the potential energy increases everywhere by a fixed positive value. how does the force magnitude change?

The final kinetic energy of the block is 308.98 J.

Let's solve the problem using the work-energy theorem.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy

W = ΔKE

Initially, the block is at rest. Therefore, its initial kinetic energy is zero.

Ki = 0

We have to find the final kinetic energy of the block. Hence, Kf = ?

Work done on the block

W = Fscosθ

Work done by the applied force,

F = 75 Ns = 8 mθ = 37°

W = Fscosθ

W = 75 × 8 × cos 37°

W = 451.27 J

Work done by the frictional force

Ff = μkFn

The normal force

Fn = mg

Fn = 6 × 9.8

Fn = 58.8 N

Here,

Ff = μkFn

Ff = 0.28 × 58.8

Ff = 16.51 J

Work of friction:

W = Ff × s

W = 16.51 × 8

W = 132.1 J

The total work done on the block,

Wtotal = W + Wfriction

Wtotal = 451.27 + 132.1

Wtotal = 583.37 J

According to the work-energy theorem,

Wtotal = ΔKE

ΔKE = Wtotal

ΔKE = 583.37 J

Final kinetic energy of the block

Kf = KEFinal

Kf = ΔKE

Kf = 583.37 J

Kf = 308.98 J

Therefore, the final kinetic energy of the block is 308.98 J.

Complete question:

A 6 kg block is pushed 8m up a rough 37 degree inclined plane by a horizontal force of 75 N. The initial speed of the block is 2.2 m/s up the plane and a constant kinetic friction force of 25 N opposes the motion. Calculate the fianl kinetic energy of the block.

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Blood flows with a speed of 30 cm/s along a horizontal tube with a cross-section diameter of 1.6 cm.The speed of blood flow in the part of the same tube that has a diameter of 0.8 cm is 15 cm/s.

To arrive at this answer, we can use the formula for the flow rate of a fluid in a pipe:
Q = A × V
where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the velocity of the fluid.
Therefore, if we substitute the values for A and V of the first section, we can calculate the flow rate for that section:
Q1 = A1 × V1
Q1 = π ×(1.6 cm/2)² × 30 cm/s
Q1 = 24.72 cm³/s
Now we can use the flow rate and the cross-sectional area of the second section to calculate the velocity of the fluid:
Q1 = A2 × V2
V2 = Q1 / A2
V2 = 24.72 cm³/s / (π × (0.8 cm/2)²)
V2 = 15 cm/s
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If a star is 11 pc away from us, its apparent visual magnitude will be lower than its absolute visual magnitude. The star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude.

This is because the apparent magnitude of a star is affected by its distance from us. As the distance increases, the star appears dimmer, and its apparent magnitude decreases.

The distance modulus formula gives us a way to calculate the difference between the apparent and absolute magnitudes of a star:

Distance modulus = 5 * log(distance in parsecs) - 5

For a star that is 11 pc away, the distance modulus is,

Distance modulus = 5 * log(11) - 5 = 1.38

This means that the star's apparent magnitude will be 1.38 magnitudes lower than its absolute magnitude.

If the same star were only 5 pc away from us, the distance modulus would be,

Distance modulus = 5 * log(5) - 5 = 0.38

In this case, the star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude. This means that the star would appear brighter and have a higher apparent magnitude when it is closer to us.

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The formula for calculating capacitance is as follows:

C = Q/V

Where,

C = capacitance (Farads)

Q = charge (Coulombs)

V = voltage (Volts)

As given,

Q = 27.0 μC

V = 9.00 V

Substituting the given values in the above equation

C = 27.0 μC/9.00 V = 3.00 μF

Therefore, the value of capacitance is 3.00 μF.

The formula for calculating charge stored is as follows:

Q = CV

Where,

Q = charge (Coulombs)

C = capacitance (Farads)

V = voltage (Volts)

As given,

C = 3.00 μF

V = 12.0 V

Substituting the given values in the above equation,

Q = (3.00 × 10⁻⁶ F) × 12.0 V = 36.0 μC

Therefore, the charge stored is 36.0 μC.

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The given statement "an object floating in a container of water and partially submerged has the same density as the water" is true.

When an object is placed in water, it sinks until the weight of the water displaced by the object equals the weight of the object.

If an object has the same density as water, it displaces an equal amount of water to its own weight. When it displaces the same amount of water that has an equivalent mass to the object, it will float partially submerged. If the object's density is greater than water, it will sink. If the object's density is less than that of water, it will float entirely above the water's surface.

Density is defined as the mass of an object per unit volume. The formula for density is mass/volume. Density is a crucial physical property that is used to define and classify materials. The density of an object is determined by its mass and volume. The unit of measurement for density is kg/m3 or g/cm3. The density of water is 1 g/cm3, which is why objects with a density of less than 1 g/cm3 float on water.

An object floating in a container of water and partially submerged has the same density as the water.

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The sound level for this noise is approximately 50 decibels.

Sound level is a logarithmic measure of the ratio between the sound pressure level of a particular sound wave and a reference level. The reference level is typically set at the threshold of human hearing, which corresponds to an intensity of 10^-12 W/m^2. The sound level (measured in decibels, dB) of a sound wave is given by,

L = 10 log10(I/I0)

where I is the intensity of the sound wave and I0 is the reference intensity, which is typically set at 10^-12 W/m^2.

So, for an intensity of 10^-7 W/m^2 in a typical classroom, we can calculate the sound level as,

L = 10 log10(I/I0) = 10 log10(10^-7/10^-12) = 10 log10(10^5) = 50 dB

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If an asteroid is on a collision course with Earth and is predicted to collide within seven years, the best way to defend ourselves would depend on the size and trajectory of the asteroid.

An asteroid is a small, rocky object that orbits the Sun. Most asteroids are found in the asteroid belt, a region between the orbits of Mars and Jupiter. Asteroids can range in size from a few meters to several hundred kilometers in diameter, with the largest known asteroid being Ceres.

Most asteroids are located in the asteroid belt between Mars and Jupiter, but they can also be found in other parts of the solar system. Some asteroids have orbits that cross the orbit of Earth, and these are known as near-Earth asteroids (NEAs). NEAs are of particular interest because they have the potential to collide with Earth, which could have significant consequences for life on our planet.

Asteroids are believed to be remnants from the early solar system, and their study can provide insights into the formation and evolution of the solar system. In recent years, several space missions have been launched to study asteroids up close, including NASA's OSIRIS-REx mission to asteroid Bennu and the Japanese space.

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It would take approximately 19.2 years for the asteroid to orbit once around the sun. But that none of the answer choices match the calculated value of approximately 19.2 years.

The period (T) of an orbit of a celestial body with semimajor axis (a) around the sun can be calculated using Kepler's third law:

T² = (4π² / GM) * a³

where G is the gravitational constant and M is the mass of the sun.

Plugging in the given value for the semimajor axis (a = 4 AU), we get:

T² = (4π² / (6.674 × 10⁻¹¹ m³/(kg s²) * 1.989 × 10³⁰ kg)) * (4 AU)³

T² = 3.652 × 10¹⁶ s²

Taking the square root of both sides, we get:

T = 6.04 × 10⁸ s

We can convert this time to years by dividing by the number of seconds in a year:

T = (6.04 × 10⁸ s) / (31,536,000 s/year)

T ≈ 19.2 years

Therefore, it would take approximately 19.2 years for the asteroid to orbit once around the sun. The closest answer choice is 16 years.

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Each player must pull with a force of 1250 N to drag the coach at a steady 2.0 m/s.

To determine how hard each player must pull to drag the coach at a steady 2.0 m/s, we need to use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration:

Fnet = m * a

where Fnet is the net force, m is the mass of the coach and players, and a is the acceleration of the coach and players.

Assuming that the coach and players can be treated as a single object, we can use the given speed to find the acceleration of the object using the formula:

a = v² / (2 * d)

where v is the speed (2.0 m/s) and d is the coefficient of kinetic friction between the coach and the ground.

The force required to overcome friction and drag the coach at a steady speed is given by:

Ffriction = friction coefficient * Fnormal

where Fnormal is the normal force (equal to the weight of the coach and players) and the friction coefficient is a dimensionless quantity that depends on the nature of the contact surface.

Assuming a friction coefficient of 0.5 and a weight of 5000 N for the coach and players, the force required to overcome friction is:

F_friction = (0.5) * (5000 N) = 2500 N

The net force required to move the coach and players at a steady 2.0 m/s is therefore:

Fnet = Ffriction = 2500 N

Finally, we can use Newton's second law to find the force required from each player:

Fnet = m * a

2500 N = (m_coach + m_players) * (v² / (2 * d))

Solving for the mass (m_coach + m_players), we get:

m_coach + m_players = (2500 N * 2 * d) / v²

Assuming a value of 0.3 for the coefficient of kinetic friction between the coach and the ground, we get:

m_coach + m_players = (2500 N * 2 * 0.3) / (2.0 m/s)² = 562.5 kg

Therefore, the force required from each player is:

Fplayer = Fnet / 2 = 1250 N

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The current that flows through the 200 Ω resistor is 1.56 A.

Given resistance values of 200 Ω, 250 Ω, and 1000 Ω are connected in parallel across a source. The source current is 6 A. We are required to find the current that flows through the 200 Ω resistor.

Recall that when resistors are connected in parallel, the current is divided among them. And the voltage across each resistor is the same. The equivalent resistance of three parallel resistors is given by;

1/Rp = 1/R1 + 1/R2 + 1/R3Rp = (R1 x R2 x R3)/(R1R2 + R1R3 + R2R3)

Put the values into the formula;

Rp = (200 x 250 x 1000)/(200×250 + 200×1000 + 250×1000)

Rp = 52.17 Ω

The total current in the circuit, It = 6 A

From Ohm's Law;

V = IR,

where V is the voltage across each resistor

V1 = V2 = V3V = I×R

Therefore; V = I×Rp

The current flowing through the 200 Ω resistor, I1 = V1/200 = I × Rp/200The current flowing through the 200 Ω resistor, I1 = (6×52.17)/200I1 = 1.56 A

Thus, the current that flows through the 200 Ω resistor is 1.56 A.

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The primary mirror of an astronomical telescope is often shaped and polished to a parabolic shape because a parabolic shape allows for the mirror to collect the most amount of light and focus the parallel rays of light to a single point for better image clarity.

The reason that the primary mirror of an astronomical telescope is often shaped and polished to a parabolic shape is to reduce spherical aberration.

What is an astronomical telescope?An astronomical telescope is an optical instrument that aids in the observation of remote objects by collecting electromagnetic radiation such as visible light. It consists of two primary components: a primary mirror or lens that gathers and focuses light, and an eyepiece or camera that magnifies and projects the image formed by the primary.

A parabolic shape is a mirror or lens that has a curve that is more curved in the center than at the edges, and it is often used in astronomical telescopes to reduce spherical aberration. Spherical aberration is an optical defect that causes the image of a point source to become fuzzy and blurred. It occurs when the rays passing through the edges of a spherical lens or mirror become focused at a different distance than those passing through the center. This causes the image to be blurred around the edges, which makes it difficult to view small or distant objects. Parabolic mirrors are used to correct this problem because they are designed to focus all incoming light to a single point, resulting in a sharper and clearer image.

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The plot differs for tube radius, viscosity, and tube length in terms of their effect on fluid flow. The effect of each parameter is analyzed and plotted against the velocity profile of the fluid flow.

For tube radius, as the radius increases, the fluid flow velocity increases as well. This can be observed in the plot where the velocity profile is a bell-shaped curve, with the peak shifting to the right as the radius increases.

For viscosity, the effect is the opposite. As viscosity increases, the fluid flow velocity decreases. This can be observed in the plot where the velocity profile is a flatter curve, with a smaller peak as the viscosity increases.

For tube length, there is a similar effect as tube radius. As the length increases, the fluid flow velocity decreases. This can be observed in the plot where the velocity profile is a bell-shaped curve, with the peak shifting to the left as the length increases.

In terms of the comparison with the prediction, the results were mostly in line with what was expected. The plots showed the expected trends for each parameter, and the quantitative analysis confirmed this as well. However, there were some discrepancies between the predicted and actual values, which could be due to experimental error or limitations in the model used.

Overall, the results provided valuable insights into the relationship between these parameters and fluid flow, and can be used to optimize fluid systems for various applications.

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The magnitude of the torque on the current loop is calculated using the formula τ=BIA sinθ, where B is the magnitude of the magnetic field, I is the current, A is the area of the loop, and θ is the angle between the magnetic field and the loop's plane. In this case, the magnitude of the torque is τ = (1.2 T)(0.5 A)(5 cm x 5 cm)sin(30°) = 7.5 x 10-3 Nm.

The torque is the rotational force that causes the loop to rotate. This is due to the fact that a force is exerted on the loop by the magnetic field when there is a current running through it. This force generates a torque on the loop, which will cause it to rotate until the angle between the plane of the loop and the magnetic field is 0°.

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The net work (W) done on the particle by the external forces during its motion can be expressed in terms of the initial (Ki) and final (Kf) kinetic energies as: [tex]W = ((1/2) \times m \times  vf^2) - ((1/2) \times  m \times  vi^2)[/tex]

To find the net work (W) done on the particle by the external forces during the particle's motion in terms of the initial (Ki) and final (Kf) kinetic energies, we will use the work-energy theorem. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

Step 1: Calculate the initial kinetic energy (Ki) and final kinetic energy (Kf).
Ki = (1/2) * m * vi²
Kf = (1/2) * m * vf²

Step 2: Calculate the change in kinetic energy (ΔK) as the difference between Kf and Ki.
ΔK = Kf - Ki

Step 3: According to the work-energy theorem, the net work (W) done on the particle by the external forces during its motion is equal to the change in kinetic energy (ΔK).
W = ΔK

Step 4: Substitute the expressions for Ki and Kf from step 1 into the equation for W from step 3.
W = ((1/2) * m * vf²) - ((1/2) * m * vi²)

In conclusion, the net work (W) done on the particle by the external forces during its motion can be expressed in terms of the initial (Ki) and final (Kf) kinetic energies as:  W = ((1/2) * m * vf²) - ((1/2) * m * vi²)

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Complete Question:

Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf. Work done on the particle by the external forces during the particle's motion. To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement is given by s=xf −xi.

The process described in the question is known as scenario planning. It is a strategic planning method that involves generating multiple plausible scenarios of future conditions and analyzing the potential impact of each scenario on an organization or a system.

Scenario planning is a useful tool for decision-making, risk management, and identifying opportunities in an uncertain or rapidly changing environment.

By developing a range of scenarios, decision-makers can anticipate potential challenges and opportunities and develop strategies to respond effectively to each situation.

This approach allows organizations to be better prepared and more resilient in the face of future uncertainties. Scenario planning can be applied to various fields, including business, economics, environmental planning, and public policy.

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The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is: 0.001 newtons per tesla

The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is given by the formula F = (μI) / 2πr,

where μ is the permeability of free space, (4π x 10-7 N/A²)

I is current, and r is the radius of the loop.

In this case, the force is (4π x 10-7 x 5.5) / (2π x 0.1) = 0.001 N/T.

In other words, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla.

The formula for the force generated per unit field strength on a loop is derived from the fact that the force is a result of the magnetic field generated by the current flowing in the loop.

The magnitude of the magnetic field generated is proportional to the current and inversely proportional to the radius of the loop. Since the force is a product of the current and the magnetic field, it is proportional to the square of the current and inversely proportional to the square of the radius of the loop.

In summary, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla, given by the formula F = (μI) / 2πr, where μ is the permeability of free space (4π x 10-7 N/A²), I is current, and r is the radius of the loop.

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True. An incandescent bulb may have a higher initial cost than other types of lightbulbs, but it uses less energy over its lifetime and thus reduces energy costs.

For an incandescent bulb, the given statement is true. In candescent bulbs are traditional bulbs, which use a filament to create light. These bulbs are less efficient, as they waste most of the electricity they use as heat rather than light. As a result, the bulbs are less cost-effective in the long run.

They use up more energy than modern alternatives such as CFLs (compact fluorescent lights) or LEDs (light-emitting diodes). Despite their low initial cost, incandescent bulbs are not recommended for long-term use. They consume more electricity and thus have a greater impact on the environment. Therefore, it is not true that the energy costs of an incandescent bulb will be low over its life time.

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The electric field stored between two square plates of 9.3 cm on a side and separated by a 2.5 mm air gap is 1110 N/C. This can be calculated using Coulomb's law and the given information.

Coulomb's law states that the electric field is equal to the charge (Q) divided by the permittivity of free space (ε₀) multiplied by the distance (d) squared:

E=Q/(ε₀*d²).
Plugging in the given information,

E=(13 nC)/(8.85 x 10⁻¹² * 0.0025²) = 1110 N/C.
This answer uses Coulomb's law to calculate the electric field stored between two square plates, given the plates' side lengths, air gap width, and charge magnitude.

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a..... 1.04 x 10⁻⁴ Vi

b.... 9500 V

A) Determine the capacitance (to the nearest 10 μF).

First, we should identify the formula that we will use to solve the problem.

The formula that relates to capacitance is:

C = 2E / V². Where C is the capacitance in farads, E is the energy stored in joules, and V is the voltage across the capacitor in volts.

Converting the energy to joules, we have: E = 250J.

Now we know that the voltage needs to drop to half of its initial value in 10 ms. We can use the following formula to calculate the capacitance: C = R x t / ln(Vi / Vf) where R is the resistance in ohms, t is the time in seconds, Vi is the initial voltage, and Vf is the final voltage, which is half of the initial voltage.

B) Plugging in the given values, we get:

C = 48 x 0.01 / ln(Vi / (Vi / 2))Simplifying and solving for capacitance, we get:

C = 1.04 x 10⁻⁴ ViNow we can use the energy formula to solve for Vi:Vi = √(2E / C)

Plugging in the given values, we get:Vi = √(2 x 250 / 1.04 x 10⁻⁴)Simplifying and solving for Vi, we get:Vi = 9469 V

Therefore, the capacitance that meets these specifications is 100 μF and the initial capacitor voltage that meets these specifications is 9500 V, to the nearest 100 V.

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The projectile will be moving at a speed of 57.21 m/s when it reaches the top of its trajectory.

When a projectile is fired at a speed of 138 and an upward angle of 65 degrees, the speed at the top of the trajectory can be calculated. To solve this problem, you need to understand some basic physics concepts. Here's how you can solve this problem:
1. First, identify the given values and write them down:
Initial velocity (u) = 138 m/s
Angle of projection (θ) = 65 degrees
Acceleration due to gravity (g) = 9.81 m/s²
2. Now, break down the initial velocity into its horizontal and vertical components:
Initial velocity in the horizontal direction = u cos θ
Initial velocity in the vertical direction = u sin θ
3. Use the equation of motion to calculate the time taken by the projectile to reach the top of its trajectory:
u sin θ = gt/2
t = 2u sin θ/g
4. Use the time obtained in step 3 to calculate the velocity at the top of the trajectory:
v = u cos θ
Where,
v = final velocity
u = initial velocity
θ = angle of projection
5. Substitute the given values in the equation to get the final answer:
v = u cos θ
v = 138 cos 65
v = 57.21 m/s
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Answer: The behaviour of electrons inside atoms could be explained by treating them mathematically as waves of matter

Explanation:

Erwin Schrödinger proposed the quantum mechanical model of the atom, which treats electrons as matter waves.

Answer:

[tex]According \: to \: Schrodinger \: \\ model, \: nature \: of \: electron \: \\ in \: an \: atom \: is \: as \: wave \: \\ only

[/tex]

The angular velocity assuming a constant speed for a track-star who runs 400 m circular track in 45 s is 0.139 radians/s.

To calculate the angular velocity first the circumference of the track is 400 meters.

This means that the angular displacement of the track star during the race is:

θ = s / r

where θ is the angular displacement,

s is the distance traveled by the track star (which is equal to the circumference of the track), and

r is the radius of the circular track.

2.) Since the radius of the circular track is half of its diameter, we have:

r = 400 m / 2 = 200 m

Plugging this into the equation for angular displacement, we get:

θ = 400 m / 200 m = 2π radians

3.) Next, we can use the formula for angular velocity:

ω = θ / t

where ω is the angular velocity and

t is the time it takes for the track star to complete the race.

4.)Plugging in the values we have:

ω = θ / t

ω = 2π radians / 45 s

Therefore, the angular velocity of the track star is:

ω = 0.139 radians/s (rounded to three significant figures)

Therefore, the track star's angular velocity assuming a constant speed is approximately 0.139 radians/s

The angular displacement of the track star is equal to one complete revolution around the circular track, which is equal to 2π radians.

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The given statement " A series circuit is a current divider and a parallel circuit is a voltage divider circuit " is True

In a series circuit, the electric current is the same through each component, and the total current is equal to the sum of the currents through each component. Therefore, the current is divided among the components.

In a parallel circuit, the potential voltage across each component is the same, and the total voltage is equal to the sum of the voltages across each component. Therefore, the voltage is divided among the components.

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(a) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in series is approximately 4.2017 µF.

(b) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in parallel is 12.70 µF.

When two capacitors are connected in series, the equivalent capacitance is given by the formula,

1/Ceq = 1/C1 + 1/C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

1/Ceq = 1/4.20 µF + 1/8.50 µF

1/Ceq = 0.238 µF^-1

Ceq = 1 / (0.238 µF^-1)

Ceq = 4.2017 µF (rounded to four significant figures)

When two capacitors are connected in parallel, the equivalent capacitance is given by the formula,

Ceq = C1 + C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

Ceq = 4.20 µF + 8.50 µF

Ceq = 12.70 µF

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Final velocity of Jeff's car is 7.133 m/s south. The direction is 59.3° south of east.

In this issue, we can utilize preservation of energy to track down the last speed and course of Jeff's crash mobile after the impact with Julia's. Before the impact, the energy in the x-heading is zero, and in the y-course, it is 60 kg × 4 m/s = 240 kg⋅m/s north. Julia's force is 45 kg × 6 m/s = 270 kg⋅m/s west.After the crash, the energy in the x-course is rationed. The absolute energy in the x-course is as yet zero, as Julia's force that way is likewise zero. In the y-heading, the absolute force after the crash is 60 kg × vj + 45 kg × 2 m/s sin 15°, where vj is Jeff's last speed in the y-course.Utilizing protection of energy, we can compare the force when the crash in the y-heading:

60 kg × 4 m/s + 45 kg × 6 m/s = 60 kg × vj + 45 kg × 2 m/s sin 15°

Working on this situation, we get:

240 kg⋅m/s + 270 kg⋅m/s = 60 kg × vj + 12.19 kg⋅m/s

Addressing for vj, we get:

vj = (240 kg⋅m/s + 270 kg⋅m/s - 12.19 kg⋅m/s)/60 kg

vj = 7.133 m/s south

Consequently, Jeff's last speed is 7.133 m/s south. To find the course, we can utilize geometry. The point of Jeff's last speed concerning the x-pivot is given by:

θ = tan^-1(vj/4 m/s)

θ = 59.3° south of east

Accordingly, the last speed and heading of Jeff's amusement cart are 7.133 m/s at a point of 59.3° south of east.

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Answer:

Explanation:

Because by Wien's Law, they emit strongest in infrared and human eyes cannot see infrared radiation

Each aircraft must be moving at a speed of 85.75 km/hr towards each other to hear a pitch that is 2 times the emitted frequency.

What is frequency ?

Frequency is a physical quantity that describes the number of occurrences of a repeating event per unit of time. It is often measured in Hertz (Hz), which represents the number of cycles or vibrations per second.

In the context of waves, such as sound waves or electromagnetic waves, frequency refers to the number of complete cycles of the wave that occur in one second. A high frequency wave has more cycles per second than a low frequency wave.

Frequency is also an important concept in physics, particularly in the study of oscillations and waves. It is used to describe the behavior of systems that oscillate or vibrate, such as a simple pendulum or a guitar string. In these cases, the frequency of the oscillation is related to the natural frequency of the system, which is determined by its mass, stiffness, and other properties.

When two aircraft are moving towards each other, the sound waves from each aircraft are compressed, leading to a higher pitch than the emitted frequency. The pitch heard by the pilots of the aircraft can be calculated using the following formula:

Pitch heard = Emitted frequency * (Speed of sound + Speed of observer) / (Speed of sound - Speed of source)

Since the two aircraft are flying towards each other at the same speed, we can assume that the speed of one aircraft is x km/hr, and the speed of the other aircraft is also x km/hr. Therefore, the relative speed between the two aircraft is 2x km/hr.

Substituting the values given in the formula, we get:

2 * Emitted frequency = Emitted frequency * (343 + 2x) / (343 - x)

Simplifying this equation, we get:

686 - 2x = 343 + 2x

4x = 343

x = 85.75 km/hr

Therefore, each aircraft must be moving at a speed of 85.75 km/hr towards each other to hear a pitch that is 2 times the emitted frequency.

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The time it takes the object to fall through the change in speed using the impulse-momentum theorem is 0.62 seconds.

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse exerted on it.

To calculate the time it takes the object to increase it speed using the  impulse-momentum theorem, we use the formula below.

Formula:

Recall that F/m = acceleration. Therefore,

Where:

From the question,

Given:

Substitute these values into equation 1 and solve for t

Hence, the time it takes the object to fall is 0.62 seconds.

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