what kind of image is created between the center of curvature and the focal point by a concave mirror?

Answers

Answer 1

Between the centre of curvature and the focus point, a concave mirror produces a virtual, upright, and magnified image. A virtual image is the term used to describe this kind of image.

The centre of curvature in a concave mirror is situated on the same side as the object. The image created is virtual, which means it cannot be projected onto a screen when the object is positioned between the centre of curvature and the focus point. The picture is bigger than the mirror and appears to be behind it.

The position of the item in relation to the focus point and centre of curvature determines the precise features of the image, including its size and distance from the mirror.

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Related Questions

What is the total energy that the ball has when the launcher is in the ""ready to launch"" position with the spring fully compressed?

Answers

Work is done when a spring is extended or compressed . Elastic potential energy is stored in the spring hope that helpss.

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

Answers

   After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

   Here we will apply the principle of impulse and momentum.

 Horizontal component:

          -mv₀ cos30⁰ + RxΔt  =0

                                 RxΔt = mv₀sin30⁰

                                          = 0.001941 × 1400sin30⁰

                                   RxΔt = 1.3587 lb.s

         Vertical component:

                           -mv₀cos30⁰  + RyΔt =  -mv'

                                     RyΔt = m( v₀cos30⁰-v')

                                     RyΔt = 0.001941(1400cos30⁰ - 14.968)

                                              = 2.32 lb.s

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An 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of 30o with the horizontal. The force pushing the crate is parallel to the slope. If the speed of the crate increases at a rate of 1.5 m/s2, find the work done by the force. Group of answer choices 260 J 200 J 1 J 200 kJ 140 J

Answers

The work done by the force is 140 J.

Calculation:

It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of [tex]30^\circ[/tex] with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5 [tex]\text{ m/s}^2[/tex].

It is required to find the work done by the force.

The free-body diagram is shown below,

Here, W=80 N, a=-1.5 [tex]\text{ m/s}^2[/tex]

We can write the net force equation as,

[tex]\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}[/tex]

Here W is the weight of the object, m is the mass of the body, and F is the applied force.

Thus,

[tex]\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}[/tex]

It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.

Thus the work one is,

[tex]\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}[/tex]

Thus the last option is the correct one.

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A box of mass 8 kg slides across a frictionless surface at an initial speed 1.5 m/s into a relaxed spring of spring constant 69 N/m.

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Time for which box will remain in contact with spring is 1.1 sec

Given:

mass of box = 8 kg

initial speed of box = 1.5 m/s

spring constant = 69 N/m

To Find:

time for which box will remain in contact with spring

Solution: The time taken to complete one oscillation is called the time period. where m is the mass, and k is the spring constant.

Time period is given as

t = π √m/k

t = 3.14 √8/69

t = 1.1 sec

So time for which box will remain in contact with spring is 1.1 sec

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Which statement about income is not true?
A. Income should increase over your career.
B. Income is part of a budget.
C. Income can come from multiple sources.
D. Income only includes salary.

Answers

Answer: d) income only includes salary

Explanation:

i just took a test for economics and personal finance and got it correct

The statement that "Income only includes salary " is not true.

Income is the money that an individual or organization receives in return for their services or goods. 

Depending on the context—such as taxation, financial accounting, or economic analysis—income may have a variety of definitions.

For the majority of people, their total earnings include their wages and salaries, investment returns, pension payments, and other receipts. 

For businesses, income refers to the money made from selling goods and services as well as any interest or dividends paid on the company's cash holdings and reserves.

So, the amount of money, property, and other value transfers received over a predetermined period in return for goods or services is often referred to as "income."

That's why income does not include only salary.

Hence Statement (D) about the income is not true.

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What occurs when a temperate glacier meets its pressure-melting point?

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Answer:

a whole new glacier will start because the other one is melting

A brick and a feather fall to the earth at their respective terminal velocities. Which objectexperiences the greater force of air friction?

Answers

Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero.  Thus, neither object is accelerating.

Recall Newton's second law: [tex]\sum {\vec {F}}=m \vec {a}[/tex]

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, [tex]F_{gravity}=mg[/tex], so the Force of Gravity acting on an object is directly proportional to the object's mass.  (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.  

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from the ground, which path would the bowling ball most closely follow after leaving the airplane?

Answers

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Path of the bowling ball

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

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When the Sun runs out of hydrogen in its core, it will become larger and more luminous because Choose one: A. energy balance no longer holds, and the star just drifts apart. B. infalling material rebounds off the core and puffs up the star. C. it starts fusing hydrogen in a shell around a helium core. D. it starts fusing helium in a shell and hydrogen in the core.

Answers

It starts fusing hydrogen in a shell around a helium core

Eventually, a star will run out of hydrogen fuel in its core, nuclear fusion will stop, and the star will enter a new stage of its lifetime.

What happen to the Sun when it runs out of hydrogen ?

Faster hydrogen burning means the star will grow larger and more luminous to allow the extra energy created to escape but since the temperature of the core remains constant the star's surface temperature will drop.

When our Sun runs out of hydrogen fuel in the core, it will contract and heat up to a sufficient degree that helium fusion can begin

Once that mass/temperature threshold is crossed, the star begins fusing hydrogen into helium, and will encounter one of three different fates.

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PLEASE HELP QUICK! Why does thermal energy increase on a roller coaster? Especially when it goes up the second hill?

Answers

as friction does negative work, the mechanical energy decreases. The work done by friction transforms mechanical energy into heat, a non-mechanical form of energy. As a result, the temperature of the track and the wheels increase.

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf

Answers

The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

The peak emf produced in a rotating loops= N×B×A×wN= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequencySo, B = emf/(N×A×w)What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/sNow, B= 24/(300×0.00261×42)

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

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The light ray is now incident on a glass-air interface, with an angle of incidence in the glass of 45 degrees. What is the angle of light ray emerging from the glass into air

Answers

The ray will not emerge into the air medium from glass medium.

To find the answer, we need to know about the critical angle.

What's critical angle of glass?Critical angle of a medium can be determined from the relation as sinФ = 1/n, n = refractive index of that medium.As glass has refractive index of 1.5, so Critical angle = sin⁻¹(1/1.5) = 42 °Why does the light incident at 45° inside a glass not emerge to the air medium?As we got the critical angle of glass is 42°, so the light incident at 45° which is greater than 42° will reflect back into the glass medium instead of emergence into the air medium.

Thus, we can conclude that the light will not emerge into air medium.

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What is the radial distance between the 500 v equipotential surface and the 1000 v surface?.

Answers

Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.

Explanation: To find the answer, we have to know more about the equipotential surfaces.

What are equipotential surfaces?An equipotential surface is the locus of all points which have the same potential due to the charge distribution.Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.We have the equation for electric potential as,

                            [tex]V=\frac{Qk}{r}[/tex]  , where k is equal to 1/(4π∈₀) =[tex]8.99*10^9[/tex] .

equation for radial distance will be,

                          [tex]r_d=r_1-r_2[/tex]

How to solve the problem?For the first surface, we can write the equation of potential as,

                   [tex]500V=\frac{Qk}{r_1} \\thus, r_1=\frac{Qk}{500V}= (1.79*10^7 *Q) m[/tex]

For the second surface, we can write the equation of potential as,

                  [tex]1000V=\frac{Qk}{r_2}\\ r_2= \frac{Qk}{1000V} =(8.99*10^6*Q) m[/tex]

Thus, the radial distance will be,

                  [tex]r_d=1.79*10^7Q-8.99*10^6Q=(8.91*10^6*Q)m[/tex]

Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.

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Two 1-d waves of amplitude a and period t and constant phase 0 are superposed. If wave 1 travels 13. 5 wavelengths farther than wave 2, will the interference be:________

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Two 1-d waves of amplitude a and period t and constant phase 0 are superposed. If wave 1 travels 13. 5 wavelengths farther than wave 2,  the interference is 3/2π.

Interference of light is the phenomena of multiple light waves interfering with each other under certain circumstances, causing the mixed amplitudes of the waves to both boom or lower.one of the exceptional examples of interference is demonstrated by the light contemplated from a film of oil floating on water.

The phenomenon of addition or superposition of mild waves which produces an increase in intensity at some factors and a decrease in depth at a few other points are known as interference of light.

The intensities of light at exceptional factors of the medium are distinctive. At a few factors of the medium, depth is most and at a few different points intensity is maximum ( about 0) This phenomenon is called the interference of mild waves, the interference is known as negative interference.

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Assignment 1.4
VOLUME OF SPHERE
The radius of sphere 'r' is measured with vernier callipers as
(r+Ar)= (2.25 +0.01) cm. Calculate the volume of sphere. (47.7+0.6) cm³.
PHYSICS

Answers

you can use the graph to estimate the value of x when y is required to start a doubles match with the sun and the sky very well from her childhood and a kind young women who lives with her wearing a black and a veil and shows to be in that position or they have a bad past or bad life that they want to take revenge on every one by murder goes on by them in the streets of St and St Mary's Church and its followers are known as Sikhs and the other children claim she

If there is no change in the velocity of the object then it is known to be in ______

a) constant speed
b) uniform acceleration
c) uniform motion
d) non-uniform motion

Answers

OPTION (C) IS CORRECT

ANSWER - UNIFORM MOTION

If there is no change in the velocity of the object then it is known to be in UNIFORM MOTION.

EXPLORE MORE:-

EXAMPLE - A CAR COVERS A DISTANCE OF 15 KM WE'D EVERY 2 HOURS

FOR AN UNIFORM MOTION, ACCELERATION IS ZERO , AS THERE IS NO CHANGE IN VELOCITY....

-THANKS.!!


5.1 What is the nucleus composed of ?

Answers

Answer:

5.1 The DNA double helix is the most recognizable nucleic acid structure, but these are ribozymes.

An airplane traveling at speed, 180 m/s, emits sound at a frequency of 2 000 Hz. What is the change in frequency (in Hz) heard by a stationary listener as the plane approaching and passing by

Answers

Answer:

As the plane moves toward the listener, the apparent frequency of the plane would be [tex]4250\; {\rm Hz}[/tex] ([tex]2250\; {\rm Hz}[/tex] higher than the frequency at the source.)

As the plane moves away from the listener, the apparent frequency of the plane would be approximately [tex]1308\; {\rm Hz}[/tex] (approximately [tex]692\; {\rm Hz}[/tex] lower than the frequency at the source.)

Assumption: the speed of sound in the air is [tex]340\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Crests of this sound wave travel toward the listener at a constant [tex]v = 340\; {\rm m\cdot s^{-1}}[/tex]. Since there is a pause of [tex]t = 1 / f = (1/2000)\; {\rm s}[/tex] between every two consecutive crests of this sound wave, the distance between each pair of consecutive crests would be:

[tex]\begin{aligned}\lambda &= \frac{v}{f} \\ &= \frac{340\; {\rm m \cdot s^{-1}}}{2000\; {\rm s^{-1}}} \\ &= 0.17\; {\rm m} \end{aligned}[/tex].

Hence, if the aircraft wasn't moving, the first crest would have a head start of [tex]\lambda = 0.17\; {\rm m}[/tex] relative to the second one. This head start would ensure that the first crest arrive [tex]t = \lambda / v = 0.17\; {\rm m} / (340\; {\rm m \cdot s^{-1}}) = (1/2000)\; {\rm s}[/tex] earlier than the second crest.  

However, at a speed of [tex]v_{\text{s}} = 180\; {\rm m\cdot s^{-1}}[/tex], the aircraft would have travelled an additional [tex]v_\text{s}\, t = 180\; {\rm m\cdot s^{-1}} \times (1/2000)\; {\rm s} = 0.09\; {\rm m}[/tex] within that [tex]t = (1 / 2000)\; {\rm s}[/tex].

If the aircraft was travelling towards the listener, the head start of the first crest over the next one would be reduced to [tex]\lambda - v_\text{s}\, t =[/tex][tex]0.17\; {\rm m} - 0.09\; {\rm m} = 0.08\; {\rm m}[/tex]. The first crest would arrive earlier than the second one by [tex](\lambda - v_{\text{s}}\, t) / (v) = (0.08\; {\rm m}) / (340\; {\rm m\cdot s^{-1}}) \approx 0.000235\; {\rm s}[/tex].In contrast, if the aircraft was travelling away from the listener, the head start of the first crest over the next one would be increased to [tex]\lambda + v_\text{s}\, t = 0.17\; {\rm m} + 0.09\; {\rm m} = 0.26\; {\rm m}[/tex]. The first crest would arrive earlier than the second one by [tex](\lambda + v_{\text{s}}\, t) / (v) = (0.26\; {\rm m}) / (340\; {\rm m\cdot s^{-1}}) \approx 0.000765\; {\rm s}[/tex].

In other words, if the aircraft was moving towards the listener, the period of the sound would appear to the listener to be approximately [tex]0.000235\; {\rm s}[/tex]. in contrast, if the aircraft was moving away from the listener, the period of the sound would appear to the listener as approximately [tex]0.000765\; {\rm s}[/tex].

Therefore:

When the aircraft moves toward the listener, the listener would hear a frequency of [tex]f = 1 / t \approx 1 / 0.000235\; {\rm s} = 4250\; {\rm Hz}[/tex].When the aircraft moves away from the listener, the listener would hear a frequency of approximately [tex]f = 1 / t \approx 1 / 0.000765\; {\rm s} \approx 1307\; {\rm Hz}[/tex].

A laser is shone through a double slit and a particular interference pattern is observed on a screen some distance away. If the separation between the openings is increased, the distance between the fringes will

Answers

If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

What is Interference fringe ?

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

Distances between laser fringes are short which is due to light wavelength.The interference fringes can be estimated by knowing slit separation and wavelength.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

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How many watts of power are needed to exert a force of 30 n over a distance of 100 m for 1 minute?

Answers

Power require in Watts is 50 watts

Given:

Force = 30 N

Distance = 100m

Time for which force applied = 1 min = 60 sec

To Find:

Power

Solution: A watt is a unit of power. It is the amount of energy an item needs to function; the rate at which energy is consumed. One watt is equivalent to electricity flowing at a rate of one joule (unit of energy) per second

P = W/t

where W is work done by force and t is the time for which force is applied

W = F x d

W = 30 x 100 = 3000

P = 3000/60 = 50 watts

So Power require in Watts is 50 watts

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A speed is a scalar physical quantity. Why?​

Answers

Answer:

A speed is a scalar quantity because it only has magnitude.

Answer:

what is electronic configuration

Compared with a car moving at some original speed, how much work must the brakes of a car supply to stop a car that is moving twice as fast? how will the stopping distances compare?.

Answers

It takes four times the work and four times the stopping distance.

Given:

original speed of car = x

brakes applied so final speed = 0

moving speed of another car = 2x

To Find:

work done to stop the car

Solution: Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

Work done is given by

W = KE(final) - KE(initial)

W = 1/2m(v)^2 - 1/2m(u)^2

W = 1/2m(0 - (x)^2)

W = 1/2m(4x^2)

W = 4(1/2mx^2)

W = 4W

So, work done will be 4 times initial work

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Upon inspiration, what is the name of the air in the conducting zone that is not available for gas exchange?.

Answers

Dead space is the name of the air in the conducting zone that is not available for gas exchange.

What is conducting zone?

The conducting zone consists of all of the structures that provide passageways for air to travel into and out of the lungs.

The conducting zone is part of the respiratory system. It serves as a route for incoming and outgoing air and filtering debris and pathogens from the incoming air.In order to compensate for the reduction in gas exchange in the respiratory system, ventilation will increase

The volume of air that is inhaled that does not take part in the gas exchange is dead space. Because it either remains in the conducting airways or reaches alveoli that are not perfused or poorly perfused.

Dead space volume is air in the conducting airways that does not participate in gas exchange between the alveoli and blood.

Hence,

Dead space is the name of the air in the conducting zone that is not available for gas exchange.

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In the NEC Code, what does 210.8 indicate

A. Article 210, Section 8
B. Chapter 210, Paragraph 8
C. Page 210, Chapter 8
D. That Article 210 contains 8 paragraphs

Answers

In the NEC Code, 210.8 indicates article 210, section 8 and is denoted as option A.

What is NEC code?

This is referred to the standards which are required for the safe installation of electric wires and equipment in USA.

This comprises of several articles covering different topics and sections which gives it a more elaborate meaning and understanding.

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The north-pole end of a bar magnet is held near a stationary positively charged piece of plastic. How is the plastic affected?.

Answers

When the north-pole end of a bar magnet is held near a stationary positively charged piece of plastic, it remains unaffected.

What is a bar magnet?

A bar magnet is made of ferromagnetic substance that produces

magnetic field.

What are the properties of a bar magnet?

Magnets attract ferromagnetic materials like iron, cobalt and nickel.Like poles repel each other and unlike poles attract each other.The strength of both poles is always the same.When a magnetic material is kept in contact with a bar magnet it acquires the magnetic property.Some materials like gold, silver and water get repelled by magnets.

Since plastic is not a ferromagnetic material or any other material like gold and silver ,they remain unaffected in the presence of magnetic field.

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I need to know how to solve: finding magnitudes of forces related to a sum of three vectors

Answers

The solution is (v, u) = (- 585.709, 593,034). Please notice that the value of v only means that the direction of the real vector is antiparallel to the supposed one.

What are the magnitudes of two vectors to get the zero vector by vector sum?

According to the definition of vector sum and vectors in rectangular form, we must solve the following vector equation:

(0, 0) = 205 · (cos 23°, - sin 23°) + v · (- cos 75°, sin 75°) + u · (- cos 55°, - sin 55°)

(0, 0) = (188.703, 80.100) + v · (- 0.259, 0.966) + u · (- 0.574, 0.819)

(- 188.703, - 80.100) =  v · (- 0.259, 0.966) + u · (- 0.574, 0.819)

Then, we must solve the following system of linear equations:

- 0.259 · v - 0.574 · u = - 188.703

0.966 · v + 0.819 · u = - 80.100

The solution is (v, u) = (- 585.709, 593,034). Please notice that the value of v only means that the direction of the real vector is antiparallel to the supposed one.

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A 2. 0mc charge in an external field of 20n/c, north (direction) will experience a force of: ___0. 04____________ newtons, the direction of the force is __north___________

Answers

2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north

Force in an electric field  = charge * electric field

given

charge =  2* [tex]10^{-3}[/tex] C

electric field  = 20 N/C

Force  in an electric field   = 2* [tex]10^{-3}[/tex] * 20

                                            = 0.04 N

since , external field is in north direction then the force must be in north direction because the direction of an electric field at a point is the same as the direction of the electric force acting on a positive test charge

hence,  2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north

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Which types of decay will cause the mass number a of a nucleus to change?.

Answers

Nuclear decay results in the change in the mass number of a nucleus. In precise it is alpha decay.

What is nuclear decay?

The process through which an unstable atomic nucleus loses energy through radiation is known as radioactive decay, also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration.

What is alpha decay?A nuclear process known as "alpha decay" releases a particle made up of two protons and two neutrons when an unstable nucleus transforms to another element. The helium nucleus that is being ejected is known as an alpha particle. Positive charge and a sizable mass characterize alpha particles.

How is mass number of nucleus is changed by alpha decay?

When a nucleus emits an alpha particle, these changes happen:

the mass number decreases by 4. the atomic number decreases by 2.

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(10%) Problem 2: The frequency range for AM radio is 540 to 1600 kHz. The frequency range for FM radio is 88.0 to 108 MHz.

Answers

188m - 556m is the wavelength range for AM radio.

2.77 - 3.40m is the wavelength range for FM radio.

1) For AM radio

f(max) = 1600 × 10³ Hz

f(min) = 540 ×10³Hz

c = fλ

λ = c/f where,

c = speed of light

f = frequency

λ= wavelength

So,

λ(min) = 3 × [tex]10^{8}[/tex] / 1600 × 10³ = 188 m

λ(max) = 3 × 10⁸/540 ×10³ = 556 m

So wavelength range of AM radio is 188 m - 556m

2) For FM radio

f(max) = 108 × 10⁶ Hz

f(min) = 88 × 10⁶ Hz

λ(min)  = 3 × 10⁸ / 108 × 10⁶ = 2.77 m

λ(max) = 3 × 10⁸ / 88 × 10⁶ = 3.40 m

So wavelength range of FM radio is 2.77m - 3.40m

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A sound wave traveling through dry air has a frequency of 16 hz, a
wavelength of 22 m, and a speed of 350 m/s. when the sound wave passes
through a cloud of methane, its wavelength changes to 28 m, while its
frequency remains the same. what is its new speed? (the equation for the
speed of a wave is v= f * 1.)

Answers

Answer:

[tex]\huge\boxed{\sf v = 448\ m/s}[/tex]

Explanation:

Given data:

Frequency = f = 16 Hz

Wavelength = λ = 28 m

Required:

Speed = v = ?

Formula:

v = fλ

Solution:

Put the givens.

v = (16)(28)

v = 448 m/s

[tex]\rule[225]{225}{2}[/tex]

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