3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.0100 s. What is the force exerted by his glove on the ball? Assume the ball slows down with constant acceleration.

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

Answer:

Terminal voltage = 8.36 V

Current = 1.82 A

Explanation:

E.M.F of battery = 12V

Internal resistance of battery (r) = 2Ω

Resistance of resistor (R) = 4.6Ω

Now the formula for terminal voltage across the battery is;

V = ε - Ir

Where ε is EMF and I is electric current

Using ohms law, we know that V = IR and I = V/R.

Thus, let's put V/R for current in the potential difference equation;

V = ε - r(V/R)

Thus, lets make V the subject of the formula ;

V + (rV/R) = ε

V(1 + r/R) = ε

So, V = ε/(1 + r/R)

V = 12/(1 + (2/4.6))

V = 12/(1 + 0.4348)

V = 12/1.4348 = 8.36 V

Thus from V=IR, we can find current. So 8.36 = I(4.6)

I = 8.36/4.6 = 1.82 A

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given that,

The radius of earth r = 6.3781×10^6 m

The angular speed of earth w = 7.27*10^-5 rad/s

The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

The mass of earth m = 5.972 × 10^24 kg

The inertia of sphere I = 2/5 * m* r^2

Solution:

angular acceleration of the earth from rest to w is given by α:

                               α = w / t

                               α = (7.27*10^-5) / (1.57784760 * 10^8)

                               α = 4.60754*10^-13 rad/s^2

The required torque τ is given by:

                               τ = I*α

                               τ = 2/5 * m* r^2 * α

 τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

 τ = 4.47746 * 10^25 N-m

Power required P to turn the earth to the speed w is:

                          P = τ*w

                          P = (4.47746 * 10^25)*(7.27*10^-5)

                          P = 3.25511*10^21 W

Energy E required is :

                          E = P / t

                          E = (3.25511*10^21) / (1.57784760 * 10^8)

                          E = 2.06301 * 10^13 J

Answer:

Orbital period of satellite is 5.83 x 10³ s

Explanation:

The orbital period of satellite revolving around Earth is given by the equation :

[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex]      .....(1)

Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.

In this problem,

Height of satellite, h = 500 km = 500 x 10³ m

Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).

 [tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]

T = 5.83 x 10³ s

Answer:

1600 cal

Explanation:

The formula

q=mΔ[tex]_{fus}[/tex][tex]H^{0}[/tex]

is used to calculate the heat required to melt a solid where q=amount of heat, m=mass and Δ[tex]_{fus}[/tex][tex]H^{0}[/tex] is the enthalphy of fusion

now we substitute, m=20g, Δ[tex]_{fus}[/tex][tex]H^{0}[/tex]=80cal/g

Therefore, q=20g x 80cal/g =1600 cal

I hope you find this information useful and interesting! Good luck!

The heat released when 20.0 g of water at 0°C is frozen to ice is 1600 calories.

When water freezes, it releases heat energy. The quantity of heat energy needed to convert a substance from a solid to a liquid or vice versa without affecting its temperature is known as the heat of fusion.

In this case, we are asked to find the number of calories of heat released when moment 20 g of water freezes to ice at zero degrees.

To calculate this, we can use the formula:

Heat released = mass of water * heat of fusion

Substituting the given values:

Heat released = 20.0 g * 80. cal/g = 1600 cal

Therefore, 1600 calories of heat are released when 20.0 g of water freezes to ice at 0 °C.

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Answer:

so we have a good place to live at.

Explanation:

Environment is understood to be the set of natural and human factors that surround man in his daily life. Thus, landforms, natural resources, buildings, etc., are part of it.

Taking these definitions into account, a regular environment is one in which all its conditions and components are found with the fewest possible alterations, or with human alterations that do not negatively affect its natural conditions.

In this way, an environment that has not definitively consumed its resources or that has not significantly affected the natural status of the region is considered regular. This characteristic is important because it allows the environment to not be negatively affected, allowing a normal development of human life.

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Answer:

45 C.

Explanation:

Given:

Time, t = 5.0 h

= 18000 s

Current, I = 2.5 mA

= 0.0025 A

Potential difference, V = 9 V

Q = I × t

= 18000 × 0.0025

= 45 C.

Final answer:

The force of gravity and air resistance do work on a simple pendulum, with gravity doing positive and negative work as the pendulum moves, and air resistance dissipating energy from the system. Tension in the string does not do work since it is always perpendicular to the pendulum's motion.

Explanation:

In the context of a simple pendulum, the forces capable of doing work are the force of gravity and air resistance. The force of gravity does work on the pendulum as it swings back and forth because it has a component along the direction of the pendulum's movement. Specifically, as the pendulum swings, gravity pulls it downward, causing it to accelerate towards the lowest point of its path. Upon reaching this lowest point, the pendulum has maximum kinetic energy because the potential energy due to its elevated position has been converted into kinetic energy. The pendulum then slows down as it climbs against gravity until it stops momentarily at the highest point of its swing, and gravity starts doing negative work, converting kinetic energy back into potential energy. The tension in the string, although a force present in the pendulum system, does no work because it acts perpendicular to the direction of the pendulum's motion at all times.

Air resistance does work, although typically very small, by opposing the motion of the pendulum and thus removing energy from the system, mostly in the form of thermal energy due to the friction between the air and the pendulum bob. In simple pendulum motion, air resistance is usually considered negligible, but it does contribute to the eventual stopping of the pendulum due to energy dissipation.

The force of gravity does work in both directions, the tension in the string does no work, and air resistance does work by dissipating the pendulum's energy.

Force of Gravity (Weight): The force of gravity does work on the pendulum bob. Gravity acts downward and provides the restoring force that accelerates the pendulum back toward its equilibrium position. As the pendulum swings, gravity causes the bob to move, thereby doing positive work as the bob descends and negative work (taking energy away from the pendulum's motion) as the bob ascends.

Tension in the String: The tension in the string does no work on the pendulum bob. This is because the tension force always acts perpendicular to the direction of the bob's instantaneous motion. Work is defined as the force component in the direction of displacement times the displacement itself. Since there is no displacement in the direction of the tension force, it does no work.

Air Resistance: Air resistance does work on the pendulum bob, but in the opposite sense to gravity. Air resistance acts against the direction of motion of the bob, causing it to lose energy over time. This dissipation of energy due to air resistance results in damping, gradually reducing the amplitude of the pendulum's oscillation.

Answer:

lead

Explanation:

Lead was once widely used in the United States as a gasoline additive.

Addition of lead is in the form of tetra ethyl lead(II).

It helps to improve the octane rating of gasoline and to produce more useful energy via each combustion step.

Answer:

Center of mass

Explanation:

The Doppler technique is a good method for discovering exoplanets. It uses the Doppler effect to analyze the motion and properties of the star and planet. Both the planet and the star are orbiting a common center of mass. This means that the star and the planet gravitationally attract one another, causing them to orbit around a point of mass central to both bodies.

If we use the Doppler method to measure the period with which a star alternately moves toward and away from us due to an orbiting planet, then we also know the planet's orbital period

When utilizing the Doppler method to measure the periodic shifts in a star's spectral lines caused by an orbiting planet, not only do we discern the star's radial velocity variations, but crucially, we also ascertain the orbital period of the planet. This period signifies how long it takes for the planet to complete one orbit around its host star.

The Doppler effect causes a star's light to shift towards the blue end of the spectrum as it moves closer to Earth and towards the red end as it moves away. By analyzing these shifts, astronomers can deduce the star's motion induced by the gravitational pull of its companion planet. The periodicity of these shifts corresponds directly to the planet's orbital period. This information is fundamental in understanding the planet's characteristics, including its distance from the star and its mass, contributing significantly to our knowledge of exoplanetary systems in the vast cosmos.

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Both the objects travel equal distance before stopping.

Explanation:

Given-

Mass of object 1, m₁ = 4kg

Speed of object 1, v₁ = 2m/s

Mass of object 2, m₂ = 1kg

Speed of object 2, v₂ = 4m/s

Force₁ = Force₂ = F

Distance, s = ?

We know,

[tex]v^2 - u^2 = 2as\\\\[/tex]

Where, v is the final velocity

            u is the initial velocity

            a is the acceleration

            s is the distance

When the brake is applied, the object comes to rest and the final velocity, v becomes 0. So,

[tex]s = \frac{u^2}{2a}[/tex]

We know,

[tex]a = \frac{F}{m}[/tex]

The stopping distance becomes,

[tex]s = \frac{u^2m}{2F}[/tex]

For object 1:

[tex]s = \frac{(2)^2 X 4 }{F}[/tex]

[tex]s = \frac{16}{F}[/tex]

For object 2:

[tex]s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}[/tex]

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.

Answer:

The hang time is 0.95 s

Explanation:

Spud Webb has a height of 5 feet 8 inches and is one of the shortest basketball players to play in the NBA.

He could make an amazing vertical leap of 1.1 m off the ground. For such a leap to calculate the hang time(the time spent in the air after leaving the ground and before touching down again), we use the formula:

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Where S is the distance traveled, u is the initial velocity, t is the time taken and g is the acceleration due to gravity.

Given that:

g = 9.8 m/s²

S = [tex]y_{f} -y_{i} =0-1.1=-1.1[/tex]

u = 0

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Substituting values:

[tex]-1.1=(0)t -\frac{1}{2}(9.8)t^{2}\\-1.1=0-4.9t^{2} \\-1.1=-4.9t^{2}[/tex]

Dividing through by -4.9 we get:

[tex]\frac{-1.1}{-4.9} =\frac{-4.9}{-4.9}t^{2}[/tex]

[tex]t^{2}=0.2245\\ t=\sqrt{0.2245}=0.474[/tex]

t = 0.474 s

The hang time = 2t = 2 × 0.474 = 0.95 s

The hang time is 0.95 s

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

[tex]F=6.00*10^{-6}N[/tex]

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

[tex]F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}[/tex]

Substitute the given values

[tex]F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N[/tex]

Answer:

A=0

Explanation:

I got 100% on this assignment

This question involves the basic concept of displacement.

The frog's displacement is "0 cm".

In order to calculate the displacement of the frog, we must consider both the magnitude and direction of the movements made by the frog. Here, we will take the forward direction as positive and the backward direction as negative. Hence, the displacement of the frog will be as follows:

Displacement = 18 cm - 6 cm - 12 cm

Displacement = 0 cm

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The attached picture shows the difference between the displacement and the distance.

Answer:

B. OSMOTIC PRESSURE WILL BE LOWER IN THE ARTERIOLE END OF THE CAPPILLARY BED COMPARED TO THE VENOUS END.

Explanation:

This is true for filtration to take place in the cappillary bed. Osmotic pressure is the net pressure that drives movement of fluid from the interstitial fluid back into the capillaries. Osmotic pressure increase favors reabsorption as water moves from region of higher water concentration in the interstitial fluid to the lower region of water concentration in the capillaries.

At the ends of a capillary bed, the difference in the hydrostatic and osmotic pressures provides a net filtration or reabsorption ratio. At the arteriole end of the capillary bed, hydrostatic pressure is greater than the osmotic pressure allowing movements of fluid to the interstitial fluid (filtration) while as the blood moves to the venous end, the osmotic pressure becomes greater than than hydrostatic pressure.

Osmotic pressure is usually higher at the arteriole end of the capillary bed than at the venous end (Option C). This happens because plasma proteins remain in the capillary causing water to move back into the capillary.

In general, it is expected that osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end (option C). This is because during capillary exchange, fluids and solutes are filtered out at the arteriole end of capillaries due to higher blood pressure, and then reabsorbed at the venous end due to higher osmotic pressure. This helps maintain fluid balance and prevent edema.

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Answer:

Explanation:

Given

Power output [tex]P=5\ kW[/tex]

efficiency [tex]\eta =25\ \%[/tex]

Engine expels [tex]Q_r=8\times 10^3\ J[/tex]

Efficiency is given by

[tex]\eta =1-\dfrac{Q_r}{Q_s}[/tex]

where [tex]Q_s[/tex]=Heat supplied

[tex]0.25=1-\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]0.75=\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]Q_s=\dfrac{8\times 10^3}{0.75}[/tex]

[tex]Q_s=10.667\ kJ[/tex]

Work Produced by  cycle

[tex]W=Q_s-Q_r[/tex]

[tex]W=10.667-8[/tex]

[tex]W=2.667\ kJ[/tex]

Time interval for which power is supplied

[tex]P\times t=W[/tex]

[tex]t=\dfrac{W}{P}[/tex]

[tex]t=\dfrac{2.667}{5}[/tex]

[tex]t=0.5334\ s[/tex]  

The energy taken in each cycle of this particular heat engine is 2.0 x 10^4 J, and the time interval for each cycle is four seconds.

The efficiency of a heat engine (e) is derived from work output (W) divided by the energy input (Qin). Given the mechanical power output of the engine and its efficiency, we can use this formula to determine the energy input and the time interval for each cycle.

(a) Energy taken in during each cycle: Since efficiency e = Wout/Qin, then Qin = Wout / e = 5,000W (or 5 x 103 J/s) / 0.25 = 2.0 x 104 J. Where Wout is 5000W converted to Joules per second.

(b) Time interval for each cycle: The energy balance for one cycle is given by Qin = Wout + Qexhaust, where Wout is work output and Qexhaust is exhaust energy. The time for one cycle t = Qin / W = 2.0 x 104 J / 5,000 J/s = 4 seconds. The time interval for each cycle is four seconds.

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Answer:

[tex]E_a=1124.83 J/mol[/tex]

Explanation:

Given that second order equation

K₁ = 0.0796 L/mol-s , T₁= 737⁰C

T₁ = 737 + 273 K = 1010 K

K₂ = 0.0815 L/mol-s , T₂=947°C

T₂=947+273 K= 1220 K

The activation energy given as follows

[tex]\ln\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]

Now by putting the values we can get

[tex]\ln\dfrac{0.0815}{0.0796}=\dfrac{E_a}{8.314}\left ( \dfrac{1}{1010}-\dfrac{1}{1220} \right )[/tex]

[tex]0.023=0.00017\times \dfrac{E_a}{8.314}[/tex]

[tex]E_a=0.023\times \dfrac{8.314}{0.00017}[/tex]

[tex]E_a=1124.83 J/mol[/tex]

Therefore the activation energy will be 1124.83 J/mol

Complete Question:

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.  

The bodies have the masses shown in this table (refer attached figure).  

Which motor and body should Devon use to build the car with the greatest acceleration?

motor 1, with body 1

motor 1, with body 2

motor 2, with body 1

motor 2, with body 2

Answer:

Devon should build the car with motor 2 and body 1 for having the greatest acceleration.

Explanation:

As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.

It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,

          [tex]acceleration =\frac{\text { Force }}{\text { mass }}[/tex]

It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.

The car with the greatest acceleration will be one that has a higher power-to-weight ratio, provided by a light yet powerful motor, and a lighter, aerodynamically efficient body. Both factors--lightweight and power-- are crucial for achieving high acceleration.

In deciding what motor and body to use, Devon must consider factors like the power to weight ratio, the torque of the motor, and the aerodynamics of the body. A higher power-to-weight ratio generally translates to greater acceleration. Therefore, Devon should choose a motor that is powerful yet light. When considering the body, Devon should go for a lighter body as heavy bodies slows down a car's acceleration. Besides weight, a body whose design is aerodynamically efficient will enhance acceleration because it reduces air resistance.

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Answer:

Angular displacement=2π/3.9 rad

Explanation:

Given data

Time t=3.9s

Required

The angular displacement during a 1.0 s time interval

Solution

In 3.9 second the child covers a full circle=2π rad

Angular displacement after 1.0 second is given as:

[tex]=\frac{2\pi }{3.9} rad[/tex]

Angular displacement=2π/3.9 rad

The child's angular displacement on the merry-go-round during a 1.0s time interval would be approximately 1.609 radians.

To find the child's angular displacement on the merry-go-round, we first need to know the rate at which the merry-go-round is turning. This is called the angular velocity and is measured in radians per seconds (rad/s). If it takes 3.9 seconds for the merry-go-round to make one full revolution, this equals 2π radians. Therefore, the angular speed of the merry-go-round is 2π/3.9 rad/s.

Now, if we want to know how much the child displaces in 1.0 second, we simply multiply the angular speed by the time interval. So the angular displacement is (2π/3.9 rad/s)*1.0s = 1.609 rad.

Therefore, the child's angular displacement during a 1.0s time interval would be approximately 1.609 radians.

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Answer:Hydrogen lines will be weak if the star is too hot or too cold.

Explanation:

At higher temperatures, the hydrogen atom ionizes due to the atomic collisions. M spectral class stars are mainly the main sequence and red stars. They are in a temperature range of 3000 K which means that these stars have maximum ionized calcium lines. In this star, hydrogen atoms have electrons in the lower energy state and it is difficult to absorb photons. These stars do not have enough temperature for absorption and undergo fusion.

A negative test charge moves toward regions of higher electric potential and lower electric potential energy, which is the opposite direction of the electric field defined by positive charges.

A negative test charge will accelerate toward regions of higher electric potential and lower electric potential energy. This happens because a negative charge moves oppositely to the electric field direction, which is defined from high to low potential. When a negative charge, such as an electron, moves toward a higher potential, it is moving towards a region where it would have a lower potential energy if it were positive; however, since it is negative, its potential energy actually decreases as it moves in this direction.

Understanding electric fields and potentials, consider that a positive test charge is repelled by positive charges and attracted to negative charges. Since the field lines point away from positive charges and toward negative charges, a negative test charge would move in the direction opposite to the field lines, meaning it moves from lower to higher potential but in doing so, it lowers its electric potential energy.

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = [tex]\sqrt{(\frac{v_x}{\omega })^2+y^2 }[/tex]  = [tex]\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }[/tex]  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

Full Question:

What is the Lewis structure of the covalent compound that contains one nitrogen atom, one hydrogen atom, and one carbon atom?

Explanation:

This covalent compound is the hydrogen cyanide, HCN.

The following steps are used to obtain it;

Step 1. Draw a skeleton structure of the compound based on the elements

Put the least electronegative atom C in the middle with H and Cl on either side.

H-C-N

Step 2. Count the valence (outermost) electrons you can use

H + C + N = 1 + 4 + 5 = 10

Step 3. Add these electrons to give every atom an octet

You have to put a triple bond between C and N.

Note: Each bond is made up of two electrons.

The lewis structure is given in the image below;

Answer:

The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

Explanation:

Given:

Distance moved in the East direction (d) = 45 km

Distance moved in the North direction (D) = 110 km

Displacement is defined as the difference of final position and initial position.

Let us draw a diagram representing the above situation.

Point A is the starting point and point C is the final position of RV.

So, the displacement of RV in two days is given as:

Displacement = Final position - Initial position = AC

Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.

Using Pythagoras theorem, we have:

[tex]AC^2=AB^2+BC^2\\\\AC=\sqrt{AB^2+BC^2}[/tex]

Plug in the given values and solve for AC. This gives,

[tex]AC=\sqrt{(45\ km)^2+(110\ km)^2}\\\\AC=\sqrt{2025+12100}\ km\\\\AC=\sqrt{14125}\ km\\\\AC=118.85\ km[/tex]

Now, the direction of displacement with the east direction is given as:

[tex]\theta=\tan^{-1}(\frac{BC}{AB})\\\\\theta =\tan^{-1}(\frac{110}{45})=67.75^\circ[/tex]

Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

The displacement of the rv over the two days is 118.85 km.

The given parameters;

The displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;

[tex]c^2 = a^2 + b^2\\\\c = \sqrt{a^2 + b^2} \\\\c = \sqrt{(45)^2 + (110)^2} \\\\c = 118.85 \ km[/tex]

Thus, the displacement of the rv over the two days is 118.85 km.

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Answer:

C

Explanation:

Both technicians are correct.

Cheers

Technician B is correct because the ECT and IAT should read similarly after a car has been sitting for a while, reflecting the same ambient temperature. TSBs can address a variety of issues, not necessarily involving a specific DTC as Technician A suggests.

Technician B is correct. Technical Service Bulletins (TSBs) typically address widespread problems or issues found in a particular model of a vehicle. While these bulletins may include information about specific Diagnostic Trouble Codes (DTCs), they do not always involve a specific stored DTC, as Technician A suggests. TSBs can cover various issues, including non-DTC-related performance, noise, and other iterative improvements that are not related to any stored codes.

Technician B is accurate in saying that the Engine Coolant Temperature (ECT) sensor and the Intake Air Temperature (IAT) sensor readings should be close to the same after the vehicle has been sitting for several hours, typically overnight. This is because both the coolant in the engine and the air in the intake manifold would have reached ambient temperature, reflecting similar temperatures if the vehicle's sensors are functioning correctly.

Answer choice B, which credits Technician B, is the correct option.

Answer:

0.28 m/s

Explanation:

From the equation of a moving wave,

V = λf.................. Equation 1

Where V = speed of the water wave, λ = wave length of the water wave, f = frequency of the water wave.

Given: λ = 7.0 cm = 0.07 m, f = 4 Hz.

Substitute into equation 1

V = 0.07(4)

V = 0.28 m/s.

Hence the speed of the water wave = 0.28 m/s

Final answer:

The force exerted by the toy locomotive on the caboose is 3.08 N, which is calculated by adding the force required to overcome friction and the force necessary for acceleration according to Newton's second law.

Explanation:

To determine the magnitude of the force exerted by the toy locomotive on the caboose, we can use Newton's second law, which states that the force is equal to the mass times the acceleration (F = ma).

However, we also need to account for the frictional force acting on the caboose. The total force required to accelerate the caboose includes the force to overcome friction and the force needed for acceleration:

Step 1: Calculate the force needed to overcome friction, which is already given as 0.48 N.

Step 2: Determine the force needed to accelerate the caboose using the formula F = ma, where m is the mass of the caboose (1.0 kg) and a is the acceleration (2.6 m/s2).

Step 3: Add the force to overcome friction to the force required for acceleration to get the total force exerted by the locomotive on the caboose.

Step 2 in detail: F = ma = 1.0 kg × 2.6 m/s2 = 2.6 N.

Step 3 in detail: The total force exerted by the locomotive on the caboose is 2.6 N (to accelerate) + 0.48 N (to overcome friction) = 3.08 N.

Therefore, the force exerted by the toy locomotive on the caboose is 3.08 N.

Answer:

Technician A

Explanation:

The stator is a STATIONARY component of a rotary system. The stator is a stationary component in the alternator, therefore, technician B saying that it is a rotating component is WRONG.

Also, slip rings allows the flow of electrical signals and electrical power from a stationary structure to a rrotating structure. Therefore, technician A is correct in his take that the slip rings allow current to flow from the stationary end frame to the moving rotor.

Final answer:

Technician A is correct about slip rings allowing current flow to the rotor, while Technician B is incorrect as the stator is the stationary part of an alternator.

Explanation:

Technician A is correct in saying that slip rings allow current to flow from the stationary part to the moving rotor. Slip rings are made of a ring and brushes; the ring is attached to the rotating part of a machine (like an alternator's rotor), and the brushes are attached to the stationary part (such as the end frame), facilitating electrical contact as the rotor spins.

Technician B is incorrect; the stator is the stationary part of an alternator or electric motor that contains conductors where the electric current is induced by the motion of the rotor. In an alternator, the rotor is the rotating component, not the stator.

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

   mass per unit length ([tex]\mu[/tex]) = [tex]\frac{M}{l}[/tex] = 1 kg/m

      Tension = 4 N

    Speed (v) = [tex]\sqrt{\frac{F}{\mu}}[/tex]

So, when F is doubled then change in value of F will be as follows.

               F = 4 + 4 = 8 N

Therefore, speed will be calculated as follows.

           v = [tex]\sqrt{\frac{8 N}{1 kg/m}}[/tex]

              = 2.8 m/s

Thus, we can conclude that the wave speed is 2.8 m/s.