A disk of radius R = 7. 52 cm is centered at the origin and lies along the y – z plane. The disk has a surface charge density σ = 5. 88 × 10 − 6 C / m 2. Evaluate the electric field produced by this disk along the x axis at point (P = 1. 01 m, 0. 00 m). The Coulomb force constant k = 1 / ( 4 π ϵ 0 ) = 8. 99 × 10 9 N ⋅ m 2 / C 2

Answer:

B. Atoms are altered in both processes

Explanation:

Answer: B. Atoms are altered in both processes

Explanation:

When cyclooctatetraene is treated with excess potassium metal dipotassium cyclooctatetraene dianion compound is formed.

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The wavelength, as a fraction of the length of a box, of the second excited state of a one-particle one-dimensional particle in a box is equal to $2/3$.

For a particle in a box, the wavelength of the particle is given as , where L is the length of the box and n is the quantum number. In this case, the quantum number is 2, since it is the second excited state. In addition, the fraction of the wavelength as compared to the length of the box is given as.

Substituting the values we have, Thus, for the second excited state (n = 2), Therefore, the wavelength is equal to the length of the box for the second excited state. However, to get the fraction of the wavelength as compared to the length of the box, we divide the wavelength by the length of the box to get .Hence, the wavelength, as a fraction of the length of the box, of the second excited state of a o

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The work done in pumping all the liquid out of the tank is 2500π lb-ft.

The potential energy of an object is given by the formula:

PE = mgh

Given:

Density of oil = 50 lb/ft^3

Diameter of the tank = 10 ft ( r = 5 ft)

To calculate the work done, we need to find the mass of the oil and the height it is lifted. Let's start by finding the mass:

Volume of oil in the tank:

The volume of a hemisphere = (2/3) * π * (r^3)

Volume = (2/3) * π * (5^3) = (2/3) * π * 125 = 250π/3 ft^3

Mass of the oil:

Mass = Volume * Density = (250π/3) * 50 lb

Now, we need to find the height of the tank. For a hemisphere, the height is equal to the radius (5 ft).

Finally, we can calculate the work done:

Work = PE = mgh = (250π/3) * 50 lb * 5 ft

Work = (250π/3) * 50 * 5 lb-ft

       = 2500π lb-ft

So, the work done in pumping all the liquid out of the tank is approximately 2500π lb-ft.

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Vertical distance from the point of observation on the Earth's surface to the point being measured. Altitude: Vertical distance from mean sea level to the point being measured.

gravitational potential energy = mass * acceleration due to gravity * height

gravitational potential energy = 55 * 9.8 * 370

gravitational potential energy = 539 * 370

gravitational potential energy = 199,430

Hence the total change in the skier's gravitational potential energy is 199,430Joules.

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Explanation:

Even though the man exerts force on the wall, the wall doesn't move . Work done= Force applied* displacement and since the displacement of the wall is zero., work done is also zero.

Answer:

Some electrons move from the neutral object to the positively charged object. Both objects become positively charged, but the object that started positively charged is less positively charged than it was initially.

Explanation:

This is because since protons do not move from the nucleus, but, electrons on the surface move, the electrons from the neutral object move to the positively charged object until there is no net movement of electrons between both objects. At this point, both objects carry the same charge and the positively charged object becomes less positively than it was initially.

Answer:

D: 9.5

Explanation:

2 seconds divided by 1 is 1 so 19 divided by 1 is 9.5

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

The balance of the seesaw would be affected if it were brought to the moon. On Earth, the balance of the seesaw depends on the force of gravity acting on the children and the distribution of their weights. The seesaw is designed in a way that when one child goes up, the other child goes down, maintaining balance.

However, on the moon, the force of gravity is only about 1/6th of what it is on Earth. This means that the children would experience a much weaker gravitational force. As a result, the seesaw would not be able to balance perfectly on the moon without some adjustments.

To balance the seesaw on the moon, the position of the children would need to be adjusted. The child with less weight would need to move closer to the center of the seesaw, while the child with more weight would need to move towards the end of the seesaw. By adjusting their positions, they can compensate for the weaker gravitational force and maintain balance on the seesaw.

In summary, the children would not be able to balance perfectly on the seesaw if it were brought to the moon. They would need to adjust their positions to account for the weaker gravity in order to maintain balance.

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Answer:

The answer is thermometer

Explanation:

I took the test and got it right

225 mins long will a bus take to travel 150 km at an average speed of 40 km h

Since the average speed is 40km/h it means

40 km can be travelled by it in 1 hour

1 km can be travelled by it in 1/40 hour

So 150km can be travelled by it in 150/40 hours or 225 mins.

By formula time=distance/speed=150/40=3.75hours

Speed = Distance/Time – This tells us how slow or fast an object moves. It describes the distance travelled divided by the time taken to cover the distance. Time = Distance / Speed, as the speed increases the time taken will decrease and vice versa.

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The crate moves with velocity vecv in the positive X-direction, as shown in the FBD above. Forces acting upon that crates are A 65 kilogramme crate is being kept from sliding to across warehouse floor by a force pf 250 N.

Writing equal forces in their component parts using Newton's second law

0 for Sigma F x.

0 Rightarrow F-F friction

----(1) \s\

0 for Sigma F y.

-mg=0 rightarrow

——(2)

Equation (1) yields F friction=F.

Rightarrow ___k=F ___k

Rightarrow mu _k=frac F N

Rightarrow mu k=frac fmg

——-from Equation (2)

Rightarrow _k=frac 250*65*9.8=0.39

fr = Fr/N

The resistive frictional force (Fr) divided more by normal of perpendicular energy (N) pushing adjacent objects together yields the friction coefficient (fr), which is a numerical value. Equation fr = Fr/N serves as a representation of it.

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The crate moves with velocity vecv in the positive X-direction, as shown in the FBD above. Forces acting upon that crates are A 65 kilogramme crate is being kept from sliding to across warehouse floor by a force pf 250 N.

0 for Sigma F x.

0 Rightarrow F-F friction

----(1) \s\

0 for Sigma F y.

-mg=0 rightarrow

——(2)

Equation (1) yields F friction=F.

Rightarrow ___k=F ___k

Rightarrow mu _k=frac F N

Rightarrow mu k=frac fmg

——-from Equation (2)

Rightarrow _k=frac 250*65*9.8=0.39

fr = Fr/N

The resistive frictional force (Fr) divided more by normal of perpendicular energy (N) pushing adjacent objects together yields the friction coefficient (fr), which is a numerical value. Equation fr = Fr/N serves as a representation of it.

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Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

\(R_{t} = 341.7 \: \Omega\)

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

\( R' = R_{1} + R_{2} + R_{3} \)

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.  

Entonces:

\( R' = 3R \)

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

\( \frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}} \)

\( R'' = \frac{3}{4}R \)

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

\( R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R \)

Esta resistencia está ahora en paralelo con R₇, entonces:

\( \frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}} \)

\( R'''' = \frac{11}{15}R \)

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

\(R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega = 341.7 \: \Omega\)

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

\(V-i_{1}R-i_{3}R=0\) (1)

Malla 2

\(-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0\) (2)

Malla 3

\(-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0\) (3)

Recordemos tambien que:

\(i_{1}=i_{2}+i_{3}\) (4)

\(i_{2}=i_{4}+i_{5}\) (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

\(i_{1}=3/13\: A\)

\(i_{2}=4/65\: A\)

\(i_{3}=11/65\: A\)

\(i_{4}=1/65\: A\)

\(i_{5}=3/65\: A\)

Finalmente:

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Answer:833

Explanation:

Answer:

Mechanical waves are A. Longitudinal and C. Transverse.

Explanation:

I hope this helps you.^_^

Answer:

209.3 m

Explanation:

x component will be  253 cos 55.8

y component will be  253 sin 55.8 = 209.3 m

Using the concepts of motion, we got that 0.816m is the maximum vertical height the box will reach if its initial velocity at the bottom of the ramp is 4.0 m/s.

We use Newton`s third equation of motion,

v² -u²=2aS---------(eq1)

where

v is the final velocity, and u is the initial velocity and a is the acceleration of the object and S is the displacement of the object.

We know, that at the maximum height, final velocity will be =0

Acceleration will be acceleration due to gravity(g)=9.8m/sec² .

Putting the values in (eq1)

=>0-(4)² = 2×(-9.8)×S

=>2×9.8×S = 16

=>S = [(16)/(2×9.8)]

=>S = 8/9.8

=>S =0.816m

Hence, the the maximum vertical height the box will reach if its initial velocity at the bottom of the ramp is 4.0 m/s is 0.816m

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The key observation enabled Edwin Hubble to prove this by observing individual Cepheid variable stars in Andromeda and applying the period-luminosity relation.

American astronomer Edwin Powell Hubble lived from November 20, 1889, through September 28, 1953. He was a significant contributor to the development of observational cosmology and extragalactic astronomy.

Hubble established that many objects formerly believed to be "nebulae" or clouds of gas and dust, were actually galaxies outside the Milky Way. He scaled galactic and extragalactic distances using the strong direct correlation between a classical Cepheid variable's luminosity and pulsation time, which was found in 1908 by Henrietta Swan Leavitt.

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Explanation:

he is very keen on taking up and is the age

Answer:To determine the minimum force required to move the block, we need to consider the static friction acting on the block. The static friction force can be calculated using the formula:

f_s = μ_s * N

where f_s is the static friction force, μ_s is the coefficient of static friction, and N is the normal force.

In this case, the normal force N is equal to the weight of the block, which is given as 50 N.

Substituting the values into the formula, we have:

f_s = 0.74 * 50

f_s = 37 N

Therefore, the minimum force required to move the block is 37 N.

Answer:

80 J

Explanation:

PE = mgh

PE = (4 kg)(9.8 m/s^2)(2 m)

PE = 78.4 J and with sig figs, it would be 80 J

Complete question:

A hydraulic jack is used to lift a car by applying a force of 120N at the pump piston, if the area of the ram and pump piston are 100cm squared and 1m squared respectively. What is the weight of the car?

Answer:

the weight of the car is 1.2 N

Explanation:

Given;

applied force, F₁ = 120 N

area of the effort (pump), A₁ = 1 m²

area of the load (ram), A₂ = 100 cm² = 1 x 10⁻² m²

let the weight of the car = F₂

The applied pressure is constant and the following equations can be used to calculate the weight of the car.

\(P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\F_2 = \frac{F_1 \times A_2}{A_1} \\\\F_2 = \frac{120 \times (1\times 10^{-2})}{1} \\\\F_2 = 1.2 \ N\)

Therefore, the weight of the car is 1.2 N

Answer:

4 m/s²

Explanation:

v = v₀ + at

80 = 0 + 20a

80 = 20a

a = 4 m/s²

Explanation:

1. By Newton's 3rd Law, both forces are equal and opposite. Hence both forces are of the same magnitude.

2. The reaction force is the force of arrow on bowstring.

3. Forward force of fish on water and vice versa?

4. The mass of the person is very small as compared to mass of earth. Hence the force exerted is very small amd hence the recoil is not noticed.

5. Reaction force is the force of ball against the bat. By Newton's Third Law, they form an action reaction pair. Hence, they are equal in magnitude and opposite in direction.

6. The reaction force will also decreases as both forces are equal in magnitude.

7 im nt sure sorry Ahahah

Answer:

a)5.925

ii)149.5 or383.5

Explanation:

ii)3.9×105-2.5×104

409.5-260

the zero in 260 may or may not be significant since it is after two significant figures thus we have two options

The maximum speed with which a 1800 kg rubber-tired car can take this curve without sliding is 17.89 m/s (rounded to two decimal places).

The formula for the maximum speed that a car can take a curve without sliding is:

v = √(rgμ)

Where:

v is the maximum speed (in m/s)

r is the radius of the curve (in m)

g is the acceleration due to gravity (9.81 m/s²)

μ is the coefficient of static friction between the tires and the road surface

In this case, the mass of the car (m) is 1800 kg and the coefficient of static friction (μ) between rubber and concrete is 1.0.

Therefore, the maximum speed of the car can be calculated as follows:

Let's say that the radius of the curve is 50 m. Then:

v = \(\sqrt{rg}\)μ

v = \(\sqrt{(50) (9.81) (1.0)}\)

= 22.14 m/s

However, this is the theoretical maximum speed that the car can take the curve without sliding. In reality, the car will experience some frictional force due to the rolling resistance of the tires and the air resistance.

Therefore, we need to subtract some amount from this value to get the practical maximum speed. Let's say that we subtract 20% from the theoretical value.

Then:

v = 0.8 × 22.14v

= 17.71 m/s

Rounding this value to two decimal places, we get:

v ≈ 17.89 m/s

Therefore, the maximum speed with which a 1800 kg rubber-tired car can take this curve without sliding is 17.89 m/s (rounded to two decimal places).

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Answer:

high sunlight intensity

high wind speed

Explanation:

The environmental conditions in the UK on 7th June 2017 are improved because the electricity generation in the UK is done by the renewable sources of energy.

A renewable resource,  is a type of natural resource that will refill to replace the fraction destroyed by consumption  in a finite length of time in a human timescale, either naturally or other periodic processes.

The environmental conditions in the UK are improved on 7th June 2017 due to following reasons ;

1. The renewable energy sources create comparatively low amounts of Gas (GHG) emissions and traditional pollution,

2. During manufacture and transportation. They do emit some emissions and pollutants.

3. Using fossil fuels to generate electricity that creates no greenhouse gas emissions and lowers some forms of air pollution.

4. Increasing energy diversity and minimizing reliance on foreign fuels.

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Answer:

The diagram is showing to bar magnets .....the magnet you force him on something sharp or something here you can use the magnet to control the object he can control the magnets and Maltese Cross .... it's so nice to watch the magnet move as much as the magnet then you attach it to a magnet eh move you move it

Explanation:

brainliest me pls

Answer:

I don't know the answer

Explanation:

can you brainliest me pls