If vector P was fixed but the direction of Vector Q could be changed, for which value of tetha will the resultant force have a maximum value ? ​

Florida typically has a much smaller temperature swing than a dry desert environment, where the climate will typically show wide temperature swings on a daily basis, from very cold to very hot because of the water surrounding Florida.

The correct option is B.

The State of Florida is a peninsula, which means it is surrounded on three sides by water: the Gulf of Mexico, the Atlantic Ocean, and the Straits of Florida.

Hence, because of the waters surrounding Florida, it has fewer temperature swings.

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Answer:

Explanation:

d

Answer:

t = 0.995 s, 5.127 s

Explanation:

Δx = v₀t + 1/2at²

25.0 m = (30 m/s)t + 1/2(-9.8 m/s²)t²

-4.9t² + 30t - 25 = 0   solve this quadratic equation for t (2 solutions)

t = 0.995 s, 5.127 s

It reaches a height of 25 m in 0.995 s, then again 5.127 s later on its way back down

Every 13,000 years, Polaris and Vega switch places as the North Star. Today, Polaris, the brightest star in the constellation Ursa Minor, lies within one degree of the Earth's axis.

Currently, Polaris is practically exactly where the Earth's rotation axis points. However, the North Star in the year 3000 BCE was a star called Thuban (also known as Alpha Draconis), and due to the precession of the rotation axis, the bright star Vega will become the North Star in roughly 13,000 years.

The North Star around 3000 BC was a slender star called Thuban in the constellation of Draco. The North Star did not replace Polaris.

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The final velocity of object A is equal to the final velocity of object B, which is 5.0 m/s.

If the collision between objects A and B is elastic, then both momentum and kinetic energy are conserved.

Let the initial velocity of object A be [tex]v_A[/tex] and the initial velocity of object B be [tex]v_B[/tex]. The mass of object A is  and the mass of object B is [tex]m_B[/tex].

Conservation of momentum:

[tex]m_A v_A + m_B v_B = m_A v' + m_B v'[/tex]

where v' is the final velocity of both objects.

Conservation of kinetic energy:

[tex]\dfrac{1}{2} m_A v_A^2 + \dfrac{1}{2} m_B v_B^2 = \dfrac{1}{2} m_A v'^2 + \dfrac{1}{2} m_B v'^2[/tex]

Substituting v' in terms of [tex]v_A[/tex] and [tex]v_B[/tex] using the momentum conservation equation,

[tex]v_A = \dfrac{m_A - m_B}{m_A + m_B}v_B + \dfrac{2m_B}{m_A + m_B}v_B[/tex]

[tex]v_A = \dfrac{(m_A - m_B + 2m_B)v_B}{m_A + m_B}[/tex]

[tex]v_A = \dfrac{(m_A + m_B)v_B}{m_A + m_B}\\= v_B[/tex]

Therefore, the final velocity of object A is equal to the final velocity of object B.

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Answer:

Explanation:

WHEN WE PROJECT A BALL FROM THE ROOFTOP, THE BALL travels down and the ball have greatest speed at the moment strike on the earth.  ball have smallest speed at the moment of release.. the acceleration is constant.

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On the surface of Mars, the gravitational field strength is roughly [tex]3.71 m/s^2.[/tex]

The influence that a large body has on the space surrounding it, exerting force on another large body, is described by the gravitational field. The following formula can be used to determine the strength of the gravitational field (g) on the surface of a planet:

g = [tex]\frac {GM}{ r^2}[/tex]

where G is the gravitational constant [tex]6.67 \times 10^{-11} Nm^2/kg^2[/tex], M is the mass of the planet, and r is the radius of the planet.

Plugging in the values for Mars:

g =  [tex]6.67 \times 10^{-11} Nm^2/kg^2[/tex] x [tex]\frac{(6.37 \times 10^23 kg)}{(3.43 \times 10^6 m)^2}[/tex]

g= [tex]3.71 m/s^2.[/tex]

Consequently, the strength of the gravitational field on the surface of Mars is about. [tex]3.71 m/s^2.[/tex]

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Projectile motion exhibits negligible horizontal acceleration if air resistance is disregarded. As a result, the particle moves horizontally at a constant speed.

A projectile is an item on which gravity is the only force at work. Projectiles come in a variety of forms. A fallen object is referred to as a projectile motion (provided that the influence of air resistance is negligible).A projectile is something that is thrown vertically upward (provided that the influence of air resistance is negligible). In addition, any object sent upward at an angle to the horizontal is considered a projectile (provided that the influence of air resistance is negligible). Any object that, after being launched or dropped, keeps moving due to its own inertia and is solely impacted by the gravitational force acting downward is considered a projectile.

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The required elapsed time between spots when frequency and time for an experiment is specified is calculated to be 100 ms.

We know the relation between frequency and time as,

f = 1/T

where,

f is frequency

T is time period

It is given that f₁ = 40 hz

Time T₁ = 25 ms

If the timer is set to f₂ = 10 hz, time T₂ = ?

From the above relation, the equation can be written as,

f₁ T₁ = f₂ T₂

T₂ = f₁ T₁/f₂ = 40(25)/10 = 100 ms

Thus, the elapsed time between spots is calculated to be 100 ms.

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The ball was rolling at a speed of 4.44 m/s when it left the table.

What is speed?

Speed is a scalar quantity that describes the magnitude of an object's velocity, or the rate at which an object is moving. It is defined as the distance traveled by an object divided by the time it takes to travel that distance. The unit of speed is usually meters per second (m/s) or kilometers per hour (km/h).

We can use the conservation of energy principle to solve this problem. The initial potential energy of the ball when it was at the edge of the table is equal to its weight multiplied by its height above the floor:

PEi = mgh

PEi = m * 9.8 m/s² * 1.0 m = 9.8 N * m

At the moment the ball strikes the floor, all of its potential energy has been converted into kinetic energy:

KEf = 1/2 * m * v²

where v is the velocity of the ball just before it strikes the floor. We can equate the initial potential energy to the final kinetic energy and solve for the velocity of the ball when it left the table:

PEi = KEf

9.8 N * m = 1/2 * m * v²

v²= 2 * PEi / m = 2 * 9.8 N * m / m = 19.6 m²/s²

v = √(v²) = √(19.6 m²/s²) = 4.44 m/s

So, the ball was rolling at a speed of 4.44 m/s when it left the table.

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The rock which undergoing free fall spends approximately 1.02 seconds more in the air if the rock lands 20.0 m below where it was thrown upward.

The rock undergoing free fall on planet x is thrown upward at 5.0 m/s and it undergoes constant acceleration vertically downward. When the rock landing the velocity was 15.0 m/s.

The time spent in the air by the rock can be calculated using

v = u + at

where v is the final velocity 15 m/s downward, u is the initial velocity 5 m/s upwards, a is the acceleration which is constant here and equal to acceleration due to gravity ⇒ 9.8 m/s² and finally t is the time in the air.

v = u + at  ⇒ v - u = at

15.0 m/s - 5.0 m/s = g × t

10.0 m/s = g × t

t = 10.0 m/s / g  ⇒ t = 10.0 m/s / 9.8 m/s²  = 1.02 s

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Average power provided by the tension in cable when 5 skiers where uplift is 176.04watt.

We can characterize power as the speed of taking care of business, it is the work done in unit time. The SI unit of force is Watt (W) which is joules each second (J/s). Once in a while the force of engine vehicles and different machines is given concerning Strength (hp), which is roughly equivalent to 745.7 watts.We can characterize normal power as the complete energy consumed separated by the all out time taken. In basic language, we can say that typical power is the typical measure of work done or energy changed over per unit of time.

We know that power is the ratio of the work done to the respective time taken

here,work done is the potential energy at height of 115m which is equla to mgh.

Now,weight of 5 skiers is =65×5kg=325kg.

Now,potential energy of 5 skiers is 325×65=21125Joules.

Total time taken =2min=2×60=120sec.

Therefore,power provided by the tension in cable is 21125/120=176.04Watt.

Hence,power provided by the tension in cable is 176.04watt.

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Nuclear energy production generates several byproducts, including spent nuclear fuel and high-level radioactive waste. These byproducts pose a significant challenge for safe disposal and management.

There are several options for disposing of nuclear waste, including:

Deep geological disposal: This involves burying the waste in deep geological formations, such as salt mines or clay deposits, where it will be isolated from the biosphere for a very long time.

Reprocessing: This involves separating reusable materials from spent nuclear fuel, with the remaining waste being further processed and prepared for disposal.

Interim storage: This involves temporarily storing waste in secure facilities until a permanent disposal solution can be found.

Ocean dumping: This involves disposing of waste in the deep ocean, which is currently banned under international agreements.

The most widely accepted method of disposing of high-level radioactive waste is deep geological disposal. However, the implementation of this method requires significant investment, public acceptance, and technical expertise, and is currently only being carried out in a few countries. As the amount of nuclear waste continues to grow, it is important for countries to develop safe and sustainable disposal solutions to protect human health and the environment.

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There is Covalent bond between Carbon and Hydrogen.

In chemistry, a covalent bond is an interatomic coupling formed when two atoms share an electron pair. The electrical attraction of their nuclei for the identical electrons is what causes the binding.

When the linked atoms' combined energies are lower than those of widely spaced atoms, a covalent connection is created. To learn more, go to chemical bonding: ionic bonding. The inorganic elements hydrogen, nitrogen, chlorine, water, and ammonia (H2, N2, Cl2, H2O, NH3) as well as all organic compounds are examples of molecules with covalent connections.

Covalent bonds are denoted as solid lines joining atom pairs in structural representations of molecules.

Therefore, There is Covalent bond between Carbon and Hydrogen.

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The magnitude of total momentum of the two-puck system is 7.0 kg m/s.

The total momentum of the two-puck system before the collision is the vector sum of the momentum of each puck. The momentum of each puck can be found using the equation:

p = mv, where m is the mass and v is the velocity.

The momentum of the 1.5 kg puck is:

p1 = m1 * v1 = 1.5 kg * 2.0 m/s = 3.0 kg m/s

The momentum of the 4.0 kg puck is:

p2 = m2 * v2 = 4.0 kg * 1.0 m/s = 4.0 kg m/s

The total momentum of the two-puck system before the collision is:

p_before = p1 + p2 = 3.0 kg m/s + 4.0 kg m/s = 7.0 kg m/s

Since the collision is elastic (no friction), the total momentum of the two-puck system after the collision will be the same as before the collision, meaning the magnitude of the total momentum after the collision will be 7.0 kg m/s.

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The woman's average velocity during her entire motion is 0.823m/s.

Distance traveled in the south =  60×60×2= 7200m

Distance traveled in the North = 3000m

Net Displacement = 7200-3000= 4200m

Average velocity= Net Displacement /Time = 4200m / (60+25)×60= 0.823m/s.

The pace is the directional speed of an item in motion as a demonstration of its price of alternate in the role as observed from a specific body of reference and as measured via a particular widespread of time (e.g. 60 km/h northbound). speed is a fundamental concept in kinematics, the branch of classical mechanics that describes the motion of our bodies.

Velocity is a bodily vector amount; each significance and route are needed to define it. The scalar absolute value (significance) of speed is known as velocity, being a coherent derived unit whose quantity is measured inside the SI (metric device) as meters in step with second (m/s or m⋅s−1). as an example, "5 meters in step with 2d" is a scalar, whereas "5 meters consistent with 2nd east" is a vector. If there is a trade-in velocity, route, or each, then the object is stated to be present process an acceleration.

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Answer:

0.024

Explanation:

no expansivity ok but I can do it

linear expansivity=1.2*10^-5

=2a

=2*1.2*20^-5

=2.4*10^-5

∆A =?

A1 =20

2.4*10^-5 =∆A/A1(100-50)

∆A =2.4*10^-5*1000

=0.024

The buoyant forces acting on the two balls will be the sam. This is because buoyant force is determined by the volume of the object and not the depth at which it is submerged.

The buoyant force is the same for any object of the same volume regardless of the depth.

The buoyant force is an upward force exerted on a submerged object by the surrounding liquid. The magnitude of this force is equal to the weight of the liquid displaced by the object. This force is commonly used to explain why objects float in water.

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A mechanical wave is a vibration of matter, a wave responsible for the transmission of energy through a medium.

His four examples of mechanical waves?

Creeping waves, water waves, stadium waves, and jump rope waves are other examples of mechanical waves. Everyone needs a medium to exist. A Slinky wave requires a Slinky turn. Water waves needs the water. Stadium waves need stadium fans. A jump rope wave requires a jump rope.

What are the sources of mechanical waves?

Mechanical waves are created by interactions between adjacent particles in a medium. This interaction transfers energy and momentum from one particle to the next, and the net effect is to pass these quantities from the source to the receiver.

What is important in mechanical waves?

A mechanical wave is a wave that requires a medium to carry energy from one place to another. Mechanical waves cannot travel through particle-free regions of space because they rely on particle interactions for energy transport.

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To calculate the time required to increase the speed of an electric wagon from 79.4 m/s to 119.5 m/s at an average rate of 11.2 m/s2, you can use the following equation:

This gives a result of 4.03 seconds.

When the electric wagon is accelerating, the driver should be mindful of the acceleration rate to ensure that the vehicle is accelerated safely. If an excessive rate of acceleration is used, the vehicle may end up going too fast and this could be dangerous. Therefore, it is important to be aware of the acceleration rate and to keep an eye on the speedometer to ensure that it does not exceed the desired speed.

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The velocity at the bottom of the arc with the ball of 500kg at the end of a 30 m cable suspended will be 8.88 m/s.

Velocity can be defined as the displacement of the object in a particular time period. Velocity is a vector quantity because it has both the magnitude and direction. The SI unit of velocity is m/s.

We have given mass, m = 500 kg

Length of cable, l = 30 m

Angular displacement of the ball, θ = 30°

Gravitational potential energy of the ball at the displaced position is equal to,

U = mgl (1 - cosθ)

Maximum kinetic energy of the ball, Ke = 1/2 mv²

According to energy conservation,

1/2 mv² = mgl (1 - cosθ)

v = √2gl(1 - cosθ)

v = √2 × 9.8 × 30(1-cos30)

v = 8.875 m/s

Therefore, the velocity of the arc is 8.88 m/s.

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At the point when waves change starting with one medium then onto the next, the frequency of the wave changes relying upon the properties of the two mediums.

Infrared to Radio waves: Frequency increments

Radio waves to Microwaves: Frequency diminishes

Bright to Gamma beams: Frequency diminishes

Gamma beams to X-beams: Frequency diminishes

Infrared to Bright: Frequency diminishes

Yellow to blue: Frequency diminishes

Green to orange: Frequency increments

Infrared to red: Frequency increments

Bright to violet: Frequency diminishes

At the point when waves change starting with one medium then onto the next, the frequency might change relying upon the refractive lists of the two media.

Assuming the refractive list of the subsequent medium is more noteworthy than the first, then, at that point, the frequency will diminish. In the event that the refractive record of the subsequent medium is lower than the first, then, at that point, the frequency will increment.

The spectrum of all EM radiation is known as the electromagnetic (EM) spectrum. The visible light from your house's lamp and the radio waves from a radio station are both examples of electromagnetic radiation, which is energy that moves and spreads out.

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The mass which when hung at the 85cm mark will balance the ruler horizontally​ is  600 g.

The mass which when hung at the 85cm mark will balance the ruler horizontally​ is calculated as follows;

( 150 g x 60 cm ) = ( 85 cm - 60 cm) m

( 150 g x 60 cm ) = ( 15 cm) m

where;

The mass which when hung at the 85cm mark is calculated as

m = ( 150 g x 60 cm ) / ( 15 cm )

m = 600 g

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It is observed that both the pennies will land at the same time. This is because second penny has only horizontal velocity.

Projectile motion occurs when an object moves through two or more dimensions simultaneously, usually under the influence of gravity. The kinematic equations regulating projectile motion apply independently in each dimension, which is a crucial aspect of the motion.

Both pennies will land at the same time. The second penny's velocity is fully horizontal, therefore its vertical velocity is still zero, and this is the explanation for it. As a result, it will have the exact same kinematic equation in the y-dimension as the first penny and will fall to the ground at the same moment. The time till impact will not alter regardless of the second penny's horizontal speed.

The question is incomplete. The complete question is 'If one penny is balanced on the edge of the table and you place your index finger on a second penny then flick the second penny so that it ravels off the table, while the first penny is gently nudged off the edge, which penny will land first, the penny moving horizontally or the one that simply drops off the table. Please explain the answer.'

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Answer:

[tex]20\; {\rm m\cdot s^{-1}}[/tex], assuming that friction is negligible.

Explanation:

If an object of mass [tex]m[/tex] moves at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of this object would be [tex]p = m\, v[/tex].

if friction is negligible, the sum of the momentum should be the same before and after the collision.

Before the collision:

In other words, the sum of momentum was [tex]40\; {\rm kg\cdot m\cdot s^{-1}}[/tex] before the collision.

After the collision:

The sum of the momentum stays unchanged at [tex]40\; {\rm kg\cdot m\cdot s^{-1}}[/tex]. Subtract the momentum of the ice skater from the sum to find the momentum of the snowman:

[tex]40\; {\rm kg\cdot m\cdot s^{-1}} - 0\; {\rm kg\cdot m\cdot s^{-1}} = 40\; {\rm kg\cdot m\cdot s^{-1}}[/tex].

Divide the momentum of the snowman by its mass to find its velocity:

[tex]\begin{aligned}\frac{40\; {\rm kg\cdot m\cdot s^{-1}}}{2\; {\rm kg}} = 20\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In other words, the collision of the snowman would be [tex]20\; {\rm m\cdot s^{-1}}[/tex] after the collision.

According to the graph the blood move 8 cm during one beat.

To find how far does the blood move during one beat, we need to find the area under the velocity curve.

The blood displaces [tex]\triangle x =x_f-x_i[/tex] between t = 0.1 s and t = 0.3 s is the area under the curve from t = 0.1 s to t = 0.3 s.

In this case, the curve is a triangle, thus the area of a triangle is, area of the triangle between t = 0.1 s to t = 0.3 s.

[tex]\dfrac{1}{2} \times 0.2 \times 0.8[/tex] = 8cm

Therefore, blood moves 8 cm during one beat.

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Answer:

Below

Explanation:

The block does not slide, so the force of friction equals the down the slope force.  Ff = 26 n     the Fn  is given as 42 N

Definition of  static coefficient of friction =  Ff/Fn = 26/42 = .62

The degree of change in a thermometer's thermometric characteristic for a unit temperature change is known as sensitivity. When something fluctuates, it undergoes rapid, unpredictably changing change.

Thermal fluctuations are arbitrary departures from the average state of a system that take place in an gravity system, according to statistical mechanics. All thermal fluctuations grow in size and frequency as temperature rises and diminish as temperature falls closer to absolute zero.

That must be "temperature fluctuation," I suppose. Low pressure and high pressure are two separate sorts of pressure, which is why that is the case. Low pressure systems frequently bring inclement weather, which may include wind, rain, snow, or hail. As also, in those.

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The number of equations required to relate the position of a point on one cord to a point on another cord in a system of pulleys with more than one cord is equal to twice the number of cords.

In a system of pulleys with more than one cable, the number of equations needed to connect the location of a point on one cord to a point on another chord is equal to twice the number of cords. This is because for each cord, an equation is needed to describe its position as a function of time, and another equation is needed to describe its velocity as a function of time.

In a system of pulleys with more than one cable, the number of equations needed to connect the location of a point on one cord to a point on another chord is equal to twice the number of cords. This is due to the necessity for two equations: one to describe each cord's position as a function of time and another to explain its velocity as a function of time.

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The complete question is

if a system of pulleys contains more than one cord, relate the position of a point on one cord to a point on another cord. how many equations are required? group of answer choices

a. one one per cord

b. twice the number of cords

c. thrice the number of cords

d. none of the above

As for the formula, yr=yc +(I/(A*yc)), is to find the force acting on a vertical plate (plate perpendicular to the surface of the water)

Assume a flat plate is present in a tank of water, at a depth of h from the water's surface, and that the plate is parallel to the water's surface.

The force acting on the plate will be equal to the weight of the water present above it.

The volume of the water present above it is A*h, where A is the area of the flat plate and h is depth.

kg/m3 is the density of water in rho.

rho*h*A gives the mass of the water above the plate.

Thus by multiplying g, we get the weight of the water above the plate.

In the above experiment, the force acting on the quadrant is balanced by slotted weights and you will be able to obtain the force.It would have been better if you had showed the simplified diagram.

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