A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after being released?
A.
49 joules
B.
98 joules
C.
1.0 × 102 joules
D.
1.1 × 102 joules

Answer:

D. Electrons mover from one to another

Explanation:

Static electricity occurs when electrons are transferred from one object to another.

Answer:

b

Explanation:

Answer:

\(V_y = 16.44\ m/s\)

\(V_y = 7.32\ m/s\)

Explanation:

Given

\(\theta = 24\)

\(Velocity\ (V) = 18m/s\)

Required

Determine the vertical and horizontal components

The vertical (Vy) and horizontal (Vx) components is calculated as thus:

\(V_x = Vcos\theta\)

\(V_y = Vsin\theta\)

Calculating Vertical Components:

\(V_y = Vsin\theta\)

\(V_y = 18 * sin24\)

\(V_y = 18 * 0.40673664307\)

\(V_y = 7.32125957526\)

\(V_y = 7.32\ m/s\) --- Approximated

Calculating Horizontal Components:

\(V_x = Vcos\theta\)

\(V_x = 18 * cos24\)

\(V_x = 18 * 0.91354545764\)

\(V_x = 16.4438182375\)

\(V_x = 16.44\ m/s\) --- Approximated

The electric potential at a point due to two charges can be determined by adding the electric potentials from each charge separately using the equation V = k * q / r, where V is the electric potential, k is the electrostatic constant, q is the charge, and r is the distance from the charge to the point.

The electric potential at a point due to two charges can be calculated by summing the electric potentials due to each charge separately. The electric potential, also known as voltage, is a scalar quantity that represents the amount of electric potential energy per unit charge at a given point.

To find the total electric potential at point P, 1.0 cm away from q₂, we need to consider the electric potentials due to both charges. The electric potential due to a point charge is given by the equation V = k * q / r, where V is the electric potential, k is the electrostatic constant (9 x 10⁹ Nm²/C²), q is the charge, and r is the distance from the charge to the point.

Let's denote the charges as q₁ and q₂. Since point P is 1.0 cm away from q₂, we can use the equation to calculate the electric potential due to q₂. Then, we can sum it with the electric potential due to q₁ to find the total electric potential at point P.

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Range of  v0( 0 in subscript) is 2.67 m/s to 4.67 m/s range for marble peed to land in the hole.

There is no vertical component of velocity here, as per the problem.

h = 2.75 m

But horizontal distance x = v0t (v in subscript)

h = 1/2 gt2(t square)

⇒ t =\(\sqrt{ 2h/g}\)

Putting this in the 1st equation,

we get v0 = x/t =x\(\sqrt{9/2h}\)

Since the hole is 2m away and 1.5m wide ,

x lies between 2m and 3.5m

Lower bound of v0 = 2 × \(\sqrt{9.8/2 * 2.75}\)  = -2.67 m/s

Upper bound of v0 = 3.5 × \(\sqrt{9.8/2 * 2.75}\) = -4.67 m/s

Hence, the range of v0 is 2.67 m/s to 4.67 m/s.

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When the ball is halfway to the top of its flight after being thrown vertically upwards with an initial velocity v and kinetic energy Ek, its velocity is zero, and its kinetic energy is also zero.

As the ball moves upwards, it experiences a decrease in velocity due to the gravitational force acting in the opposite direction. At the halfway point of its flight, the ball momentarily comes to a stop before changing direction and descending.

At this point, its velocity is zero. Since kinetic energy is proportional to the square of velocity, when the velocity is zero, the kinetic energy is also zero. Therefore, when the ball is halfway to the top of its flight, its velocity is zero, and its kinetic energy is also zero.

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Take the direction of motion to be the positive direction. The crate slows to a rest from 30.0 m/s in a matter of 5.0 s, so it has acceleration a such that

0 = 30.0 m/s + a (5.0 s)   →   a = -6.0 m/s²

At the moment its speed is 0, the crate has a net force of s + f acting in negative direction, where s and f denote the magnitudes of the stopping force (s = 250 N) and the friction force, respectively. By Newton's second law, we have

(-s) + (-f) = (100 kg) (-6.0 m/s²)

250 N + f = 600 N

f = 350 N

The friction force is proportional to the normal force by a factor of µ, the coefficient of kinetic friction. There is no movement in the up- and downward directions, so Newton's second law says

(-w) + n = 0

where w is the weight of the crate and n is the magnitude of the normal force. So

n = w = (100 kg) (9.80 m/s²) = 980 N

Then

f = µ n

350 N = µ (980 N)

µ = (350 N) / (980 N) ≈ 0.357

Answer:

A straight line with no slope.

Answer:

C on edge 2021

Explanation:

Given data

*The given complete set has a mass is m = 3467 kg

*The speed of the cart with a horizontal velocity relative to the floor is v_1 = 0.02 m/s

*The speed of the librarian is v_2 = 2.40 m/s

The formula for the librarian's mass (m_l) is given by the conservation of momentum as

Substitute the known values in the above expression as

Hence, the librarian's mass is m_l = 28.89 kg

The eccentricity of the object's orbit can be determined by using the ratio of its maximum angular speed to its minimum angular speed.

Let's denote the maximum angular speed as ω_max and the minimum angular speed as ω_min. We are given that the ratio of these two speeds is n:

n = ω_max / ω_min

The angular speed (ω) is related to the angular momentum (L) and the moment of inertia (I) of the object by the equation:

L = Iω

Since the object moves in an inverse square centripetal force field, the angular momentum (L) is conserved. Therefore, we can write:

L_max = L_min

Iω_max = Iω_min

The moment of inertia (I) can be expressed as the product of the mass (m) and the square of the distance (r) from the object to the axis of rotation:

I = mr^2

Substituting this into the equation above, we get:

m(r^2)ω_max = m(r^2)ω_min

Canceling out the mass (m) and the square of the distance (r^2), we obtain:

ω_max = ω_min

This implies that the maximum and minimum angular speeds are equal, contradicting the given ratio n = ω_max / ω_min. Therefore, there must be an error in the question or the provided information.

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Answer:

yes

Explanation:

as speed is a scalar quantity and velocity is a vector quantity. Which means speed only depends on the magnitude at which it travels but velocity depends on the speed at which travel and also the direction to where the object travels.

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The landing velocity of the object is 31 m/s.

The given parameters;

height from which the object is released, h = 49 m

The landing velocity of the object is calculated by applying third equation of motion as shown below;

v² = u² + 2gh

where;

The landing velocity of the object is calculated as;

v² = 0 + 2(9.8)(49)

v² = 960.4

v = √960.4

v = 30.99 m/s

v ≈ 31 m/s

Thus, the landing velocity of the object is 31 m/s.

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Answer:

what scenarios?  but kinetic is energy in motion (like a ball rolling) and potential is stored energy (ball laying on ground)

example (kinetic to potential) : baseball flying and and lands on the ground then stops rolling

roller coaster rolling down a hill then stopping (kinetic to potential)

roller coaster on top of a hill and then going down (potential to kinetic)

Explanation:

To determine the position u of the mass at any time t, we can use the equation of motion for a mass-spring system without damping:n m * u''(t) + k * u(t) = 0

m = 2 lb / (32.2 ft/s^2) = 0.062 lb·s^2/ft

The spring constant k can be determined using Hooke's law:

k = F / x

where F is the force exerted by the mass and x is the displacement. In this case, the force F is the weight of the mass, and the displacement x is 6 in:

k = (2 lb) / (6 in) = (2 lb) / (6 in) * (1 ft/12 in) = 0.111 lb/ft

The equation of motion now becomes:

0.062 * u''(t) + 0.111 * u(t) = 0

To solve this second-order linear homogeneous differential equation, we assume a solution of the form u(t) = A * cos(ωt + φ).

Substituting this assumed solution into the equation of motion, we get:

-0.062 * A * ω^2 * cos(ωt + φ) + 0.111 * A * cos(ωt + φ) = 0

-0.062 * ω^2 + 0.111 = 0

Solving for ω, we get:

ω = sqrt(0.111 / 0.062) = 2.258 rad/s

From ω, we can determine the frequency f and period T:

f = ω / (2π) = 2.258 / (2π) ≈ 0.359 Hz

T = 1 / f ≈ 2.786 s

The amplitude A is determined by the initial conditions. When the mass is pulled down an additional 3 in (0.25 ft) and released, it reaches its maximum displacement, so A = 0.25 ft.

Therefore, the position u of the mass at any time t is given by:

u(t) = 0.25 * cos(2.258t)

To draw the graph of u(t), plot the position u on the y-axis and time t on the x-axis, using the equation u(t) = 0.25 * cos(2.258t). The graph will be a cosine wave with an amplitude of 0.25 ft.

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The impact of forest fires on wood production is expected to result in an increase in the equilibrium price and a decrease in the equilibrium quantity of wooden desks.

Forest fires may reduce wood production, which is needed to make wooden desks. Thus, wooden desk equilibrium price and quantity would change:

Equilibrium Price: Less wood means higher desk production costs. Wooden desks would cost more with greater production expenses. Thus, wooden desk equilibrium prices should climb.

Equilibrium Quantity: Wood shortages make it harder to make wooden workstations. Wooden desks may become scarcer due to supply constraints. Thus, wooden desk equilibrium quantity should decrease.

In conclusion, forest fires will reduce wood production and raise the equilibrium price of wooden workstations. Wood, a key ingredient in wooden desks, is scarcer.

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Becoming an effective technician requires a combination of technical and soft skills. By following these procedures, technicians can develop the skills they need to excel in their roles and deliver high-quality results.

1. Acquire the necessary technical knowledge: To become an effective technician, you need to have a strong foundation in the technical knowledge required for your job. This may involve pursuing formal education or training, reading technical manuals, attending seminars, or participating in on-the-job training.

2. Develop problem-solving skills: Effective technicians are skilled problem-solvers who can quickly identify issues and find solutions. To develop these skills, technicians should practice critical thinking, analyze problems systematically, and explore alternative solutions.

3. Improve communication skills: Technicians must communicate clearly and effectively with clients, colleagues, and supervisors. They should practice active listening, use clear and concise language, and ask questions to clarify information.

4. Foster teamwork skills: Technicians often work in teams to complete complex projects. To become an effective team player, technicians should build positive relationships with colleagues, contribute to team discussions, and collaborate to achieve common goals.

5. Practice time management skills: Effective technicians are able to manage their time efficiently, prioritize tasks, and meet deadlines. They should create schedules, set goals, and stay focused on their work.

In summary, becoming an effective technician requires a combination of technical and soft skills. By following these procedures, technicians can develop the skills they need to excel in their roles and deliver high-quality results.

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Answer:

1) Mass of Egg Chamber → 35 g = 35 g × 1 kg/(1000 g) = 0.035 kg

2) Average Mass of Grade A Egg       →    56.7 g = 0.056 kg

3) Calculated Mass

Drop Chamber + Egg → (35 + 56.7) g = 91.7 g = 91.7 g × 1 kg/(1000 g) = 0.0917 kg

4) Time of Drop → 1.5 seconds

5) Maximum Velocity of Drop Chamber (v = g·t) → v = 9.8 m/s² × 1.5 s = 14.7 m/s

6) Maximum Momentum of Drop Chamber

(P = m·v) (right before hitting the ground) → p = 0.0917 kg × 14.7 m/s = 1.34799 kg·m/s

7) Final End Momentum of Drop Chamber → After landing, velocity v = 0 m/s

(p = m·v after landing)   \({}\)                              ∴ p = m·v = m × 0 m/s = 0 kg·m/s

8) Impulse on Impact            →   \({}\)             (0 - 1.34799) kg·m/s = 1.34799 kg·m/s

(Δm·v = Final - Initial)

9) Force on Impact    \({}\)           →                 F·t = Δm·v = 1.34799 N·s

(F·t = Δm·v)

Explanation:

Given:

The mass of the ball is

The initial height of the ball is

The initial speed of the ball is

To find:

the impact speed of the ball when it strikes the ground

Explanation:

The initial potential energy of the ball is

The initial kinetic energy is

The final energy of the ball is fully kinetic energy. Let the final impact speed of the ball is

We can write, using the energy conservation principle that

Hence, the final impact speed of the ball is 46.1 m/s.

The bat should fly about 0.8065 m/s in order to hear a beat frequency of 8 Hz.

The shift in a wave's frequency caused by an observer moving away from the wave source is known as the Doppler effect or Doppler shift. The equation for this is given by,

\(f=\frac{v \pm v_0}{v\mp v_s}f_s\)

where f is the observed frequency, v is the speed or velocity of the sound waves, v₀ is observer velocity ("+" if it's towards and "-" if it's away from the source), vs is source velocity("-" if it's towards and "+" if it's away from the observer), and fs is the actual frequency of the sound waves.

The velocity of sound waves in the air is 344 m/s, v₀ is zero for the incident, v is the source velocity towards the wall, and fs is 1700 Hz.

Then, the incident frequency is given by,

\(f_{incidence}=\frac{344-0}{344-v}\times f_s\)

The wall is stationary, therefore, f(incidence)=f(reflected). Then, the new frequency the bat hears is,

\(\begin{aligned}f_{new}&=\frac{344+v}{344}\times f_{reflected}\\&=\frac{344+v}{344}\times \frac{344-0}{344-v}\times f_s\\&=\frac{344+v}{344-v} \times f_s\end{aligned}\)

To hear a beat frequency of 8 Hz, the velocity of the bat should be,

\(\begin{aligned}f_{new}-f_s&=8\\\left(\frac{344+v}{344-v} \right)f_s-f_s&=8\\\frac{344+v}{344-v}-1&=\frac{8}{f_s}\\\frac{344+v-344+v}{344-v}&=\frac{8}{1700}\\2v&=0.0047(344-v)\\2v&=1.6168-0.0047v\\2.0047v&=1.6168\\v&=\mathrm{0.8065\;m/s}\end{aligned}\)

The answer is 0.8065 m/s.

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Magnetite is a special kind of rock. It sticks to magnets. Magnetite is made of a metal called iron. The iron is what sticks to magnets.

Pieces of the mineral magnetite that have been magnetized naturally are lodestones. They are magnets that are present in nature and may draw iron. Using lodestones, the phenomenon of magnetism was first discovered in antiquity. The best magnets for this test are rare earth magnets, which are inexpensive, widely available, and effective. However, there are earth rocks that can also hold a magnet, and to make matters worse, these rocks are fairly typical. Hematite and magnetite-containing rocks and minerals are the two types of rocks that are most frequently misinterpreted. For collectors and researchers, meteorites are highly valuable in terms of money and science. From a few dollars to hundreds of thousands of dollars, meteorite values can vary. Sedimentary, metamorphic, and igneous rocks all contain magnetite.

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If the momentum of the protons is doubled then the de Broglie wavelength of protons will reduce by a factor of 2.

The de Broglie equation is one of the equations that is widely used to define the wave properties of matter.

It basically illustrates how an electron acts as a wave.

Electromagnetic radiation's dual nature as a wave and a particle with motion (expressed in frequency, wavelength).

Microscopically small electron-like particles were used to demonstrate the existence of this dual nature feature.

The formula is provided by

λ = h/mv = h/p

where

Momentum (p) = mv

It is evident from the equation that the relationships between wavelength and momentum are inverse.

So if the momentum of the protons is doubled then the de Broglie wavelength of protons will reduce by a factor of 2.

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The Moon illusion is a phenomenon where the Moon appears larger when it's near the horizon compared to when it's high in the sky. This illusion can best be explained in terms of the relationship between perceived distance and perceived size.

When the Moon is near the horizon, it appears to be farther away than when it's high in the sky, which leads our brain to assume that it's larger. This is known as the size-distance illusion. Our brain is used to perceiving objects as smaller when they're farther away, but the Moon is an exception because it's always far away. Therefore, when it's near the horizon, our brain overcompensates and perceives it as larger than when it's high in the sky. This illusion has been studied by scientists for centuries and continues to be a topic of interest in psychology and neuroscience. Overall, the Moon illusion is an example of how our brain interprets visual information and creates illusions based on our past experiences and knowledge.

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Answer:

608

Explanation:

Trust me

Answer:

608

Explanation:

I only answer after making sure the answer is correct and I get the green tick, not the red cross (Acellus)

The mirror's radius of curvature is approximately -28.4 m. This is using the mirror equation.

According to the mirror equation,

1/f = 1/do + 1/di,

where f is the focal length, do is the object distance, and di is the image distance.

In this case, the object is a car 19.0 m away and the image is formed at the same distance behind the mirror since the mirror is concave and the image is virtual.

Therefore,

do = di = -19.0 m.

We also know that the magnification,

M = -di/do, is equal to 0.35.

Substituting these values into the mirror equation and solving for f, we get:

1/f = 1/-19.0 + 1/-19.0
1/f = -0.1053
f = -9.496 m

The radius of curvature, R, is related to the focal length by the equation R = 2f. Therefore, R = -18.99 m, which we can round to -28.4 m to two significant figures.

The mirror's radius of curvature is approximately -28.4 m, which indicates a strongly concave mirror. The negative sign indicates that the mirror is concave, and the magnitude of the radius suggests a high degree of curvature. The fact that cars appear smaller than normal suggests that the mirror is magnifying the image, which is consistent with a concave mirror.

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The resistance of a circuit that contains two 50.0 Ω resistors connected in series with a 12.0 V battery is 100 Ω. The resistance of the circuit is equal to the sum of the individual resistors when they are connected in series.

In this case, two 50.0 Ω resistors are connected in series, resulting in a total resistance of 100 Ω. Resistance can be defined as the obstruction that electricity faces when it travels through a conductor. The greater the resistance, the less current is allowed to flow through the circuit. The following equation is used to calculate resistance: R = V / IWhere R is resistance, V is voltage, and I is current.

Resistance can also be calculated using Ohm's Law if we know the values of voltage and current. When resistors are connected in series, the total resistance of the circuit is equal to the sum of the individual resistors. This is expressed by the following equation:RT = R1 + R2 + ... + RNWhere RT is the total resistance of the circuit, and R1, R2, and RN are the individual resistors. In this case, the circuit contains two 50.0 Ω resistors connected in series.

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Answer:

1.25kgm^-3

Explanation:

The difference in the mass of the container is the mass of the air

m(air) =0.03-0.02

=0.01kg

Density = 0.01/0.008

=1.25kgm^-3

Answer:

In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm's law and is equal to the voltage divided by the equivalent resistance. The potential drop across each resistor can be found using Ohm's law.

Convection differs from radiation in the sense that in radiation, matter is not involved. Option D.

The term heat transfer has to do with the movement of heat from one point another. Recall that heat transfer is one of the things that does occur in nature. There are three kinds of heat transfer that occur in nature and these are;

Let us recall that in the process of convection, there is the movement of heat from one point to the other via the actual movement of the particles of the substance.  On the other hand, radiation occur when heat is transferred without an movement or contact between the objects  between which the heat is transferred. Thus, when air masses are moving, it is a process of convection taking place.

Hence, the difference between conduction and radiation is that matter is involved in convection but not in radiation.

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