Explain how fusion and fission are similar to and different from each other at the atomic level. Explain which
process is currently used at most nuclear power plants around the world and which process occurs on the sun.

Please helpp

Answer:

Explanation:

If generator really means alternator as used in cars, the following can be said:

The rotor is fed with a variable voltage ranging from 0 volt to 12 volts. The 2 brushes (fixed, not rotating, like the stator) make contact with two rings that are rotating with the rotor. The rotor acts like a fixed magnet, but the magnetic field strength is constantly updated to provide a constant output voltage, even if the load current change or the motor speed change.

The stator is basically 3 coils and 6 diodes that convert the 3 phases of AC output to an almost ripple-free DC.

Note that as more energy is extracted from the alternator (for example, adding 10 amps by turning on the headlights) the shaft resists more rotation. This can be easily tested by holding an alternator in your hand, and trying to rotate the shaft by hand with no load (rotate easily, keep rotating for a few seconds), then short-circuiting the output (much harder to rotate, stop rotating immediately). The quest for magical perpetual energy where a motor is powered by an alternator is a child's dream.

Answer:

It formed from the same cloud of dust that our entire galaxy is made from

the fossil informed us about the climatic and geographical changes that occurs in the environment over time.

We can infer about the climate and geographical changes in the environment over time at the mystery fossil dig site because the fossil provides us information about the climatic condition as well as the geography of land when this fossil organism were alive.

So we can conclude that the fossil informed us about the climatic and geographical changes that occurs in the environment over time.

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The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

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Answer:

The angle of the Sun's rays:

At about June 21 the sun is directly overhead at about 23 deg north of the equator

On about Dec 21 the sun is directly overhead at about 23 deg south of the equator

It is this tilt 23 deg to the ecliptic (path of earth about sun) that causes the major temperature changes on earth

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

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Answer:

B

Explanation:

The answer is B, because the waves are farther apart then the others. And it also looks like radio waves, which carries the least energy.

Answer: The third one

Explanation: there isnt many waves in that one les waves = less energy and i took the quiz/test.

Answer:

a. the light diverges when it passes through the lens.

Explanation:

All the light rays after passing through the concave lens diverge and when produced backwards appear to meet at a point on the principal axis of the lens. This point is known as principal focus of a concave lens.

Answer:

Valence electrons

Explanation:

I don't exactly understand the question. However, I do know that the main commonality between elements of the same group (column) is their valence electrons. Valence electrons are very important when it comes to bonding, ionization, etc.

Answer:

B. The image can either be virtual or real.

Explanation:

Concave mirrors can either produce real or virtual images. The image is upright given that it is virtual or real given that it is in front of the mirror. The images produced by the concave mirrors can also be reduced (in height), enlarged (in terms of size) or having same size as that of the object.

I hope this helps you understand about the images formed by concave mirrors.

Have a nice studies.

Answer:

ioinl

Explanation:

Answer:

20s

Explanation:

t=d/(V2-V1)

t= 80000m/(6000-2000)

t= 20s

By Newton's second and third laws, we have the following net forces at

[tex]\sum F_x = -T_1 \cos(60^\circ) + T_2 \cos(60^\circ) = 0[/tex]

[tex]\sum F_y = T_1 \sin(60^\circ) + T_2 \sin(60^\circ) - 600 \,\mathrm N = 0[/tex]

Together, these equations tell us [tex]\boxed{T_1=T_2\approx 346 \,\mathrm N}[/tex].

[tex]\sum F_x = T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5 = 0[/tex]

[tex]\sum F_y = T_3 \sin(20^\circ) - T_2 \sin(60^\circ) = 0[/tex]

[tex]\sum F_x = -T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]

[tex]\sum F_y = T_4 \cos(20^\circ) - T_1 \sin(60^\circ) = 0[/tex]

Combining the horizontal force equations for points B and C and using the fact that [tex]T_1=T_2[/tex] gives

[tex](T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5) + (-T_4  \cos(20^\circ) + T_1 \cos(60^\circ) + T_5) = 0 + 0[/tex]

[tex]\implies (T_3 - T_4) \cos(20^\circ) = 0 \implies T_3=T_4[/tex]

Then combining the vertical force equations for B and C, we find

[tex](T_3 \sin(20^\circ) - T_2 \sin(60^\circ)) + (T_4 \sin(20^\circ) - T_1 \sin(60^\circ)) = 0 + 0[/tex]

[tex]\implies 2 T_3 \sin(20^\circ) - 2 T_1 \sin(60^\circ) \implies T_3 = \dfrac{\sin(60^\circ)}{\sin(20^\circ)} T_1[/tex]

so that [tex]\boxed{T_3 = T_4 \approx 877 \,\mathrm N}[/tex]

Lastly solve for [tex]T_5[/tex] using either horizontal force equation for B or C.

[tex]-T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]

[tex]\implies T_5 = \left(\dfrac{\sin(60^\circ)}{\sin(20^\circ)} - \cos(60^\circ)\right) T_1 \implies \boxed{T_5 \approx 651\,\mathrm N}[/tex]

Answer:

V = λ f    (wavelength * frequency)

λ  = V / f = 343 m/s / 262 / s = 1.3 m

Answer:

[tex]9.8\; {\rm J}[/tex], assuming that the gravitational field strength is [tex]g = 9.8\; {\rm N \cdot kg^{-1}}[/tex].

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be [tex]0[/tex]. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let [tex]m[/tex] denote the mass of this block. It is given that [tex]m = 1\; {\rm kg}[/tex]. The weight of this block would be:

[tex]\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}[/tex].

Hence, the force that the boy applies on this block would be upward with a magnitude of [tex]F = 9.8\; {\rm N}[/tex].

The mechanical work that a force did is equal to the product of:

The displacement of this block (upward by [tex]s = 1\; {\rm m}[/tex]) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

[tex]\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}[/tex].

Answer:

maybe

Explanation:

young rocks may be found near volcanoes

The power generated or output power of the turbine is determined as 42.96 W.

The power generated or output power of the turbine is calculated as follows;

eff = 0utput power/1nput power x 100%

13 = 0utput power/(0.62 x 533)

0.13 = 0utput power/330.46

0utput power = 0.13 x 330.46

0utput power = 42.96 W

Thus, the power generated or output power of the turbine is determined as 42.96 W.

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The depth of immersion of the uniform cylindrical rod is 0.63m.

This is defined as the distance in which a body reaches when it is submerged.

Depth of immersion = Volume/Cross-sectional area

Volume = Cross-sectional area × length

               = 0.1m² × 0.63m = 0.063m³

Depth of immersion = 0.063m³ / 0.1m²

                                  = 0.63m

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From the diagram The value of cos C × sin A = [tex]\frac{3}{4}[/tex]

First step : determine the values of cos C and sin A

cos C = adjacent / hypotenuse

          = a / b

          = [tex]\sqrt{3} / 2[/tex]

          = √3/2

sin A = sin 60⁰

         = √3/2

Therefore the numerical value of cos C * sin A = [tex]\frac{3}{4}[/tex]

In conclusion From the diagram The value of cos C × sin A = [tex]\frac{3}{4}[/tex]

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The maximum speed the car can move without skidding is determined as 28 m/s.

The maximum speed of the car moving in a circular path is provided by static frictional force.

Fs = μFₙ = mv²/r

μFₙ = mv²/r

mv² = μFₙr

v² = μFₙr/m

[tex]v_{max} = \sqrt{\frac{\mu_s F_n r}{m} } \\\\v_{max} = \sqrt{\frac{\mu_s (mg) r}{m} } \\\\v_{max} = \sqrt{\mu_sgr} \\\\v_{max} = \sqrt{(0.8)(9.8)(100)}\\\\v_{max} = 28 \ m/s[/tex]

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Velocity after 20s for both

Distance traveled for car

Distance traveled by truck=

#b

Distance from Q

for car

For truck

The truck will meet the car after 75/21≈4s

Then for 6s the car keeps meeting because after 7 s Truck will archive 22m/s velocity

We have the relation

[tex]\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}[/tex]

where [tex]v_{A \mid B}[/tex] denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

[tex]v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}[/tex]

[tex]v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}[/tex]

Let's assume the river flows South-to-North, so that

[tex]\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath[/tex]

and let [tex]-90^\circ < \theta < 90^\circ[/tex] be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

[tex]\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)[/tex]

Then the velocity of the boat relative to the Earth is

[tex]\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath[/tex]

The crossing is 153.0 m wide, so that for some time [tex]t[/tex] we have

[tex]153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s[/tex]

which is minimized when [tex]\theta=0^\circ[/tex] so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

[tex]\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}[/tex]

The boat's position [tex]\vec x[/tex] at time [tex]t[/tex] is

[tex]\vec x = \vec v_{B\mid E} t[/tex]

so that after 30.0 s, the boat's final position on the other side of the river is

[tex]\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath[/tex]

and the boat would have traveled a total distance of

[tex]\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}[/tex]

If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.

The momentum is the product of its mass and velocity.

This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.

Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

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Answer:

Currently 1,915

Explanation:

I don't know if this is still exactly accurate to date, but when I answered this question, an astronomer had been tracking that many constallations

The total energy is transferred to the liquid is = 217809.9 J

It is given in the question that—

weight of the liquid= 0.3kg

specific heat capacity=22001/kg C

temperature changes between 20°C to 53°C

we can easily solve this problem by using the formula of the calorimetry principle.

The formula is —

Q= m×s×t [Q=total energy is transferred, m= mass of the substance, s= specific heat, t= final temperature - initial temperature]

by putting the value on what is given in this question —

Q= 0.3×22001×(53-20)

= 217809.9 J

So energy transferred to the liquid is=217809.9 J

The universe may indeed have an end one day according but it is yet to be shown in science.

While there is certainly no way a person will live long enough to know if the universe will end or not, nevertheless certain scientific calculations are helpful.

From astronomical calculations, the universe is anticipated to have existed for about 13.8 million years. Thus, the universe may indeed have an end one day according but it is yet to be shown in science.

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The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

Use the formula:

Power × time = mass × specific heat

Given mass = 1. 5kg

Specific latent heat of vaporization = 4000000 J/ Kg

Power = 2500 W

Substitute the values into the formula

Power × time = mass × specific heat

2500 × time = 1. 5 × 4000000

Make 'time' the subject

time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds

Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

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The distance, van travels while coming to rest after brake applied is 6.4m

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Given is a delivery van  whose kinetic energy K.E = 80000J and the force applied to brake F = 12500 N.

The brakes are applied to stop the van, so, final velocity is zero. Then, change in kinetic energy will be

According to work energy theorem, work done is

W = ΔK.E = K.Ef - K.Ei

W = -  80000J

Work done is also defined as the product of force applied and the distance it travelled.

W = Fs

80000 = 12500 x s

s = 6.4 m

Thus, the distance, van travels while coming to rest after brake applied is 6.4m

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The electric force between two charged particles can be increased by decreasing the distance between the two particles.

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.

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