A converging lens can be defined as __________


a lens that causes parallel light rays to bounce off the surface

a lens that allows parallel light rays to pass without changing direction

a lens that causes parallel light rays to separate from each other

a lens that causes parallel light rays to focus at a specific location

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.  

The reaction is

[tex]C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy[/tex]

Here some energy is released in form of neutrino.

Answer:

313kelvin

Explanation:

40 degree celcius plus 273=313K

Solution :

Let us consider the Gaussian surface that is in the form of a cylinder having a radius of r and a length of A which is [tex]$\text{coaxial with the charged cylinder}$[/tex].

The charged enclosed by the cylinder is given by,

[tex]$q=\rho V$[/tex]       (here, V = [tex]$\pi r^2l$[/tex]  is the volume of the cylinder)

  [tex]$=\pi r^2lp$[/tex]    

If [tex]$\rho$[/tex] is positive, then the electric field lines moves in the radial outward direction and is normal to Gaussian surface which is distributed uniformly.

Therefore, total flux through Gaussian cylinder is :

[tex]$\phi=EA_{cyl}$[/tex]

   [tex]$=E(2\pi rl)$[/tex]

Now using Gauss' law, we get

[tex]$2\pi \epsilon_0rlE = \pi r^2lp$[/tex]

or [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Therefore, the electric field is [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Hence, option (d) is correct.

Answer:

Explanation:

[tex]h=-v^2 /2g[/tex]

[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]

[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]

Answer:

Dynamo

Explanation:

Dynamo started to rotate which is known as kinetic energy.When dynamo is in running it produces electricity.dynamo specially used for generating electricity.

A fixed mass of gas has a volume of 25 [tex]cm^3[/tex], the pressure of the gas is 100 kPa, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

PV = nRT (P= pressure of the gas, V =volume, n = number of moles of gas, R = gas constant, and T =temperature of the gas in kelvin)

Suppose the gas is an ideal gas and that the temperature is constant,

P1V1 = P2V2

Here P1 = 100 kPa, V1 = 25 [tex]cm^3[/tex], V2 = 10 [tex]cm^3[/tex],

100 kPa x 25 [tex]cm^3[/tex] = P2 x 10 [tex]cm^3[/tex]

P2 = (100 kPa x 25 [tex]cm^3[/tex]) / 10 [tex]cm^3[/tex]

P2 = 250 kPa

the change in pressure of the gas is,

ΔP = P2 - P1 = 250 kPa - 100 kPa = 150 kPa

The reason is that when the volume of a fixed mass of gas is decreased, the pressure of the gas increases proportionally, so here assuming that the temperature is constant it is calculated.

Hence, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

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Answer:

vận tốc bằng quãng đường chia thời gian

Explanation:

v=s/t

Explanation:

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Answer:

[tex]g=16.31\ m/s^2[/tex]

Explanation:

Given that,

The length of the pendulum, l = 2 m

The period of the pendulum, T = 2.2 s

The formula for the time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

or

[tex]T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi ^2l}{T^2}\\\\g=\dfrac{4\pi ^2\times 2}{(2.2)^2}\\\\g=16.31\ m/s^2[/tex]

So, the free fall acceleration is [tex]16.31\ m/s^2[/tex].

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

Answer:

The correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

Explanation:

The inter-stimulus interval (ISI) is the temporal interval between two successive stimuli, measured from the offset of the first stimulus to the commencement of the second.

A cell's refractory period is the time during which it is unable to replicate an action potential. Therefore, the absolute refractory period is the amount of time it takes for a second action potential to be initiated, regardless of how large a stimulus is applied repeatedly.

A second action potential is generated when the gap between the stimuli decreases until the interval reaches the absolute refractory period.

Therefore, the correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

That is, when the interval between the stimuli decreases, a second action potential is generated until the interval reaches the absolute refractory period.

Answer:

3. ultraviolet and x rays

The two forms of electromagnetic energy are used to produce the most spectacular fluorescence when placed in darkness are ultraviolet and x rays.

Electromagnetic radiation is waves of the electromagnetic field, propagating through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, light, ultraviolet, X-rays, and gamma rays.

Fluorescence is the process in which a substance absorbs light at a high energy, short wavelength and emits light at a lower energy, usually visible wavelength.

Ultraviolet rays is a type of electromagnetic waves in which the wavelength is shorter than visible rays. It is responsible for 10% of sunlight and causes sun tan. It is used to purify water in water purifiers. It kill germs.

X rays lies beyond ultraviolet rays. It is used to diagnose in medical field. It can destroy living tissues so excessive exposure should be limited to reduce harmful effect.

The color change of fluorescent minerals is most spectacular when the minerals are placed in darkness and exposed to electromagnetic energy shorter than visible light.

Ultraviolet and x rays are shorter than visible light so they exhibit fluorescent property.

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Fundamental

forming a necessary base or core; of central importance.

"the protection of fundamental human rights"

Answer:

Acceleration is the rate of change of velocity. What is commonly said as ‘ten meter per second squared’ can bee broken down into ‘ten meter per second per second.’ This gives us the true meaning of the term acceleration.

Just like ‘ten meter per second’ means increasing the displacement by ten meters every second, ‘ten meter per second per second’ means increasing the velocity by ten meters per second every second.

So, basically if you consider the case of a free fall motion, where ‘t’ stands for time and ‘v’ stand for velocity at that instant:

At,

t=0, v=0 m/s

t=1, v=0+10 m/s;

t=2, v=0+10+10 m/s;

t=3, v=0+10+10+10 m/s;

This can also be thought as a Arithmetic Progression where common difference ‘D’ is the acceleration(a), since it adds a 10 m/s to velocity every second and the first term ‘A’ stands for the initial velocity (u). Using this approach we can derive to the first equation of motion:

v = u + at

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Explanation:

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It means that the vehicle's speed increases at the rate of 10m/s every second.

At any time, it's speed is 10 m/s faster than it was 1 second earlier.

Answer:

The water table is the upper surface of the zone of saturation. The zone of saturation is where the pores and fractures of the ground are saturated with water. It can also be simply explained as, the upper level, below which the ground is saturated.

Answer:

Speed = 1.6 m/s

Explanation:

Formula,

Speed = Distance ÷ Time

Answer:

I am not sure if this is the answer

acceleration: 2.2m/s

Explanation:

here

initial velocity(u): 11m/s

Final velocity(v): 33m/s

time taken(t): 10 s

now

a:v-u/t

or

acceleration:final velocity-initial velocity/time taken

or

a: 33-11/10

or

a:22/10, divide it

: a=2.2m/s#

Answer:

Yes

Explanation:

If there were to be a supernova of a star or a planet/meter that was directed at earth, then we would know in advance and make a plan to stop it

Answer:

Velocity = distance / time

V = 10/1

V = 10km/h

Answer:10km/h or 2.77m/s.

Explanation:

Distance =10km

Time =1h

Velocity =10/1 =10km/h

Or,

Distance =10km =10000m

Time =1h =60min = 3600s

Velocity =10000/3600 =2.77m/s

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

Answer:

4 capacitors

Explanation:

Given

[tex]n = 5[/tex] --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

[tex]c = n - 1[/tex]

So, we have:

[tex]c = 5 - 1[/tex]

[tex]c = 4[/tex]

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

Answer:

1.1 m/s

Explanation:

Applying,

v = u+at.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

From the question,

Given: u = 0.3 ms, a = 0.4 m/s², t = 2 seconds

Substitute these values into equation 1

v = 0.3+0.4(2)

v = 0.3+0.8

v = 1.1 m/s

Hence the velocity at the end of 2 seconds is 1.1 m/s

Answer: True

Explanation:

Answer:2 protons and 2 neutrons

Explanation:In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons.

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

Answer:

Explanation:

When a light bulb connects to an electrical power supply, an electrical current flows from one metal contact to the other. As the current travels through the wires and the filament, the filament heats up to the point where it begins to emit photons, which are small packets of visible light.

Answer:

Oh, umm…

Jump! (The higher, the better!)

Drop your pencil! (or pen or ruler, whichever you prefer!)

Throw a ball of paper! (and make sure you pick it back up later!)

Explanation:

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Answer:

Figure is not there

Explanation: