*9. A small loop of area 6.8 mm2 is placed inside a long solenoid that has 854 turns/cm and carries a sinusoidally varying current i of amplitude 1.28 A and angular frequency 212 rad/s . The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop

Answer:

the answer is b

Explanation:

hope it helps ...

Lithium (Li) is a metal located in Group 1 of the periodic table and is known as an alkali metal. It stands out for its low density and capacity to conduct electricity, along with exhibiting the typical properties of metals.

The element lithium (Li) is a metal, and notably, it is the first metal in the periodic table located in Group 1. It has the lowest atomic number (3) among the metals. The defining characteristics of metals include their ability to conduct electricity, possess malleability, and often exhibit a shiny luster. Lithium, being part of the alkali metals category, shares these properties. Furthermore, lithium is the least dense metal and has a unique electronic structure where the third electron occupies an outer shell, differentiating it from nonmetals.

Lithium is used in various applications including batteries and the aerospace industry due to its low density and high reactivity. The chemically similar elements to lithium include other alkali metals such as sodium and potassium, which also share similar chemical properties.

Answer:    

Compared to windshield the airbag exerts much lesser force  

Explanation:

Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time

Impulse = Impulsive force *time

I = F*Δt

If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.

Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.

During the collision, the passenger is carried towards the windshield and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.

The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.

Thus compared to the windshield the airbag exerts much lesser force.

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

Answer:

a = 40 m / s²

Explanation:

For this exercise we can use the vertical launch equations to which these are valid in any system with inertial

      v = v₀ + a t

      a = (v-v₀) / t

We calculate

     a = (90 - 10) / 2

     a = 80/2

     a = 40 m / s²

This is the value of the planet's gravity acceleration

Answer:

an inhibitor of angiotensin II

Explanation:

Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.

Answer:

Explanation:

Atomic mass of hydrogen is 1 . so 1 g of hydrogen will have 6.02 x 10²³ atoms

3 g of hydrogen will have 3 x 6.02 x 10²³ atoms . This will give 3 x 6.02 x 10²³  protons and same number of electrons

So number of protons at north pole = 3 x 6.02 x 10²³

charge on these protons = 1.6 x 10⁻¹⁹ x  3 x 6.02 x 10²³ C

= 28.9 x 10⁴ C  

similarly total charge on electrons at south pole = 28.9 x 10⁴ C

distance between them = diameter of the earth = 2 x 6356 x 10³ m

= 12712 x 10³ m

Attractive force between these charges

= k q₁q₂ / r² , q₁ ,q₂ are charges and r is distance between charges.

= 9 x 10⁹ x (28.9 x 10⁴)² / (12712 x 10³ )²

= 4.6517 x 10⁻⁵ x 10¹¹

= 4.6517 x 10⁶ N

Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

Answer:

shadows are made by blocking light

Given that,

Horizontal velocity of the object, v = 20 m/s

Height of the cliff, h = 125 m

We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,

[tex]h=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g and u = 0

[tex]h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s[/tex]

So, the object will take 5 seconds to fall to the ground from the cliff.

The time taken for the object to fall to the ground is 5 s

From the question given above, the following data were obtained:

The time taken for the object to fall to the ground can be obtained as follow:

[tex]t = \sqrt{ \frac{2h}{g}} \\ \\ t = \sqrt{ \frac{2 \times 125}{10}} \\ \\ t = 5 \: s \\ \\ [/tex]

Therefore, the time taken for the object to get to the ground is 5 s

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Answer:

ii. its distance from the other galaxy.

iii. its size compared to that of the other galaxy

Explanation:

A galaxy is a vast constituent of interstellar medium of gas, dust and billions of stars bounded by the gravitational pull. There are tens of billions of  galaxies existing in the whole universe. Various numbers of galaxies can merge to form an inter-cluster, which is a more massive and bigger galaxy.

The constituent of an individual galaxy are held together by mutual gravitational pull, but may interact with other galaxies depending on; its distance from other galaxies and its size compared to that of the other galaxy. This process would result into the formation of a giant elliptical galaxy and a number of stars.

Answer:

1.Gravity acts on existing material, shaping it into a new form.

2.a much larger galaxy, a nearby galaxy

3.low gas content, little star formation

Explanation:

i just did the test and its 100%. dont believe me? just do the test and see for your self

Answer:

120 decibels

Explanation:

we know sound in decibels is given by.

[tex]\beta =10log(\frac{I_{} }{I_{0} } )[/tex]

Where I is intensity of the sound and [tex]I_{0}=10^-12W/m^{2 }[/tex] is reference intensity.

substituting All of this in our decibel formula with I =1W/[tex]m^2[/tex]

gives us 120 decibels.

The sound intensity level of a sound is measured in decibels (dB) using a logarithmic scale that uses a reference intensity of 10^-12 W/m², which is the threshold of human hearing. The decibel level of a sound is determined by the ratio of its intensity to this reference intensity, with each increment of 10 dB corresponding to a tenfold increase in intensity.

The sound intensity level ß, measured in decibels (dB), provides a logarithmic measure of sound intensity I in watts per meter squared (W/m²), with the formula

ß = 10 log(I/I0)

where I0 = 10⁻¹² W/m² is the reference intensity, corresponding to the threshold of human hearing. The decibel level of a sound is thus derived from the ratio of its intensity to this reference intensity. A sound with the same intensity as the reference intensity (I = I0) has a decibel level of 0 dB, because log 10(1) = 0.

Interpreting decibels involves understanding that because the formula uses a logarithm, each increment of 10 dB corresponds to a tenfold increase in intensity. Therefore, a sound with an intensity of 10^-11 W/m² would be 10 dB, a sound with an intensity of 10⁻¹⁰ W/m² would be 20 dB, and so on. The decibel scale helps to compress the vast range of sound intensities that human ears can perceive into a more manageable range of numbers.

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Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, [tex]N - mg = F_c[/tex]

[tex]N-mg=\frac{mv^2}{r}[/tex]

Now substitute the given values, we get

[tex]N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}[/tex]

[tex]N=15.68+46.08[/tex]

N = 61.76  N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In this case, the magnitude of the normal force is equal to the weight of the car, which is the product of its mass and the acceleration due to gravity.

The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In a vertical circle, the net external force equals the necessary centripetal force. The only two external forces acting on the car are its weight and the normal force of the road. Since the car is not leaving the surface and the net vertical force must be zero, the vertical components of the two external forces must balance each other.

The vertical component of the normal force is equal to the car's weight, which is mg. The weight can be calculated by multiplying the mass of the car by the acceleration due to gravity. In this case, the weight is 1.60 kg multiplied by 9.8 m/s². Therefore, the magnitude of the normal force exerted on the car at point A is also 1.60 kg multiplied by 9.8 m/s².

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Answer:

energy of activation is the correct answer.

Explanation:

The activation energy is the energy needed to initiate an exergonic reaction. It is required to kickstart the process, despite the reaction being one that eventually releases energy.

The energy required to initiate an exergonic reaction is called the activation energy. This is the minimum amount of energy required for a chemical reaction to occur. In an exergonic reaction, the activation energy is critical even though these reactions release energy overall, because it's needed to start the process. To envision this, think of pushing a boulder off a hill; the initial push (activation energy) is required for the action (the exergonic reaction) to begin, after which gravity (the reaction's natural tendency) takes over and the boulder rolls down (energy is released).

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When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. The location and orientation of the reflected ray depend on the shape of the mirror. Additional rays can be traced from the base of the object to locate the extended image.

When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. For a concave mirror, the reflected ray passes through the focal point, while for a convex mirror, the reflected ray extends backward through the focal point. The location and orientation of the reflected ray depend on the shape of the mirror. To locate the extended image, additional rays can be traced from the base of the object along the optical axis. All four principal rays run parallel to the optical axis, reflect from the mirror, and then run back along the optical axis. The image of the base of the object is located directly above the image of the tip, as the mirror is symmetrical from top to bottom.

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Ray tracing involves drawing rays from an object through points of interest like the focal point of a mirror and then determining where the reflected rays intersect to locate the image. It is a crucial method in optics to determine the properties of images formed by mirrors and lenses.

To trace the path of a ray emitted from the tip of an object toward the focal point of a mirror, as well as the reflected ray, one must understand the basic principles of ray tracing with mirrors. With a concave mirror, an incident ray that travels parallel to the optical axis will be reflected through the focal point on the same side of the mirror. Conversely, with a convex mirror, a ray that travels parallel to the optical axis will reflect as if it originates from the focal point behind the mirror, forming a virtual focus. To construct the reflected ray, you draw the ray until it reaches the mirror’s surface and then redirect it according to the mirror type's ray tracing rules.

Furthermore, the intersection of the reflected rays (real or virtual) determines the location of the image. If the rays meet in real space, the image is real; if they only appear to intersect upon extension in virtual space, the image is virtual.

To achieve a complete picture of how an image is formed, one must also trace rays from another point on the object. For instance, tracing rays from the base of the object can help determine the orientation of the image. In the case of rays that are collinear with the optical axis, the image will maintain the object's vertical orientation. By tracing at least two different rays following the simple ray tracing rules, one can locate the image formed by mirrors and lenses.

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Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

[tex]p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }[/tex]

[tex]p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }[/tex]

[tex]p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }[/tex]

[tex]N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}[/tex]

[tex]N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}[/tex]

[tex]N = 1300(16) e^{-3}[/tex]

N = 1035.57

N = 1036 times

Answer:

 The neutron core is completely destroyed

Explanation:

 A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

Final answer:

The incorrect statement is that a supernova explosion results in the complete destruction of the neutron core; often, the core becomes a neutron star. Supernovae can be life-threatening to nearby planets, but the actual danger zone is closer than a few dozen light-years. They are also responsible for creating new elements heavier than iron.

Explanation:

A supernova explosion is a cataclysmic event at the end of a massive star's life cycle. During a supernova, a number of phenomena occur, including the ejection of elements the star fused during its lifetime, and the creation of new elements heavier than iron. One of the statements provided as a result of supernova explosions is incorrect: 'The neutron core is completely destroyed'. In many cases, the core does not get completely destroyed; instead, it often collapses into a densely packed neutron star. Another result is a potential threat to life on planets within a few dozen light-years due to a life-threatening dose of radiation. It is important to note, though, that planets would need to be quite close to the supernova, much closer than a few dozen light-years, to receive such a dose. Finally, supernovae also contribute to the formation of new elements, with the violent energies enabling the fusion of atoms into elements heavier than iron.

Answer: 296.1 J or 6.98 kJ

Explanation:

Given

Length of chain, l = 11 m

Mass of chain, m = 95 kg

It is worthy of note that chain has two ends. So this depends which of the two ends should end up at

the given height of 7 m.

Scenario 1. If it is the leading end:

Weight of chain raised

W = mg/l

W = 9.8×95/11

W = 84.6 N

Height raised = ½ m (which is the height of the centre of gravity of the raised portion).

Work done = 84.6 × 1/2 * 7 = 296.1 Joules

Scenario 2. If it is the trailing end:

Weight of chain raised = 9.8×95 = 931 N.

Height raised (average) = 12.5 m (5.5 m for the chain midpoint + 7 m off the ground)

Work done = 931 × 7.5 = 6.98 kJoules

Thus, the work required is either of 296.1 J or 6.98 kJ

The work required to lift one end of a 95 kg chain to 7 meters height is 3261.85 J.

To calculate the work required to raise one end of the chain to a height of 7 meters, we consider the work done against the force of gravity. The chain has a mass of 95 kilograms, and we need to move it through a vertical distance of 7 meters.

Work is defined as the product of force and the distance over which the force is applied. In this case, the force is the weight of the chain, which is the product of its mass and the acceleration due to gravity (g = 9.81 m/s²). However, the chain gets lighter as you pull it up, since a smaller length of chain is left hanging at every moment. To solve this, we find the average force, which would be the weight of half the chain, as only half of it will be lifted through the full height on average.

The chain's total weight is mass multiplied by the acceleration due to gravity so we have 95 kg × 9.81 m/s².

The weight of half the chain is therefore

0.5 × 95 kg × 9.81 m/s²

The work done to lift half the mass through the full height is the average force times the height (7 m).

The calculation will then be:

{W} = 0.5 ×95 kg × 9.81 m/s² ×7 m

W= 3261.85 J

So, we multiply these values to find the total work done in lifting one end of the chain to a height of 7 meters.

Answer:

w = I₂ / (I₁ -I₂) w₀ ,    L₂ = 2 L₁

Explanation:

This is an angular momentum exercise,

      L = I w

where bold indicates vectors

We must define the system as formed by the bicycle wheel, the platform, we create a reference system with the positive sign up

At the initial moment the wheel is turning and the platform is without rotation

The initial angle moment is

                 Lo = L₂ = I₂ w₀

L₁ is the angular momentum of the platform and L₂ is the angular momentum of the wheel.

In the Final moment, when the wheel was turned,

                [tex]L_{f}[/tex] = L₁ - L₂

                L_{f} = (I₁ - I₂) w

the negative sign of the angular momentum of the wheel is because it is going downwards since the two go with the same angular velocity

as all the force are internal, and there is no friction the angular momentum is conserved,

             L₀ =L_{f}

             I₂ w₀ = (I₁ -I₂) w

             w = I₂ / (I₁ -I₂) w₀

we can see that the system will complete more slowly

 we can also equalize the angular cognition equations

                 L₀ = Lf

                 L₁ = L₂-L₁

                 L₂ = 2 L₁

In this part we can see that the change in the angular momentum of the platform is twice the change in the angular momentum of the wheel.

This experiment with a rotating bicycle wheel demonstrates the conservation of angular momentum. Rotating the wheel changes the platform's rotational direction to maintain a balanced angular momentum. Returning the wheel to its initial position stops the platform's rotation, preserving the system's angular momentum at zero.

In this experiment, we examine the principles of conservation of angular momentum using a rotating bicycle wheel. When you and a platform are stationary, and a colleague spins the bicycle wheel, the system's initial angular momentum is zero.

Here’s a step-by-step explanation:

This experiment vividly shows how changing the direction of angular momentum through torque affects rotational motion while conserving the overall angular momentum of the system.

1) 7.5 N

2) 10 N

Explanation:

1)

We can solve this first part of the problem by using Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

[tex]F=ma[/tex]

where

F is the net force

m is the mass of the object

a is its acceleration

In this problem we have:

m = 1.5 kg is the mass of the ball

[tex]a=5 m/s^2[/tex] is the acceleration

So, the kick force on it was:

[tex]F=(1.5)(5)=7.5 N[/tex]

2)

In this case, the mass of the ball has increased to

m' = 2 kg

We can also solve this part by using again Newton's second law of motion:

[tex]F'=m'a[/tex]

where

F' is the new kick force

m' = 2 kg is the new mass of the ball

a is the acceleration

The acceleration is the same as before,

[tex]a=5 m/s^2[/tex]

Therefore, the new kick force is:

[tex]F'=(2)(5)=10 N[/tex]

Answer:

(A) [tex]\alpha =11.277rad/sec^2[/tex]

(B) [tex]\Theta =2846.30rad[/tex]

Explanation:

We have given initial angular speed

[tex]\omega _i=1150rpm=\frac{2\times 3.14\times 1150}{60}=120.366rad/sec[/tex]

[tex]\omega _f=2680rpm=\frac{2\times 3.14\times 2680}{60}=280.506rad/sec[/tex]

Time t = 14.2 sec

(a) From first equation of motion

[tex]\omega _f=\omega _i+\alpha t[/tex]

[tex]280.506=120.366+\alpha \times 14.2[/tex]

[tex]\alpha =11.277rad/sec^2[/tex]

(b) From third equation of motion

[tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]

[tex]280.506^2=120.366^2+2\times 11.277\times \Theta[/tex]

[tex]\Theta =2846.30rad[/tex]

Answer:

(a) 1058.4 J

(b) -10584 J

Explanation:

Parameters given:

Mass of astronaut, m = 72 kg

Distance moved by astronaut, d = 15 m

(a) WORK DONE BY FORCE FROM THE HELICOPTER

Work done is given as the product of Force applied to a body and the distance moved by the body:

W = F * d

The force from the helicopter is given as:

F = m * a

where a = acceleration of the astronaut due to the helicopter

Therefore, the work done is given as:

W = m * a * d

W = 72 * g/10 * 15

W = [tex]\frac{72 * 9.8 * 15}{10}[/tex]

W = 1058.4 J

(b) WORK DONE BY FORCE OF GRAVITY

W = F * d

The force of gravity is given as:

F = -m * g

where g = acceleration due to gravity

The negative sign is due to the fact that the astronaut moves in an opposite direction (upwards) to the force of gravity (Gravity acts downwards)

Therefore, the work done is given as:

W = -m * g * d

W = -72 * 9.8 * 15

W = -10584 J

Answer:

Core

Radiative zone

Convective zone

Photosphere

Chromosphere

Transient region

Corona

Ranks of layers based on their distance from the sun’s center

1st-corona

2nd-Transient region

3rd-chromosphere

4th-Photosphere

5th-convective zone

6th-radiative zone

7th-core

Answer: 128J

Explanation:

[tex]Formula: E_k=\frac{1}{2}mv^2[/tex]

[tex]E_k=\frac{1}{2}(16kg)(4m/s)^2 \\E_k=\frac{1}{2}(16kg)(16m^2/s^2)\\ E_k=\frac{1}{2}(256kg*m^2/s^2)\\ E_k=128kg*m^2/s^2\\or\\E_k=128J[/tex]

Answer:

That is correct

Explanation:

128 J

The 40 kg skater moves approximately 6.19 meters towards the 65 kg skater when they pull themselves along a 10 m pole on ice, utilizing the conservation of momentum and considering the center of mass of the system.

To solve how far the 40 kg skater moves when two skaters pull themselves along a pole on ice, we use the principle of conservation of momentum. This principle states that in the absence of external forces, the total momentum of a system remains constant. Since both skaters start from rest, their initial total momentum is zero. As they pull towards each other, their momenta are equal in magnitude and opposite in direction, keeping the total momentum of the system at zero.

Let's denote the distance the 65 kg skater moves as d65 and the distance the 40 kg skater moves as d40. Because the pole has a length of 10 m, these two distances must add up to 10 m, so d65 + d40 = 10. To find out how far each skater moves, we look at their center of mass. The system's center of mass doesn't change because there are no external forces acting on the system.

The position of the center of mass xCM can be found using the formula xCM = (m1 * x1 + m2 * x2) / (m1 + m2), where m1 and m2 are the masses of the skaters, and x1 and x2 are their respective positions. Considering the system initially (before movement), the center of mass is at the midpoint of the pole as the skaters are at the ends. Therefore, xCM = 5 m.

Because the total mass of the system is 105 kg and the center of mass is 5 m from the starting point of either skater, we find that the 40 kg skater moves closer to the center of mass than the 65 kg skater to maintain the center of mass position. By directly calculating it, d40 = 10 * (65 / 105) = 6.19 m. Thus, the 40 kg skater moves approximately 6.19 meters towards the 65 kg skater.

Answer:

she used force to swing her ball all way back

Explanation:

Final answer:

Shannon swinging her racket to hit a tennis ball illustrates key physics concepts, including momentum and energy transfer, crucial for propelling the ball forward. The energy transferred from the racket to the ball depends on the racket's motion, enhancing the ball's speed when the racket moves toward it. Physics principles, such as Newtonian mechanics, are essential for understanding and improving tennis skills.

Explanation:

When Shannon swings her racket to hit a tennis ball, sending it flying in the same direction, several principles of physics come into play, particularly related to momentum and energy transfer. First, when the racket contacts the ball, momentum is transferred from the racket to the ball, propelling it forward. According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, as the racket exerts force on the ball, the ball exerts an equal and opposite force on the racket, though the effects are more noticeable on the ball due to its much smaller mass.

Energy transfer during the collision between the racket and the ball is significant. A moving racket transfers kinetic energy to the stationary ball, which then moves with a velocity dependent on the imparted energy. If the racket is moving towards the ball, the speed of the ball post-collision is greatly increased. This results from the addition of the racket's velocity to that imparted to the ball, showcasing an efficient energy transfer.

Understanding the dynamics of hitting a tennis ball, including the effects of the racket's speed, angle of swing, and point of impact, requires knowledge of Newtonian mechanics and is essential for improving one's tennis playing skills. Moreover, accounting for the arc of the ball and its spin are crucial for predicting how the ball will move, enhancing the player's ability to plan their shots strategically.

Answer:

The electrostatic force between two charge doubles when when the amount of charge on one object doubles.

Explanation:

Force between two charge particle [tex]F=\frac{1}{4\pi\epsilon } \frac{q_1\cdot\ q_2}{r^{2} }[/tex]

Where,  [tex]\frac{1}{4\pi\epsilon } = constant[/tex]

[tex]q_1,q_2\ magnitude\ of\ charges.\\r\ is\ the\ distance\ between\ the\ particles.[/tex]

Now suppose as per the question

[tex]q_1= 2 q_1\ with\ q_2\ and\ r\ with\ same\ value.\\Hence\ force\ between\ the\ particle\ F_2 =2 F[/tex]

Hence, electrostatic force between two charge doubles when when the amount of charge on one object doubles.

The ice in the open container begins to melt after 19.82 minutes of heating, and the temperature of the system starts to rise above 0°C after a total heating time of 223.93 minutes.

To answer both parts of this question, we need to calculate the time for two processes: the heating of the ice to 0°C (melting point), and the melting of the ice into water at 0°C.

Firstly, we find the heat (Q) required to raise the temperature of the ice to 0°C using Q=mcΔT, which is 0.55kg * 2100 J/kg°C * (0 - (-15.3°C)) = 17842.5 J. As the heater adds 900J per minute, the time for this is 17842.5J ÷ 900J/min = 19.82 minutes.

Next, we find the heat required to melt ice into water at 0°C using Q=mLf, where Lf = 334000 J/kg. This is 0.55kg * 334000 J/kg = 183700J. The time for this can be found by 183700J ÷ 900J/min = 204.11 minutes.

tmelts (the time before ice begins to melt) is 19.82 minutes, and trise (the total time before the temperature begins to rise above 0°C) is 19.82 min + 204.11 min = 223.93 minutes.

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The time before the ice starts to melt (tmelts) is 19.63 minutes, and the total time before the temperature begins to rise above 0 °C (trise) is 223.74 minutes.

To calculate the time tmelts before the ice starts to melt, you firstly need to determine how much heat is necessary to raise the ice from -15.3 °C to 0 °C using the formula Q = m * c * ΔT, where Q is the heat, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature. In this case, Q would be:

Q = (0.550 kg) * (2,100 J/kg·K) * (15.3 K) = 17,661.5 J.

Since heat is supplied at 900 J/minute, the time tmelts can be calculated as:

tmelts = Q / (heat rate) = 17,661.5 J / (900 J/min) = 19.63 minutes.

To calculate the time trise until the temperature begins to rise above 0 °C, we must add the time it takes to melt the ice completely at 0 °C. The heat of fusion (∆Hfus) formula is used here: Q = m * Lf where Lf is the heat of fusion for ice. Considering that the heat of fusion for ice is 334,000 J/kg:

Q = (0.550 kg) * (334,000 J/kg) = 183,700 J.

For melting the ice at 0 °C, it will take:

trise = Q / (heat rate) = 183,700 J / (900 J/min) = 204.11 minutes.

Therefore, the total time before the temperature begins to rise above 0 °C is the sum of the time to warm the ice to 0 °C and the time to melt it completely, which will be 19.63 minutes + 204.11 minutes = 223.74 minutes.

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Answer:

I think that the liquids molecules are slowing down. Hope this helps!

Explanation:

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Answer:

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

Answer:

Second balloon hits ground Δt = 0.185 seconds sooner than first balloon

Explanation:

Given:-

- The first balloon is thrown horizontally with speed, u1 = 2.0 m/s

- The second balloon is thrown down with speed, u2 = 2.0 m/s

- The height from which balloon are thrown, si = 6.0 m (above ground)

Find:-

Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Solution:-

- We will first determine the time taken (t1) for the first balloon thrown horizontally with speed u1 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

                         si = 0.5*g*t1^2

Where,     g: The gravitational constant = 9.81 m/s^2

                        6.0 = 4.905*t1^2

                        4.905*t1^2 - 6.0 = 0

- Solve the quadratic equation:

                        t 1 = 1.106 s

- Similarly, the time taken (t2) for the second balloon thrown down with speed u2 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

                         si = u2*t2 + 0.5*g*t1^2

Where,     g: The gravitational constant = 9.81 m/s^2

                        6.0 = 4.905*t1^2 + 2*t2

                        4.905*t1^2 + 2*t2 - 6.0 = 0

- Solve the quadratic equation:

                        t 2 = 0.9208 s

- We see that the second balloon thrown down vertically hits the ground first. The second balloon reaches ground, t1 - t2 = 0.185 seconds, sooner than first balloon.

Answer:

Height above a surface

Explanation:

Gravitational potential energy is the energy which an object possesses due to its position above a surface.

It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.

It is given mathematically as:

P. E. = m*g*h

where m = mass of the body

g = acceleration due to gravity

h = height above a surface

m*g represents the weight of the object.

Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.