

To find the angular speed ω of the student-stool system after catching the mass, we use the equation L = Iω, where I is the moment of inertia of the system, and rearrange it to ω = L/I. The moment of inertia of the system remains 4.1 kg·m² since we are told to ignore any potential changes due to the mass being caught. Thus, ω = 1.449 kg·m²/s / 4.1 kg·m² = 0.354 rad/s.

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Answer : The unbalanced force necessary to produce the movement is, 117.6 N

Solution :

Force : It is defined as the force is equal to the product of the mass of an object and the acceleration of an object.

Formula used :

[tex]F=m\times a[/tex]

where,

F = force

m = mass = 98 Kg

a = acceleration = [tex]1.2m/s^2[/tex]

Now put all the given values in the above formula, we get the unbalanced force necessary to produce the movement.

[tex]F=(98Kg)\times (1.2m/s^2)=117.6\text{ Kg }m/s^2=117.6N[/tex]

Therefore, the unbalanced force necessary to produce the movement is, 117.6 N

The unbalanced force necessary to produce the movement of a 98-kilogram man accelerating forward at a rate of 1.2 m/s² is 117.6 Newtons.

The unbalanced force necessary to produce the movement of a 98-kilogram man accelerating forward at a rate of 1.2 m/s² can be calculated using Newton's second law of motion:

F = m x a

Where F is the force, m is the mass, and a is the acceleration.

Substituting the given values:

F = 98 kg x 1.2 m/s²

F = 117.6 N

Therefore, the unbalanced force necessary to produce the movement is 117.6 Newtons.

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