Identical 2.0-micro c charges are located on the vertices of a square with sides that are 2.0 m in length. determine the electric potential (relative to zero at infinity) at the center of the square

Answer:

10.21 m/s

Explanation:

∆x =( v ^2 sin(2θ))/ g

rearrange this to get

v=√Δx·(g÷ sin(2θ))

Plug it in

v=√10·(-9.8÷sin(2·35))

Solve

v=10.21

Explanation:

Newton's second law of motion gives the relation between mass, force and acceleration.

We know that,

Force, F = mass (m) × acceleration (a)

or

\(m=\dfrac{F}{a}\)

or

\(a=\dfrac{F}{m}\)

Hence, this is the required solution.

The final speed of the heavier object= -v

In a head-on elastic collision, the conservation of momentum and kinetic energy principles can be applied to determine the final velocities of the objects. Let the final velocity of the mass m be v_m and the final velocity of the mass 3m be v_3m.

The conservation of momentum can be expressed as:

m * v + 3m * (-v) = m * v_m + 3m * v_3m

The conservation of kinetic energy can be expressed as:

0.5 * m * v^2 + 0.5 * 3m * (-v)^2 = 0.5 * m * v_m^2 + 0.5 * 3m * v_3m^2

By solving these equations simultaneously, we can find the final velocity of the heavier object, 3m:

v_3m = -v

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Answer:

The answer is "\(\bold{3.8\ \frac{m}{s}}\)"

Explanation:

\(0 = \int_{cs} \overrightarrow{V} \overrightarrow{dn} \\\\= \int_{h_1} \overrightarrow{V_1} \overrightarrow{dA_1} + V_2A_2 +V_3A_3V_3A_3 \\\\ = -\int_{h_1} V_1dA_1 + V_2 dA_2 \\\\\to \int^{h_1}_{0} V_1 max \frac{y}{h_1} w dy -V_2 wh_2 \ \ \ \ \ \ \ \_{where} \ \ \ v_1=\frac{y}{h_1}\\\\\)

\(\to V_3A_3= V_1 max [\frac{y^2}{2h_1}]^{h_1}_{0} -V_2wh_2 \\\\= \frac{V_1 max}{2} \times wh_1 -V_2wh_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3 =w_3h_3\\\\\to \frac{V_3A_3}{w}= \frac{V_1 max}{2} \times h_1 -V_2wh_2 \\\\ \to V_3h_3= \frac{V_1 max}{2} \times h_1 -V_2wh_2 \\\\\to 5 \times 0.15 = \frac{V_1}{2} \times 0.5 - 1 \times 0.2 \\\\V_1 = 3.8 \ \frac{m}{s}\)

Answer:

compound

Explanation:

Answer:

6970 N

Explanation:

To calculate the average force exerted on the ball, we can use the impulse-momentum theorem, which states that the impulse on an object is equal to its change in momentum. In equation form:

Impulse = Δp

where Impulse is the force applied over a given time, and Δp is the change in momentum of the object.

We can calculate the momentum of the ball before the hit as:

p1 = m * v1

where m is the mass of the ball and v1 is its initial velocity (which we assume to be zero). Substituting the given values, we get:

p1 = (0.85 kg) * 0 m/s = 0 kg m/s

The momentum of the ball after the hit is:

p2 = m * v2

where v2 is the final velocity of the ball (82 m/s). Substituting the given values, we get:

p2 = (0.85 kg) * 82 m/s = 69.7 kg m/s

The change in momentum (Δp) is therefore:

Δp = p2 - p1 = 69.7 kg m/s - 0 kg m/s = 69.7 kg m/s

The impulse on the ball is equal to the change in momentum, so we have:

Impulse = Δp = 69.7 kg m/s

Finally, we can calculate the average force exerted on the ball using the formula:

Impulse = Force * time

Substituting the given values, we get:

69.7 kg m/s = Force * 0.01 s

Solving for Force, we get:

Force = 6970 N

Therefore, the average force exerted on the ball by the bat is 6970 Newtons.

The ball is 0.4549 meters from the center. Since the ball is stuck on the disk, its distance from the center is relative to the disk's angular displacement.

Calculate the angular velocity ω of the ball.

ω = 2π rad/s ×(5 rpm/60 seconds) = 0.2617 rad/sec

Calculate the moment of inertia of the ball and disk.

I = mr²

I = (0.406 kg)(r²)

Substitute the moment of inertia I and angular velocity ω into the equation:

Kinetic Energy = 0.5 ×I ×ω²

0.01504 J = 0. 5× (0.406 kg)(r²) ×(0.2617 rad/sec)²

Solve for r.

r = √(2(0.01504 J) / (0.406 kg)(0.2617 rad/sec)²)

r = 0.4549 m

Therefore, the ball is 0.4549 meters from the center.

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Answer:

F = 39.2 N   (hand force) and    N = 68.6 N (shoulder force)

Explanation:

In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole

Let's set the reference system at the fit point of the shoulder

     ∑ τ = 0

We will assume that the counterclockwise turns are positive

    w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0

all distances are measured from the support of the man (x₀ = 0.60 m)

    F = (w₁ 0.60 + W 0.1) / 0.4

    F = (m₁ 0.6 + m 0.1) g / 0.4

let's calculate

    F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4

    F = 39.2 N

this is the force that the hand must exert to keep the system in balance

We apply the translational equilibrium condition

    -w₁ -W + N - F = 0

     N = w₁ + W + F

     N = (m₁ + m) g + F

let's calculate

     N = (2.6 + 0.4) 9.8 + 39.2

     N = 68.6 N

Answer:

Explanation:

The distance will be the total distannce covered during the journey.

If you move 3 meters East and move 4 meters north, then the distance will be calculated as;

Distance = distnace through East+distance through north

Distance = 3m + 4m

Distance = 7m

Displacement is the distance covered in a specified direction. It is the shortest distance covered by me. This can be gotten using the Pythagoras theorem.

d² = 3²+4²

d² = 9+16

d² = 25

d = √25

d = 5m

Hence the displacement of the object is 5metres

Answer:

6 v

Explanation:

Answer:

C is radius of sphere

Explanation:

F is C/2

Answer: towards the centre of the curve/circle  

Explanation:

centripetal force always points towards the centre of the circle. the speed stays constant, but the direction of motion changes as the object moves across the curve, therefore there is an acceleration.

Answer:

See below

Explanation:

Toward the center of the curve

   think about it....when you are in a car and accelerate FORWARD, you are pushed BACK...opposite the accel

   when you are going around a curve you are pushed to the outside of the curve...opposite the accel which is toward the center of the curve

The spring stiffness, or spring constant, of the given perfect spring is approximately 137.9623 N/m. This means that for every meter of extension, the spring will exert a force of 137.9623 N.

This value was obtained by applying Hooke's Law and calculating the ratio of the change in force to the change in extension using two data points from the table.

To determine the spring stiffness, we need to calculate the spring constant (k) using Hooke's Law, which states that the force applied on a spring is directly proportional to the extension it undergoes.

Hooke's Law can be represented as F = kx, where F is the applied force and x is the extension of the spring.

In the given table, we have the applied forces (F) and corresponding extensions (x). We can use any two data points from the table to find the spring constant.

Let's choose the first and last data points from the table:

(x1, F1) = (0.43 m, 59.34 N) and (x2, F2) = (0.96 m, 132.48 N).

Using Hooke's Law, we can calculate the spring constant (k) as follows:

k = (F2 - F1) / (x2 - x1)
  = (132.48 N - 59.34 N) / (0.96 m - 0.43 m)
  = 73.14 N / 0.53 m
  ≈ 137.9623 N/m (rounded to 4 decimal places)

Therefore, the spring stiffness, or spring constant, is approximately 137.9623 N/m.

Hooke's Law is a fundamental concept in physics that describes the relationship between the force applied on a spring and the resulting extension it undergoes.

The formula F = kx represents this relationship, where F is the applied force, k is the spring constant, and x is the extension of the spring.

By using two data points from the table, we can calculate the spring constant by finding the ratio of the change in force to the change in extension.

This calculation allows us to quantify the stiffness of the spring.
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1. The data shows that amplitude, period, and tension all affect the speed of a wave. As amplitude and tension increase, the speed of the wave increases, while an increase in period results in a decrease in speed.

What is an amplitude?

Amplitude is the maximum displacement or distance moved by a wave from its resting position. In other words, it is the magnitude of the oscillation in a wave, or the height of a wave from its equilibrium position. In general, the greater the amplitude of a wave, the more energy it carries. In the context of sound waves, amplitude is associated with the loudness of the sound, while in the context of electromagnetic waves (such as light), it is associated with the brightness or intensity of the light.

2. The frequency of a wave has a direct relationship with the wave speed. As the frequency of a wave increases, the speed of the wave also increases.

3. Wave speed decreases as it moves from a medium of low density to one of high density. This is because a denser medium causes more resistance to the wave, resulting in a slower wave speed.

4. If the instrument is initially flat, the player should tighten the string. This is because tightening the string increases the tension, which in turn increases the speed of the wave, resulting in a higher pitch.

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The person who first proposed that electrons exhibit the properties of a wave was French physicist Louis de Broglie in 1924. De Broglie introduced the concept of wave-particle duality, suggesting that particles, such as electrons, could also have wave-like properties.

We have heard about the nature of light and the characters it displays. Interference, reflection, refraction and diffraction are some of the characteristics. Wave-Particle Duality helps us to understand the particle and wave nature of light. Based on the idea that light and all other electromagnetic radiation may be considered a particle or a wave nature, in 1923 physicists Louis De Broglie suggested that the same kind of duality must apply to the matter.

So, The person who first proposed that electrons exhibit the properties of a wave was French physicist Louis de Broglie in 1924.

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Answer:

A.) T = 1.25s

B.) H = 7.7m

Explanation:

Given that a football is kicked at 55o at 15 m/s.

How long is it in the air?

Let's use the formula below:

V = UsinØ - gt

V = 0 at maximum height.

0 = UsinØ - gt

UsinØ = gt

Substitute all the parameters into the formula

15sin55 = 9.8t

9.8t = 12.29

t = 12.29/9.8

t = 1.25 seconds

How far does it travel to go through the is at the same height as when it started?

That will be the maximum height.

Using the third equation of motion

V^2 = U^2 sin^2ø - 2gH

At maximum height, V = 0

(UsinØ)^2 = 2gH

(15sin55)^2 = 2 × 9.8H

(12.29)^2 = 19.6H

H = 150.97/19.6

H = 7.7 m

Answer:

Moment of the 4N force = 16Nm

Moment of the 6N force = 12Nm

Explanation:

Find the diagram attached.

Moment of a force is the product of the force and the perpendicular distance from the point of application

For the 4N force

Distance of the force from O is 4m

Moment of the 4N force = 4N * 4m

Moment of the 4N force = 16Nm

For the 6N force

Distance of the force from O is 2m

Moment of the 6N force = 6N * 2m

Moment of the 6N force = 12Nm

Answer:

4.1 Introduction. Our modern society depends heavily on thermal energy sources derived mostly from coal, petroleum, and natural gas combustion.

After considering the given data we conclude that the wavelength of visible light emitted by a hydrogen atom when its excited electron drops from n = 6 to n = 2 is approximately 102.55 nm.

To evaluate the wavelength (in nm) of visible light emitted by a hydrogen atom when its excited electron drops from n = 6 to n = 2, we can apply the Rydberg formula:
\(1/\lambda = R(1/n_1^{2} - 1/n_2^{2} )\)
Here:
λ = wavelength of the emitted light
R = Rydberg constant \((1.097 *10^7 m^{-1} )\)
\(n_1\) and \(n_2\) = initial and final energy levels of the electron
Applying substitution of the given values, we get:
\(1/lambda = (1.097 * 10^7 m^{-1} )(1/6^{2} - 1/2^{2} )\)
Evaluating for λ, we get:
λ = 102.55 nm
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A star of the apparent magnitude of +1 would appear farther away than a star of the apparent magnitude of +2. Hence, option B is correct.

In astronomy, a star is a brilliant plasma spheroid that is held together through gravity. The Sun is the planet's closest star. The human eye can see a great number of other stars at night, but due to our planet's size, they seem alike stationary points of light.

Numerous of the shining stars were given names, whereas the most notable stars have been grouped into constellations and asterisms. The known stars have been recognized and given standard designations in star inventories that astronomers have put together.

The birth of a star is caused by the gravitational attraction of a gaseous nebula made largely of hydrogen, helium, and traces of transition metals. Its overall mass is the primary factor influencing how it will develop and end up.

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Answer:

\(\boxed {\boxed {\sf E_p= 515.025 \ J}}\)

Explanation:

Potential energy is found using this formula:

\(E_p= mgh\)

The mass of the baby in the carriage is 1.5 kilograms and the height is 35 meters. Assuming this is on Earth, the gravitational acceleration is 9.81 meters per square second.

\(m=1.5 \ kg \\g= 9.81 \ m/s^2\\h= 35 \ m\)

Substitute the values into the formula.

\(E_p= (1.5 \ kg )( 9.81 \ m/s^2)(35 \ m )\)

Multiply the first numbers together.

\(E_p= 14.715 \ kg*m/s^2 (35 \ m )\)

Multiply again.

\(E_p= 515.025 \ kg*m^2/s^2\)

\(E_p= 515.025 \ J\)

The potential energy is 515.025 Joules.

A point on the edge of the wheel will have traveled approximately 3350.7 cm in 6.3 seconds

To find the distance traveled by a point on the edge of the wheel, we need to use the formula:

distance = (circumference of the wheel) x (number of revolutions)

First, we need to find the circumference of the wheel using its diameter:

circumference = pi x diameter
circumference = 3.14 x 33 cm
circumference = 103.62 cm

Next, we need to find the number of revolutions made by the wheel in 6.3 seconds. We can use the formula:

final speed = initial speed + (acceleration x time)

initial speed = 245 rpm
final speed = 370 rpm
time = 6.3 s

acceleration = (final speed - initial speed) / time
acceleration = (370 - 245) / 6.3
acceleration = 19.84 rpm/s

Using the formula:

number of revolutions = (average speed) x (time / 60)

average speed = (initial speed + final speed) / 2
average speed = (245 + 370) / 2
average speed = 307.5 rpm

number of revolutions = (307.5 rpm) x (6.3 s / 60)
number of revolutions = 32.01375 rev

Finally, we can plug in the values to find the distance traveled by a point on the edge of the wheel:

distance = (circumference) x (number of revolutions)
distance = 103.62 cm x 32.01375 rev
distance = 3312.5 cm

Therefore, a point on the edge of the wheel will have traveled approximately 3312.5 cm in 6.3 seconds.
Hi! To solve this problem, we need to calculate the initial angular velocity, final angular velocity, and the average angular velocity. Then, we'll use the average angular velocity to find the distance traveled by a point on the edge of the wheel.

1. Convert diameters to radii: radius = diameter / 2
  radius = 33 cm / 2 = 16.5 cm

2. Convert RPM to radians per second (rad/s):
  Initial angular velocity (ω1) = 245 rpm × (2π rad / 60 s) ≈ 25.66 rad/s
  Final angular velocity (ω2) = 370 rpm × (2π rad / 60 s) ≈ 38.79 rad/s

3. Calculate the average angular velocity (ω_avg):
  ω_avg = (ω1 + ω2) / 2 ≈ (25.66 + 38.79) / 2 ≈ 32.225 rad/s

4. Calculate the distance traveled by a point on the edge of the wheel:
  Distance = ω_avg × time × radius
  Distance ≈ 32.225 rad/s × 6.3 s × 16.5 cm ≈ 3350.7 cm

A point on the edge of the wheel will have traveled approximately 3350.7 cm in 6.3 seconds.

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Answer:

\(1.5\:\mathrm{J}\)

Explanation:

The elastic potential energy of a spring is given by the following:

\(PE_s=\frac{1}{2}k\Delta x^2\), where \(k\) is the spring constant and \(\Delta x\) is displacement.

The displacement of the string is \(0.50-0.40=0.10\: \mathrm{m}\).

Plugging in our given values, we get:

\(PE_s=\frac{1}{2}\cdot300\cdot0.10^2=\fbox{$1.5\:\mathrm{J}$}\).

The elastic potential energy stored in the spring is 1.5 joule.

Potential energy is a form of stored energy that is dependent on the relationship between different system components. When a spring is compressed or stretched, its potential energy increases.

If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy. It is capable of performing more work when raised. Potential energy is a characteristic of systems rather than of particular bodies or particles.

Spring constant of the spring : k = 300 N/m.

Initial length of the spring = 0.40 m

Final length of the spring = 0.50 m.

Hence, stretching  of the spring: Δx = 0.50 m - 0.40 m = 0.10 m.

So, the elastic potential energy stored in the spring is = 1/2 × k × Δx²

= 1/2 × 300 × (0.10)² joule

= 1.5 joule.

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Answer:

Hey mate....

Explanation:

This is ur answer....

The heat is transferred by the process of conduction of heat.

It is given that, Jimmy held the end of a metal bar over a fire while holding on to the opposite end.    

After some time, the end he was holding began to get very hot.

The process of traveling heat from one end of the bar to the other end is called conduction. It is the process of transfer of heat by the interaction of particles and motion of electrons.

Hope it helps!

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Answer:

JUST A MINUTE KAPENG MAINIT

Answer:

-8.4°C

Explanation:

From the principle of heat capacity.

The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

To solve this we must be knowing each and every concept related to energy. Therefore, -8.40°C is the final temperature of the ice.

In physics, energy is the capacity to accomplish work. It can be potential, kinetic, temperature, electrical, chemical, radioactive, or in other forms.

There is also heat and work—energy inside the process of being transferred through one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy.

H = m× c× (T2-T1)

35= 1× c× ( T2-(-25))

35= c× ( T2+25)

specific heat capacity of ice is =2.108 kJ/kgK

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°

T2= 16.60-25= -8.40°C

Therefore, -8.40°C is the final temperature of the ice.

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Answer:

Nuclear reactors are the heart of a nuclear power plant. They contain and control nuclear chain reactions that produce heat through a physical process called fission. That heat is used to make steam that spins a turbine to create electricity.

Answer:

A nuclear power plant's heart is made up of nuclear reactors. They contain and regulate nuclear chain reactions, which produce heat via a physical process known as fission. This heat is converted to steam, which powers a turbine to generate energy.

The moment of inertia of the pulley for rotation about the axle on which it is mounted is -0.0118 kg m².

Mass of block A, mA = 4.1 kg

Mass of block B, mB = 5.0 kg

Distance moved by block B, s = 1.80 m

Time taken to move distance s, t = 2.0 s

Radius of pulley, r = 0.080 m

We know that the work done by the tension force, T, is equal to the change in potential energy of the block B.

Here, we can make the following observations:

Block B is moving downwards and Block A is moving towards the right. So, the directions of gravitational forces acting on the two blocks are opposite.

Block A is moving on a frictionless surface. So, the work done by the friction force on Block A is zero.

In this case, we can use the work-energy principle to calculate the tension force, T.

First, let us calculate the change in potential energy of block B.

The initial position of Block B is taken to be the position where the rope is taut, but the block is still at rest.

The final position is when it moves downward by s = 1.80 m.

The direction of the gravitational force is downward.

Therefore, ∆PE = mgh= 5.0 × 9.8 × 1.80= 88.2 J

Now, according to the work-energy principle, the work done by the tension force T on Block B is equal to the change in potential energy of Block B.W = ∆PE

We know that, W = T × s = T × d

Therefore, T × s = ∆PE5.0 × g × s = 88.2T = 88.2/(5.0 × 9.8 × 1.80) = 1.00 N

Let the acceleration of Block B be a and the angular acceleration of the pulley be α.

We have, T = mBga + Iα

Here, a is the linear acceleration of the block B and is given by

a = s/t= 1.80/2.0= 0.90 m/s²

Now, consider the pulley. It has two masses that are symmetrically placed at a distance of r = 0.080 m from the axis of rotation. The masses are connected to each other and the tension force acts on them in opposite directions.

Therefore, we can use the parallel-axis theorem to calculate the moment of inertia of the pulley.

I = I cm + Md² where,

I cm is the moment of inertia of the pulley about its center of mass,

M is the mass of the pulley, and

d = r is the distance between the center of mass and the axis of rotation of the pulley.

Let the pulley rotate by an angle θ in time t, so that the angular velocity of the pulley is ω and the tangential velocity of the pulley at its rim is v.

Then,ω = αt and v = rω

Now, the acceleration of the pulley can be calculated as follows: a = rα

Thus, the tension force acting on the pulley can be written as T = Ia/r²

Here, the tension force acts in the opposite direction to the direction of motion of Block A.

Therefore, the acceleration of Block A can be calculated as follows:

F = maT = mAa => a = T/mA = 1.00/4.1 = 0.244 m/s²

Now, we can use the above equations to calculate the moment of inertia of the pulley.

I = (T/mA) × r² × (1/a) - M(r² + d²)= (1.00/4.1) × (0.080)² × (1/0.244) - (0.080² + 0.080²) × (4.1 + 5.0)= 0.00408 - 0.01592= -0.0118 kg m²

Since the pulley is positioned on an axle, its moment of inertia for rotation about the axle is -0.0118 kg m2.

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A cart of mass 4. 0 kg is being pulled with a force of 24 N. The cart accelerates at 3. 0m,then the net force on the cart (c).12N is the correct option.

A physics concept called force describes how two items interact. It is described as any force that has the potential to accelerate or deform an item. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction. The force needed to accelerate a mass of 1 kilogramme at a rate of 1 metre per second squared is known as the Newton (N) unit of force. Normal force, friction, and gravity are a few examples of common forces.

The net force on an object is given by the equation:

Net force = mass x acceleration

In this case, the mass of the cart is 4.0 kg and its acceleration is 3.0 m/s². Therefore, the net force on the cart is:

Net force = 4.0 kg x 3.0 m/s² = 12 N

Therefore, the correct answer is (C) 12 N.

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