The sum of kinetic energy and potential energy within a system is called
es
A)
thermal energy
B)
chemical energy.
electrical energy.
D)
mechanical energy.

With 20 windings, the generator will produce a higher EMF and, consequently, a larger current. This increased current will provide more power to the light bulb, resulting in a brighter illumination compared to the dim illumination observed with only 5 windings.

When you increase the number of wire windings in the generator from 5 to 20, the effect on the light bulb will be a brighter illumination. The brightness of the light bulb is directly proportional to the number of windings in the generator.

By increasing the number of windings, you are increasing the amount of wire wrapped around the magnet. This results in a higher number of turns per unit length, leading to an increased magnetic flux passing through the wire coils.

According to Faraday's law of electromagnetic induction, a change in magnetic flux induces an electromotive force (EMF) in a conductor, which in this case is the copper wire. The induced EMF causes electric current to flow through the wire, creating a flow of electrons.

The 30-W light bulb requires a certain amount of electrical power to produce its specified brightness. With 20 windings, the generator will produce a higher EMF and, consequently, a larger current. This increased current will provide more power to the light bulb, resulting in a brighter illumination compared to the dim illumination observed with only 5 windings.

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Wycleff is on block 1, Lilly is on block 4, Emilia is on block 12, and Quincy is on block 17 is the correct statement about each person standing at block.

As a source and observer move closer or farther apart, the Doppler effect, which is a phenomenon in physics, causes the frequency of sound, light, or other waves to rise or fall. When a source emits waves, those waves are compressed as they approach the observer.

Wycleff could hear a low-pitched sound the entire time, indicating that the police car was relocating away from him. He was thus in block 1.

A Doppler effect (change in pitch) was the first thing Lilly noticed. She was at block 4, so the car must have passed her first.

After Lilly, Emilia noticed a Doppler effect. She was at block 12 when the car passed Lilly before passing her.

The entire time, Quincy heard a high-pitched sound, which indicated that the police car was approaching. He was thus in block 17.

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Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time \(t\), the horizontal position \(x\) and vertical position \(y\) of the ball are given respectively by

\(x = v_i \cos(\theta_i) t\)

\(y = v_i \sin(\theta_i) t - \dfrac g2 t^2\)

and the horizontal velocity \(v_x\) and vertical velocity \(v_y\) are

\(v_x = v_i \cos(\theta_i)\)

\(v_y = v_i \sin(\theta_i) - gt\)

The ball reaches its maximum height with \(v_y=0\). At this point, the ball has zero vertical velocity. This happens when

\(v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g\)

which means

\(y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g\)

At the same time, the ball will have traveled half its horizontal range, so

\(x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g\)

Solve for \(v_i\) and \(\theta_i\) :

\(\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0\)

Since \(0^\circ<\theta_i<90^\circ\), we cannot have \(\sin(\theta_i)=0\), so we're left with (e)

\(3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}\)

Now,

\(\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}\)

\(\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}\)

so it follows that (d)

\(R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}\)

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

\(v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}\)

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

\(v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}\)

Then with \(v_y=0\), the ball's speed \(v\) is

\(v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}\)

Finally, in the work leading up to part (e), we showed the time to maximum height is

\(t = \dfrac{v_i \sin(\theta_i)}g\)

but this is just half the total time the ball spends in the air. The total airtime is then

\(2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}\)

and the ball is in the air over the interval (a)

\(\boxed{0 < t < 2\sqrt{\frac R{3g}}}\)

The frequency that the person on the train hears is calculated as 767 Hz.

Frequency is the number of waves that passes a fixed point in unit time.

The formula for the observed frequency (f') of a wave with a known frequency (f) due to the Doppler Effect is:

f' = f (v + vo) / (v + vs)

v : speed of the wave in the medium (in this case, the speed of sound)

vo : speed of the observer (in this case, the speed of the train)

vs is the speed of the source (in this case, assumed to be zero)

vo = 85.0 km/h * 1000 m/km / 3600 s/h = 23.6 m/s

f' = 725 Hz * (343 m/s + 23.6 m/s) / (343 m/s + 0 m/s) = 767 Hz

Therefore, the frequency that the person on the train hears is 767 Hz.

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Answer:

The answer is air

Explanation:

The magnitude of the resultant vector to round the answer to the nearest tenth, we look at the digit in the hundredth's place. If this digit is 5 or greater, we round up. If it is less than 5, we round down.

In the study of physics, we use vectors to represent quantities that have both direction and magnitude. It is often the case that we want to add two or more vectors together to obtain a single vector that represents the net result of these additions. The process of adding two or more vectors together is known as vector addition.The magnitude of the resultant vector is the length of the line that represents it on a scale drawing.

When we add two or more vectors together, the resultant vector is the vector that represents the net result of these additions. To find the magnitude of the resultant vector, we use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

In the case of vector addition, the hypotenuse is the resultant vector, and the other two sides are the component vectors. If we have two vectors a and b, the magnitude of the resultant vector is given by the following equation:|R| = √(ax2 + bx2)where R is the resultant vector, a and b are the component vectors, and x is the angle between the vectors.

For example, if the answer is 12.345, we would round it to 12.3.

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The activity is a radioactive material is the number of counts per second it produces. 1 count per second (cps) = 1 Becquerel (Bq).

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The voltage between the plates is 3.9 × \(10^-^3\) V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × \(10^-^1^5\) J

Charge of the electron = 1.602 × \(10^-^1^9\) coulomb

Using,

     ΔU = q × ΔV

4.1 × \(10^-^1^5\) = 1.602 × \(10^-^1^9\) × ΔV

      ΔV  = 3.9 × \(10^-^3\) V

Therefore, the voltage between the plates is 3.9 × \(10^-^3\) V.

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We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is

From the question we are told

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Generally the equation for the Block is mathematically given as

\(\sum Fy=0\)

\(981cos21.80 = R_2cos53.6\\\\R_2=1535N\)

the equation for the Wedge is mathematically given as

\(\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N\)

the equation for the Screw is mathematically given as

\(\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\\)

Therefore

\(M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m\)

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Over the time period from t₁ to just before t₃, the two-block-spring system is linear momentum does not change because there are no external forces acting on the system to change its momentum.

Explain why the linear momentum of the two-block–spring system is either constant or not during the time period between t₁ and the time just before t₃?

The linear momentum of the two-block-spring system remains constant from t₁ until just before t₃.

Prior to the collision at time t₂, block A has a momentum to the right that is Pa = ma ×va, whereas block B has a momentum that is Pb = 0 because it is at rest. Because there is no external force acting on the system, its overall momentum is conserved.

All out Ptotal = Dad + Pb = mama × va

Expect there is immaterial erosion between each block and the surface. Take into consideration the spring, block A, and the system for parts (a) to (d).

Following the collision at time t₂, block B experiences a velocity of vb to the right and comes to a stop. Because the collision is elastic, all of the system's kinetic energy is conserved, but momentum is still conserved:

The two-block system is moving to the right at a constant velocity vb from time t₂ to just before time t₃, so the momentum of the system is constant. Ptotal = ma₀ plus mbvb plus mb ×vb

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The potential difference is related to the kinetic energy of the electron by the formula,

1/2mv^2 = eV

where

m is the mass of the electron

e is the charge of the electron

V is the potential dfference

From the information given,

V = 382.11 x 10^3

m = 9.1 x 10^-31

e = 1.6 x 10^- 19

Thus, we have

1/2 x 9.1 x 10^-31 x v^2 = 1.6 x 10^- 19 x 382.11 x 10^3

45.5 x 10^- 32v^2 = 611.376 x 10^- 16

v^2 = 611.376 x 10^- 16/45.5 x 10^- 32

Taking square root of both sides,

v = 3.67 x 10^8 m/s

The velocity is 3.67 x 10^8 m/s

Answer:

Transition Metals

Explanation:

__________________________

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

From the darkest part of the object's shadow, you would be able to see a small amount of the light source. There would be a small amount of light that is visible, but it would be faint. On the other hand, from the lightest part of the shadow, you would be able to see much more of the light source. The light source would be far brighter and more visible, and you would be able to identify the source of the light.

Cube moves first, that the cube which is neutral in nature. Hence, the cube which is neutral will move.  

To solve this problem, we need to calculate the electrostatic force between the two charged cubes and compare it with the force of static friction between the cubes and the surface. The cube with the smaller force of friction will begin to move first. The parameter panel and sketch of the problem are shown below:

Parameter Panel:

Mass of cube 1 (m1) = 2.0 g = 0.002 kg

Mass of cube 2 (m2) = 4.0 g = 0.004 kg

Distance between the centers of the cubes (d) = 6.0 cm = 0.06 m

Charge rate of each cube (q) = 7.0 nC/s

Coefficient of static friction (μ) = 0.65

Sketch:

     |-----------|        |-----------|

     |     2     |        |     1     |

     |-----------|        |-----------|

           |                 |

          d=6.0 cm          d=6.0 cm

           |                 |

           |-----------------|

(a) To determine which cube moves first, we need to calculate the electrostatic force between the charged cubes and compare it with the force of static friction between the cubes and the surface. The electrostatic force between two charged objects is given by Coulomb's law:

F = (k * q1 * q2) / d^2

where k is the Coulomb constant (9.0 x 10^9 N m^2/C^2), q1 and q2 are the charges on the cubes, and d is the distance between them. For each cube, the charge is increasing at a rate of 7.0 nC/s, so the charge at any time t is given by:

q = 7.0 x 10^-9 C/s * t

At t = 0, the cubes are neutral and have no charge. At some later time t, the charges on the cubes are:

q1 = 7.0 x 10^-9 C/s * t

q2 = 7.0 x 10^-9 C/s * t

The electrostatic force between the cubes is then:

F = (9.0 x 10^9 N m^2/C^2) * (q1 * q2) / d^2

= (9.0 x 10^9 N m^2/C^2) * [(7.0 x 10^-9 C/s * t)^2 / (0.06 m)^2]

The force of static friction between each cube and the surface is:

Ff = μ * N

= μ * m * g

where N is the normal force, m is the mass of the cube, μ is the coefficient of static friction, and g is the acceleration due to gravity (9.81 m/s^2).

The normal force N is the force exerted by the surface on the cube, and is equal in magnitude to the weight of the cube:

N = m * g

Plugging in the values for each cube, we get:

Ff1 = μ * m1 * g

= 0.65 * 0.002 kg * 9.81 m/s^2

= 0.0127 N

Ff2 = μ * m2 * g

= 0.65 * 0.004 kg * 9.81 m/s^2

= 0.0254 N

Comparing the electrostatic force between the cubes and the force of static friction, we find:

F - Ff1 = (9.0 x 10^9 N)

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The types of chemical reactions are as follows:

Chemical reactions are changes in which new substances are formed.

The types of chemical reactions are as follows:

In conclusion, chemical reactions produce new substances.

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Answer:

C) The book's inertia carried it forward.

When the bus stops suddenly, the book tends to remain in motion due to its inertia. The book was at rest on the seat of the bus, and when the bus stopped suddenly, the book continued moving forward with the same speed and direction it had before the bus stopped. As a result, the book slid off the seat and onto the floor.

Answer:

They both pull the same amount. For every force there is an equal and opposite force.

Explanation:

The rate of heat energy exhausted to the river is 600.96 MW.

The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.

Given parameters:

Efficiency of the power plant; η = 31 %

Output  electric power; O = 270 MW.

We know that, Efficiency of the power plant;

η  = (Output  electric power/ input power)× 100%

⇒ input power = (Output  electric power × 100)/η

⇒ input power = (270 × 100)/31 MW

= 870.96 MW.

So, the rate of heat energy exhausted to the river that cools the plant =  Input power- output power

= (870.96 - 270) MW

= 600.96 MW.

Hence, heat energy exhausted to the river that cools the plant  is 600.96 MW.

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Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Answer:

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Explanation:

It is possible that identical rock formations can exist thousands of kilometers apart because of the rock cycle, a process that involves the continuous transformation of rocks through various stages such as weathering, erosion, deposition, and more.

To prove that the identical rock formations formed at the same time, geologists can look for clues such as the presence of the same type of minerals, the same layering or structure, and similar levels of weathering or erosion. This information can be incorporated into the model by including representations of these clues and explaining their significance in the rock cycle.

Processes that could have separated the rock formations over time include tectonic movement, erosion, and weathering. These processes can be incorporated into the model by including representations of tectonic plates and showing how they can move and collide, as well as by including examples of erosion and weathering and explaining their role in the rock cycle.

To determine how and when mountain ranges formed, geologists can study the rock formations, the types of minerals present, and the levels of weathering and erosion. This information can be incorporated into the model by including representations of different types of rock formations and explaining how they were formed through processes such as mountain building and erosion.

Weathering and erosion can cause both fast and slow changes to Earth's surface, and can affect both large and small areas. To include this information in the model, you could have two separate models to show different time and spatial scales. One model might show the slow weathering and erosion of a rock, while the second model might show the fast weathering and erosion of a mountainside during a landslide.

To model weathering and erosion, you could put rocks in a container and shake it many times to simulate weathering, or use water or a fan to model erosion by rivers or wind. You could also use a source of heat to model energy from Earth's interior, or a fan to model wind energy.

In your model, you should include processes that describe the cycling of Earth materials and the flow of energy that drives the cycling. These processes include weathering, erosion, deposition, compaction, cementation, melting, crystallization, pressure, deformation, subduction, and seafloor spreading.

In your article, you could start by introducing the rock cycle and explaining the various processes involved. You could then describe how these processes shape and change rocks on Earth's surface at different time and spatial scales, using examples to illustrate your points. You could also include information about the clues that geologists look for to determine the history of a rock formation, and how these clues can be used to understand the rock cycle. Finally, you could conclude by summarizing the key points and explaining the significance of the rock cycle in understanding the Earth's surface.

If P and Q are two longest paths in a connected graph then P and Q must have at least one vertex in common.

Let P and Q be two longest paths in a connected graph G. Assume for contradiction that P and Q have no vertices in common. Without loss of generality, let P start from vertex u and end at vertex v, and let Q start from vertex w and end at vertex x.

Since G is connected, there exists a path between u and w and a path between v and x. Let's consider the path between u and w first. By definition, both P and the path between u and w have u as a common vertex. If the path between u and w is longer than P, then it contradicts that P is the longest path. Hence, the path between u and w is shorter than P.

Similarly, the path between v and x must also be shorter than Q.

So, we have two paths of length less than the length of P and Q connecting vertices that are not on P or Q, which means that the concatenation of these two paths along with P and Q form a cycle. The length of this cycle must be longer than P and Q, which again contradicts the assumption that P and Q are the longest paths in G.

(b) Disprove: The statement is false. Consider a cycle of length k in the graph G. The two diametrically opposite vertices are two geodesics of length k, but they do not have any common vertices. This shows that the statement "If P and Q are two geodesics of length k in vertex in G, then P and Q have at least one common" is not always true.

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The net work must be done on the particle to cause it to move to the right at 47 m/s is 59.4J, The spring constant of spring is 281.75N/m and the length of spring stretched is 0.02m.

(1.)Given the mass of particle (m) = 75g = 0.075kg

The velocity of particle (Vl) towards left = 25m/s

The velocity of particle towards right (Vr) = 47m/s

The work done to move the particle from left to right = W

Here work done can be considered as change in Kinetic energy then:

W = (KEf - KEi)

\(W = 1/2*m*Vr^2 - 1/2*m*Vl^2\)

\(W = (1/2)*75 g*(47 m/s)^2 - (1/2)*75 g*(25 m/s)^2\)

\(W = (1/2)*75 g*(1584 m^2/s^2)\)

W = 59.4J

Therefore, the net work required to cause the particle to move from 25 m/s to 47 m/s is 59.4 J.

(2.) Given the length of spring = 8cm = 0.08m

The mass of block = 2.3kg

The extension in spring = 16cm = 0.16m

Let the spring constant = k

(A) We know that the force exerted in stretching the spring = F = kx where

F = mg then

mg = kx

2.3 * 9.8 = k * (0.16 - 0.08)

k = 22.54/0.08 = 281.75N/m

(B) The mass of new block = 3kg

Let the extended spring = x'

Then mg = kx' such that:

3 * 9.8 = 281.75 * x'

x' = 29.4/281.75 = 0.10m

the extension in spring = 0.10 - 0.08 = 0.02m

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Explanation:

The heat flow can be calculated as

Where k is the thermal conductivity, A is the area, T1 is the temperature of the ice water, T2 is the temperature of the room, and L is the thickness.

Replacing k = 0.01 J/(s m °C), A = 0.950 m², T1 = 0°C, T2 = 35 °C and L = 2.50 cm = 0.025 m, we get

Therefore, the heat flow through the walls is 13.3 W.

The negative sign indicates the direction of the heat flow, in this case, the heat goes to the water.

Answer:

Approximately \(11\; {\rm kg}\).

Explanation:

Look up the Gravitational Constant \(G\):

\(G \approx 6.6743 \times 10^{-11}\; {\rm N\cdot m^{2} \cdot kg^{-2}}\).

When two objects of mass \(M\) and \(m\) respectively are at a distance of \(r\) away from one another, the gravitational attraction between them would be:

\(\begin{aligned}F &= \frac{G\, M\, m}{r^{2}}\end{aligned}\).

In this question, it is given that \(F = 3.5 \times 10^{-6}\; {\rm N}\), \(M = 58\; {\rm kg}\), and \(r = 0.11\; {\rm m}\). Rearrange the equation to find \(m\), the mass of the other object:

\(\begin{aligned}m &= \frac{r^{2}\, F}{G\, M} \\ &\approx \frac{(0.11)^{2}\, (3.5 \times 10^{-6})}{(6.6743\times 10^{-11})\, (58)}\; {\rm kg} \\ &\approx 11\; {\rm kg}\end{aligned}\).

(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

Learn more about wavelength here: https://brainly.com/question/10728818

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