A clown car is traveling with a velocity of 35 m/s and has a mass of 957 kg. Identify the energy (PE or KE) and calculate the it.

equal in magnitude but opposite in direction

The near point of a human eye is about a distance of 25 cm.

The closest distance that an object may be viewed clearly without straining is known as the near point of the eye.

This distance (the shortest at which a distinct image may be seen) is 25 cm for a typical human eye.

The closest point within the accommodation range of the eye at which an object may be positioned while still forming a focused picture on the retina is also referred to as the near point.

In order to focus on an item at the average near point distance, a person with hyperopia must have a near point that is further away than the typical near point for someone of their age.

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Therefore, the magnetic force on an electron is 1.175 x 10⁻¹⁴N in the positive y-direction.

The force experienced by a moving charge in a magnetic field is given by the Lorentz force law. Since the charge on a proton is positive and that on an electron is negative, the direction of the magnetic force experienced by each is different.In this question, we need to find the magnitude and direction of the magnetic force on a proton and an electron as they travel from the Sun to the Earth.

Let's first calculate the magnetic force on a proton:

F = qvBsinθ

where q = charge of the particle

v = velocity of the particle

B = magnetic field strength

θ = angle between the velocity of the particle and the magnetic field = 90° (since the proton is moving perpendicular to the magnetic field)

Therefore,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= 1.175 x 10⁻¹⁴ N

Direction of magnetic force on a proton:The direction of the magnetic force on a proton can be found using the right-hand rule. According to this rule, if we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the negative y-direction.

Therefore, the magnetic force on a proton is 1.175 x 10^-14 N in the negative y-direction.

- Now, let's calculate the magnetic force on an electron:

Again using the Lorentz force law,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= -1.175 x 10⁻¹⁴ N (the negative sign indicates that the direction of the magnetic force is opposite to that of the proton)

Direction of magnetic force on an electron:Again using the right-hand rule, we can find the direction of the magnetic force on an electron. If we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the positive y-direction.

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Answer:

A....It is a crime sene

Explanation:

The orientation of an mRNA codon corresponding to a tRNA anticodon is best described as A: antiparallel.

An anticodon is a three-base sequence, paired with a particular amino acid, that a tRNA molecule brings to the associated codon of the mRNA during translation. The anticodon sequence is said to be complementary to the mRNA, using base pairs in the anti-parallel direction.  The tRNA anticodon 3-UAC-5 will make pair with the mRNA codon.

Thus, the orientation of an mRNA codon relative to a tRNA anticodon is said to be antiparallel.

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25mn type of focal length was primarily used in citizen kane.

The focal lengths and lenses used by 19 notable directors are analyzed in a new video essay by Wolfcrow, starting with Orson Welles' use of 25mm for Citizen Kane and 18mm for Touch of Evil.

In Orson Welles' Citizen Kane, deep focus was frequently used to provide the audience with insight into Charles Foster Kane's thoughts or to reveal more about his life. With deep focus, the foreground, middle ground, and backdrop are all sharply in focus at once.

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Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

Hope it helps! Please mark brainliest.

Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

The  example that represents a class 2 lever is: D. a screwdriver opening a paint can.

The load is situated in a class 2 lever halfway between the fulcrum and the effort. The screwdriver's resting place against the paint can's rim serves as the fulcrum in this scenario.

The paint can's lid serves as the load, and the force exerted by the hand on the screwdriver's handle serves as the effort. A class 2 lever can be identified by the load being situated between the fulcrum and the effort.

Tweezers, salad tongs, and opening and closing a car door are not examples of class 2 levers. Class 1 levers include things like tweezers and salad tongs, whereas class 3 levers include things like car doors that open and close.

Therefore the correct option is D.

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Answer:

A. It must stay the same, but kinetic energy (KE) can be transformed to PE and PE can be transformed to KE within the system.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Potential energy (PE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

\( P.E = mgh\)

Where,

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

\( K.E = \frac{1}{2}MV^{2}\)

Where;

Furthermore, the total energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Total energy = P.E + K.E

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

In this scenario, a system has both potential energy (PE) and kinetic energy (KE).

According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.

a) The tension in the rope is 123.9 N.

b) The moment of inertia of the wheel is 0.09 kg m².

c) The angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.

(a) To determine the tension in the rope, we need to analyze the forces acting on the object. There are two forces: the force of gravity pulling the object down the incline and the tension force pulling the object up the incline.

The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline.

The parallel component causes the object to accelerate down the incline, while the perpendicular component is balanced by the normal force of the incline.

The tension force is responsible for the object's acceleration down the incline, so we can set up the following equation:

T - mg sin(theta) = ma

where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, theta is the angle of the incline, and a is the acceleration of the object down the incline.

Putting in the given values, we get:

T - (12.5 kg)(9.81 m/s²)(sin(37°)) = (12.5 kg)(2.00 m/s²)

Solving for T, we get:

T = 123.9 N

Therefore, the tension in the rope is 123.9 N.

(b) To determine the moment of inertia of the wheel, we can use the following equation:

I = (1/2)MR²

where I is the moment of inertia, M is the mass of the wheel, and R is the radius of the wheel.

Putting in the given values, we get:

I = (1/2)(12.5 kg)(0.12 m)²

 = 0.09 kg m²

Therefore, the moment of inertia of the wheel is 0.09 kg m².

(c) To determine the angular speed of the wheel after 3.50 s, we can use the following equation:

ω = ω₀ + αt

where ω is the final angular speed, ω₀ is the initial angular speed (which is zero in this case), α is the angular acceleration, and t is the time.

We can find the angular acceleration using the following equation:

α = a/R

where a is the acceleration of the object down the incline (which we already know) and R is the radius of the wheel.

Putting in the given values, we get:

α = 2.00 m/s² / 0.12 m

  = 16.7 rad/s²

Putting in the values for α and t, we get:

ω = 0 + (16.7 rad/s²)(3.50 s)

   = 58.5 rad/s

Therefore, the angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.

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Answer:

Force

Explanation:

The amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

To calculate the energy required to lift a 2.5 kg object through a vertical distance, we need to consider the gravitational potential energy formula:

Potential energy (PE) = mass (m) × gravity (g) × height (h)

Where:

m = 2.5 kg (mass of the object)

g = 9.8 m/s² (acceleration due to gravity on Earth)

h = ? (height)

First, let's find the height (h) by rearranging the formula:

h = PE / (m × g)

Now, let's calculate the potential energy (PE) needed to lift the object. We are given that the power of the bulb is 0.20 kW, and we want to find the energy required for one minute. To convert kilowatts (kW) to joules (J), we multiply by the conversion factor of 3,600 (60 seconds × 60 minutes):

Energy (E) = power (P) × time (t)

E = 0.20 kW × 1 min × 3,600 J/kW

Now, we can substitute the values into the equation to find the height:

h = (0.20 kW × 1 min × 3,600 J/kW) / (2.5 kg × 9.8 m/s²)

Calculating the expression on the right side:

h ≈ 0.20 × 1 × 3,600 / (2.5 × 9.8) ≈ 29.03 meters (rounded to two decimal places)

Therefore, the amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

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Answer:

B

Explanation:

Answer:

B

Explanation:

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

d = 0.87 m

Answer:

* he new moon phase when the position is       Sun - Moon - Earth,

* have of the Full Moon when the position is     Sun - Earth - Moon,

*All the phases of the moon are governed by the movement of the Moon around the Earth.

Explanation:

In the solar system, the planets revolve around the sun, which is much more massive, in the case of the Earth it is more massive than its satellite, therefore the Moon revolves around the Earth in a period of approximately 28 days.

It is said that the moon is in the new moon phase when the position is Sun - Moon - Earth, so the moon cannot be seen

It is in the phase of the Full Moon when the position is

                  Sun - Earth - Moon, in this case the moon can be observed by the light reflected from it.

All the phases of the moon are governed by the movement of the Moon around the Earth.

Torque is a measure of the turning force applied to an object about a rotational axis. It is calculated as the product of the force applied to the object and the distance from the axis of rotation at which the force is applied.

Use the equation slope = mod to calculate the unknown mass (mo) for parcels A and C, setting d = 1.0 m, and parcels G and H, setting d = 1.5 m.

Record all data, tables, and four graphs for analysis.

The experiment demonstrates the application of torque in determining unknown masses and provides valuable insights into the concept of torque in physics.

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The electron number cannot be used instead because electrons can be removed from or added to an atom (option C)

The element of an atom is determined by its proton count, while the electron count can exhibit variability. Take, for instance, a sodium atom, which encompasses 11 protons and 11 electrons. However, it has the capacity to relinquish one electron, transforming into a sodium ion housing only 10 electrons.

This occurs due to the relatively loose binding of electrons to the nucleus, enabling their removal through the influence of an electric field or alternative mechanisms.

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Answer:

9.461 × 10^12 Km s

..............................

Answer:

D

Explanation:

Nothing can enter or leave so it remains constant

Answer: v = 0.9 m/s

Explanation:

Due to the horizontal jump, the initial velocity at y-component (Vy) is 0 m/s.

The initial velocity at x-component (Vx) is a constant velocity in the distance 1.5 meters during 1.7 second.

Therefore,

Vx = 1.5 / 1.7 = 0.9 m/s

Vy = 0 m/s

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

399.5 Watts.

Explanation:

From the question given above, the following data were obtained:

Potential difference (V) = 470 V

Current (I) = 0.85 A

Time (t) = 1 ms

Power (P) =?

Electrical power is defined by the following equation:

Power (P) = potential difference (V) × current (I)

P = IV

Using the above formula, the power can be obtained as follow:

Potential difference (V) = 470 V

Current (I) = 0.85 A

Power (P) =?

P = IV

P = 470 × 0.85

P = 399.5 Watts

Therefore, the power is 399.5 Watts.

Given:

Resistance, R = 7.17 Ohms

Capacitance, C = 2.59 mF

Inductance, L = 213.91 mH

Frequency, f = 60 Hz

To find:

The phase angle.

Explanation:

The inductive reactance can be calculated as:

The capacitive reactance can be calculated as:

The phase angle is given as:

Final answer:

The phase angle is -89.56 °.

In the case of rain, the static friction is halved, meaning the new static friction coefficient is 0.5μs, while the kinetic friction is reduced to one-third, resulting in a new kinetic friction coefficient of (1/3)μk.

To determine the maximum velocity at which you can make a turn without slipping in the rain at the intersection, we need to consider the changes in friction.

Let's assume the normal static friction and normal kinetic friction are represented by μs and μk, respectively.

In the case of rain, the static friction is halved, meaning the new static friction coefficient is 0.5μs, while the kinetic friction is reduced to one-third, resulting in a new kinetic friction coefficient of (1/3)μk.

To avoid slipping during the turn, we need to ensure that the centripetal force required for the turn is less than or equal to the maximum frictional force available.

The centripetal force is given by the equation mv²/r, where m is the mass, v is the velocity, and r is the radius of the turn.

The maximum frictional force in the rain can be calculated as (0.5μs)mg, where g is the acceleration due to gravity.

Thus, to avoid slipping, we set the centripetal force equal to the maximum frictional force:

mv²/r = (0.5μs)mg

Simplifying the equation, we find:

v = √(0.5μsgr)

By plugging in the values for μs, g, and the radius of the turn, we can calculate the maximum velocity.

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Answer: B

Explanation: ik trust me its right :)

Answer:

B)  5 gm

Explanation:

.5 g / cm^3   * 10 cm^3  =  5 gm     ( see how the 'cm^3' cancels out?)

0.3 m/s was the original speed of the bullet.

m1v1=m2(v1+v2)

v1=m2(v1+v2)/m1

v1=4.9×0.6÷9.7

v1=0.3 m/s

Speed is a scalar variable that expresses how much an object's location changes over time or how much it changes per unit of time. It is frequently abbreviated as "s." The distance traveled by an object over a period of time divided by the length of the period gives the average speed of the object over that period.

The speed-related metrics are time divided by distance. The most prevalent unit of speed in daily life is the kilometer per hour (kph), while the SI unit of speed is the meter per second (m/s).

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Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Explanation: