A student uses 800 W microwave for three seconds how much energy does a student use

Her average speed was about 4.6 km/hr, and her average velocity was about 3.3 km/hr.

The entire distance travelled divided by the total time taken is the definition of average speed. In this case, the total distance travelled was 7.0 km, and the total time taken was 1.5 hours. Hence, the average speed can be determined as follows:

Average Speed = [tex]\frac{7.0 km }{ 1.5 \ hours }= 4.6 km/hr[/tex]

The displacement divided by the whole time travelled is the average velocity. In this case, the displacement was 3.0 km (from Mary's home to Sheila's home), and the total time taken was 1.5 hours.The average velocity can therefore be determined as follows:

Average Velocity = [tex]\frac{3.0 km }{1.5 \ hours} = 3.3 km/hr[/tex]

Therefore,Her average velocity was roughly 3.3 km/hr, and her average speed was roughly 4.6 km/hr.

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Explanation:

Inertia refers to the tendency of an object to resist changes in its motion. In baseball terms, a baseball that is at rest on the ground has a high level of inertia because it is resistant to moving until an external force, such as a player's bat, acts on it.

Momentum, on the other hand, is the product of an object's mass and velocity and refers to the quantity of motion that an object possesses. In baseball terms, a baseball that is moving at a high velocity, such as when it is hit by a bat, has a high level of momentum.

To illustrate the difference between inertia and momentum in baseball, consider the scenario of a baseball that is hit by a bat. Before the bat hits the ball, the ball is at rest and has a high level of inertia. However, once the bat hits the ball, the ball gains momentum and begins to move. As the ball moves, it continues to possess momentum, but its inertia gradually decreases as it encounters external forces, such as air resistance and friction from the ground, which act to slow it down.

If F₁ has a greater magnitude than F₂, the box will accelerate backward because the net force is in the backward direction (1st option)

To obtain the direction in which the box will move, we shall determine the net force acting on the box. This is illustrated below:

Assumption:

Net force = Magnitude of force 1 (F₁) - Magnitude of force 2 (F₂)

Net force = F₁ - F₂

Net force = 40 - 25

Net force = 15 N backward

From the above illustration, we can see that the net force is 15 N backward.

Thus, we can conclude from the box will accelerate backward (1st option)

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The mass attached to the pulley is Kx/g.

option 1.

Assuming there is no friction, the tension in the string will be constant throughout, and the net force on the pulley-mass system will be the weight of the mass.

Let the mass of the hanging weight be m, then the weight of the mass is mg.

The spring will be extended by the same distance x, as the string is inextensible, and it will exert a force of Kx in the upward direction.

Since the pulley is in equilibrium, the net force on the pulley-mass system must be zero.

Therefore, the tension in the string must be equal to the force exerted by the spring:

T = Kx

Using the fact that the tension in the string is equal to the weight of the mass, we can write:

mg = Kx

Therefore, the mass attached to the pulley is:

m = Kx/g

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Real, inverted, and same size are the features of the image. when A concave mirror with a curvature of 8 is displayed on a ruler with a range of 0 to 14 cm.

The mirror formula may be used to calculate the image distance for an item located 4 cm from a 1.5 cm focal length mirror.

1/f = 1/u+1/v

f is the focal length

u is the object distance

v is the image distance

Keep in mind that the concave mirror's image distance and focal length are both positive.

Given:

u = 4cm

f = 1.5cm

1/v = 1/1.5-1/4

1/v = 0.67-0.25

1/v = 0.42

v = 1/0.42

v = 2.38cm

The picture is Genuine and INVERTED since the image distance value is positive.

We shall find its magnification and see if it is magnified or lessened. It is amplified if the magnification is larger than 1, and it is decreased if it is less.

Magnification = v/u

Magnification = 2.38/4

Magnification = 0.595 or. 0.6

The picture is reduced in size since the magnification is less than one (SMALLER).

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Answer:

Explanation:

To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.

Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.

Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively. Since the electric field is perpendicular to the faces, the flux through them is zero. So, Q_in = (σ1 - σ2) * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C

The magnitude of the electric field at point A is 2.31 x 10^5 N/C.

B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.

The direction of the electric field at point A is RIGHT.

C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C

The magnitude of the electric field at point B is 3.77 x 10^7 N/C.

D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.

The direction of the electric field at point B is LEFT.

E:

Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q

The net charge enclosed within this Gaussian surface is:

Q = σ1 × (2πrh)

where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:

Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C

Using Gauss's law, we can find the electric field at point C:

E × (2πrh) = Q/ε0

where ε0 is the permittivity of free space. Solving for E, we get:

E = Q / (2πε0rh)

Plugging in the values, we get:

E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C

Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.

To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.

The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:

qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)

where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.

Substituting the given values, we get:

qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC

Using Gauss's law, we have:

E * 2πrh = qenc/ε0

where ε0 is the permittivity of free space.

Solving for E, we get:

E = qenc / (2πrhε0) = 2.22 × 10^4 N/C

Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.

F:

The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.

The net flux of an electric field in a closed surface is directly proportionate to the charge contained, according to Gauss' equation.

To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.

Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.

Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively.

Φ_E = E * A = Q_in / ε0

E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C

The magnitude of the electric field at point A is 2.31 x 10^5 N/C.

B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.

The direction of the electric field at point A is RIGHT.

C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C

The magnitude of the electric field at point B is 3.77 x 10^7 N/C.

D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.

The direction of the electric field at point B is LEFT.

E:Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q

The net charge enclosed within this Gaussian surface is:

Q = σ1 × (2πrh)

where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:

Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C

Using Gauss's law, we can find the electric field at point C:

E × (2πrh) = Q/ε0

where ε0 is the permittivity of free space. Solving for E, we get:

E = Q / (2πε0rh)

Plugging in the values, we get:

E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C

Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.

To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.

The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:

qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)

where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.

Substituting the given values, we get:

qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC

Using Gauss's law, we have:

E * 2πrh = qenc/ε0

where ε0 is the permittivity of free space.

Solving for E, we get:

E = qenc / (2πrhε0) = 2.22 × 10^4 N/C

Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.

F:The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.

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Answer:

Explanation:

The time taken to travel from Fort Worth to Dallas is:

t = 3:23 pm - 2:41 pm = 42 minutes = 0.7 hours

The distance covered is:

d = 58 km

The average speed is:

v = d/t = 58 km / 0.7 hours = 82.86 km/h

To convert km/h to m/s, we can use the conversion factor:

1 km/h = 0.2778 m/s

Therefore, the average speed in m/s is:

v = 82.86 km/h × 0.2778 m/s/km = 23.06 m/s (rounded to two decimal places)

So the average speed is 23.06 m/s.

Answer:24 volts ÷ 4 amps = 6 ohms

Explanation:

We know that the Ohm's law has given the relationship between the current, voltage and the resistance of the wire. Mathematically, it can be written as :

Where

I is the current

R is the resistance

If current flowing is 4 amps and voltage is 24 volts. The formula to find the resistance will be :

R = 6 Ohms

Hence, the correct option is (d) " 24 volts ÷ 4 amps = 6 ohms ".

The initial mass of the rocket was 106 kg.

We can use the principle of conservation of momentum to solve this problem.

The momentum of the rocket before takeoff is zero, since it is at rest, and the momentum after takeoff is the product of the mass of the rocket and its velocity.

However, during the takeoff, the rocket ejects a mass of fuel at a certain velocity, which creates a backward force (thrust) that propels the rocket forward.

This thrust can be calculated using the equation:

Thrust = (mass flow rate) x (exhaust velocity)

mass flow rate = (mass of fuel burned) / (burn time)

The mass of the rocket at any given time can be calculated using the equation:

mass = (initial mass) - (mass of fuel burned)

Using these equations, we can solve for the initial mass of the rocket:

Calculate the thrust:

Thrust = (118 kg / 10.0 s) x 1600 m/s = 1,888 N

Calculate the mass of the rocket at the end of the burn:

mass(end) = (initial mass) - (mass of fuel burned) = (initial mass) - 118 kg

Use the principle of conservation of momentum to find the initial mass:

momentum before = momentum after

0 = (mass(end) + 118 kg) x 110 m/s

mass(end) = -118 kg / 110 m/s = -1.07 kg/s

mass(end) = (initial mass) - 118 kg

(initial mass) = mass(end) + 118 kg

(initial mass) = (-1.07 kg/s x 10.0 s) + 118 kg

(initial mass) = 106 kg

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The truck's acceleration is 3.0m/s² and the momentum of the truck is  144000 kg m/s.

It is the rate at which the speed and direction of a moving object vary over time.

We can use the following equation to calculate the acceleration of the truck:

a = (v - u) / t

where

a = acceleration

v = final velocity = 18.0 m/s

u = initial velocity = 0 m/s (the truck starts from rest)

t = time taken = 6.00 s

Substituting the values, we get:

a = (18.0 m/s - 0 m/s) / 6.00 s

a = 3.00 m/s²

Therefore, the acceleration of the truck is 3.00 m/s².

We can use the following equation to calculate the momentum of the truck:

p = m * v

where

p = momentum

m = mass of the truck = 8000.0 kg

v = final velocity = 18.0 m/s

Substituting the values, we get:

p = 8000.0 kg * 18.0 m/s

p = 144000 kg m/s

Therefore, the momentum of the truck after it has reached its final velocity is 144000 kg m/s.

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Answer:

Explanation:

The formula relating the speed of sound, frequency, and wavelength is:

speed = frequency x wavelength

Rearranging this formula to solve for frequency:

frequency = speed / wavelength

Substituting the given values:

frequency = 342 m/s / 1.8 m

frequency = 190 Hz

Therefore, the frequency of the sound wave is 190 Hz.

Answer:

15 hours

Explanation:

formula: f(a) = a(0.5)^(T/t)

fill in known values: 13=208(0.5)^(60/t)

use natural log to isolate t:    ln(13/208)=ln(0.5)(60/t)

solve for t: t=15

The statement that correctly defines an evolutionary stage of a star like the sun is that after leaving the main sequence, its core is stable due to electron degeneracy. That is option C.

The stages of the life cycle of a star include the following:

The evolutionary stage is also called the main sequence stage of the life cycle of the star.

In this stage, the core temperature reaches the point for the fusion to occur whereby the protons of hydrogen are converted into atoms of helium. This leads to the stability of the core of the newly formed start due to electron degeneracy.

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The speed of the wind is 20 miles per hour.

To solve this problem, we can use the formula:

Speed = Distance/Time

Let's assume that the speed of the wind is x miles per hour.

With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:

460 = Airspeed + x

Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:

420 = Airspeed - x

We can solve this system of equations to find the airspeed of the plane:

460 = Airspeed + x

420 = Airspeed - x

Adding the two equations gives:

880 = 2Airspeed

Dividing both sides by 2 gives:

Airspeed = 440 miles per hour

Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:

460 = Airspeed + x

460 = 440 + x

x = 20

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Answer:

0.625 meters

Explanation:

We can use the work-energy that the work done on an object is equal to the change in its kinetic energy:

Work = ΔK = Kf - Ki

Where:

Work is the work done on the object

ΔK is the change in kinetic energy of the object

Kf is the final kinetic energy of the object

Ki is the initial kinetic energy of the object (which is zero since the object is at rest)

The work done on the object is equal to the force applied to the object multiplied by the distance over which the force is applied:

Work = F × d

Where:

F is the force applied to the object (50 N)

d is the distance over which the force is applied (unknown)

So we can write:

F × d = Kf - Ki

Substituting the given values:

50 N × d = 1/2 × 10 kg × (2.5 m/s)^2 - 0

Simplifying:

50 N × d = 31.25 J

Solving for d:

d = 31.25 J / 50 N = 0.625 m

Therefore, the object was moved a distance of 0.625 meters.

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The secondary coil needs 120 turns.The power transmitted to the secondary winding is 0.155 W.

A transformer works by using electromagnetic induction to transfer electrical energy between two circuits. The voltage changes between the primary and secondary coil based on the ratio of the number of turns in each coil. In a step-up transformer, the voltage is increased from the primary to the secondary coil, while in a step-down transformer, the voltage is decreased.

Transformers are commonly used in electronic devices to convert voltage levels, isolate circuits, and match impedances. They are often used in power supplies to step down the voltage from the wall outlet to a level that can be used by the device. They are also used in audio amplifiers to match the impedance of the output to the speaker, and in radio and television receivers to tune in to different frequencies.

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The light waves reflecting off a red stop sign would be longer than the light waves reflecting off a violet-colored jacket. This is because red light has a longer wavelength than violet light.

Light waves, like all waves, are characterized by their wavelength. The wavelength of a wave determines its color, with shorter wavelengths appearing as blue and violet, and longer wavelengths appearing as red and orange.

Because the wavelength of red light is longer than the wavelength of violet light, the light waves reflecting off the red stop sign would be longer than the light waves reflecting off the violet-colored jacket.

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The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

We can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.

The potential energy stored in the spring is given by:

PE = (1/2)kx²

where k is the spring constant and x is the distance the spring is compressed.

PE = (1/2)(266 N/m)(5 m)² = 3325 J

This potential energy is then converted into kinetic energy:

KE = (1/2)mv²

where m is the mass of the stuntman and v is his velocity.

3325 J = (1/2)(55 kg)v²

v² = (2*3325 J) / 55 kg

v²  = 121 m²/s²

v = √(121 m²/s²) = 11 m/s

Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

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Answer: b. force per unit area.

Explanation:

The answer is:

D. 7.6m

Answer : 0.00885 farads or 8.85 microfarads

Explanation: To calculate the capacitance required, we can use the following formula:

C = 1 / [2 * pi * f * ((V^2 - Vlamp^2)/P)]

where:

C = capacitance in farads (F)

pi = 3.14159...

f = frequency in hertz (Hz)

V = voltage in volts (V)

P = power in watts (W)

Vlamp = voltage of the lamp in volts (V)

Using the given values, we have:

C = 1 / [2 * pi * 60 * ((230^2 - 100^2)/750)]

C = 1 / [2 * 3.14159 * 60 * ((230^2 - 100^2)/750)]

C = 1 / [113.09724]

C = 0.00885 farads (F)

Therefore, the capacitance required is approximately 0.00885 farads or 8.85 microfarads.

Answer:

To solve this problem, we can use the formula:

d = (v^2)/(2μg)

d = distance traveled

v = speed of the car

μ = coefficient of kinetic friction

g = acceleration due to gravity

First, let's calculate the distance traveled when the car is traveling at 23 m/s:

d = (23^2)/(2*0.7*9.81) ≈ 67.97 meters

Now, let's calculate the distance traveled when the car is going twice as fast (46 m/s):

d = (46^2)/(2*0.7*9.81) ≈ 271.88 meters

Therefore, the car would travel approximately 271.88 meters if it were going twice as fast.

Answer:

The speed of the ball just before it hits the crossbar is 7.22 m/s.

Explanation:

We can use the conservation of energy to solve this problem.

At the moment the player kicks the ball, the ball has only kinetic energy, since it was at rest on the ground before being kicked. When the ball hits the crossbar, it has both kinetic energy and potential energy, since it is at a height above the ground. We can set the initial kinetic energy equal to the sum of the final kinetic and potential energy:

(1/2)mv^2 = mgh

where:

m = mass of the ball (0.500 kg)

v = initial speed of the ball (15.0 m/s)

g = acceleration due to gravity (9.80 m/s^2)

h = height of the crossbar above the ground (2.6 m)

We want to solve for the speed of the ball just before it hits the crossbar, which we can do by rearranging the equation:

v = sqrt(2gh)

v = sqrt(29.802.6) = 7.22 m/s (rounded to two decimal places

Mechanical energy to heat energy is collision,force x distance is work done,rubbing energy friction, stewardship is using energy wisely and nuclear energy can cause heat pollution.

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Answer:

Explanation:

Assuming that air resistance is negligible, we can use the following kinematic equations to solve for the peak height:

v_f^2 = v_i^2 + 2ad

where v_f = 0 m/s (at the peak height) and a = -9.8 m/s^2 (acceleration due to gravity)

and

d = v_i t + (1/2)at^2

where d is the displacement or the peak height we want to find, v_i is the initial velocity, t is the time it takes to reach the peak height.

First, we need to resolve the initial velocity into its vertical and horizontal components:

v_i_x = v_i cos(30°) = 121.1 m/s

v_i_y = v_i sin(30°) = 70.0 m/s

Next, we can use the vertical component of the initial velocity to find the time it takes to reach the peak height:

v_f = v_i_y + at

0 m/s = 70.0 m/s + (-9.8 m/s^2)t

t = 7.14 s

Finally, we can use the time we found and the kinematic equation for displacement to find the peak height:

d = v_i_y t + (1/2)at^2

d = (70.0 m/s)(7.14 s) + (1/2)(-9.8 m/s^2)(7.14 s)^2

d = 247.5 m

Therefore, the peak height of the football is 247.5 meters.

It takes 0.1311 seconds to hear the echo of the stone.

First we need to determine the time it takes for the sound wave to travel from the stone to the bottom of the mine shaft and back up to our ears.

Let's start by finding the time it takes for the sound wave to reach the bottom of the mine shaft. We can use the formula:

time = distance / speed

The distance is the depth of the mine shaft, which is 15 meters. The speed of sound is 343 m/s, as given in the problem. Therefore, the time it takes for the sound wave to reach the bottom of the mine shaft is:

time = 15 m / 343 m/s

time = 0.0437 s

Now, we need to find the time it takes for the sound wave to travel back up to our ears. Since the sound wave travels at the same speed, 343 m/s, the distance it needs to cover is twice the depth of the mine shaft, or 30 meters. Therefore, the time it takes for the sound wave to travel back up to our ears is:

time = 30 m / 343 m/s

time = 0.0874 s

Finally, to find the total time it takes to hear the echo, we add the time it takes for the sound wave to reach the bottom of the mine shaft to the time it takes to travel back up to our ears:

total time = 0.0437 s + 0.0874 s

total time = 0.1311 s

Therefore, it takes 0.1311 seconds to hear the echo of the stone.

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Answer:

Pulse transfers a single disturbance, while wave is a continuous disturbance that transfers energy.

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Answer:

- 0.33 m/s

Explanation:

An illustration is shown above,

In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,

m1u1 - m2u2 = (m1 + m2)v

Where,

m1 = mass of the first object

u1 = initial velocity of the first object

v1 = final velocity of the first object

m2 = mass of the second object

u2 = initial velocity of the second object

v2 = final velocity of the second object

Therefore,

(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v

12 - 16 = 12v

-4 = 12v

Divide both sides by 12,

-4 / 12 = 12v / 12

-1 / 3 = v

v = -0.33 m/s

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Why the ocean near Christchurch is a different temperature than we'd expect for its latitude (distance from the equator)? Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes.

Changes to prevailing winds affect ocean currents. Changes to ocean currents affect how much energy is brought to (or taken away from) a location. In El Niño years, the prevailing winds that normally drive a warm current from the Equator past New Zealand are disrupted and may stop or even reverse.