J.J. Thomson discovered the electron by noticing that: A. molecules with the same atoms exhibited the same chemical properties. B. a beam of particles could be influenced by an electric or magnetic force. C. chemical compounds contained various combinations of at least two atoms. D. some particles in a beam were deflected when passed through a sheet of gold foil.​

 J.J. Thomson Discovered The Electron By Noticing That: A. Molecules With The Same Atoms Exhibited The

Answers

Answer 1

J.J. Thomson discovered the electron by noticing that a beam of particles could be influenced by an electric or magnetic force.. That is option B.

What is an electron?

An electron can be defined as the part of an atom that is negatively charged and is found revolving round the nucleus of an atom.

J.J. Thomson was the scientist that discovered electrons through subjecting two oppositely-charged electric plates around the cathode ray.

He noticed that the particles where deflected by both the magnetic and electric fields.

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Related Questions

The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800. Select the correct answer.

Answers

The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg

Data obtained from the questionYear 1800 percent = 0.028%Year 2013 percent = 0.040%Mass of air = 500 KgDifference =?

How to determine the mass of CO₂ in 500 Kg in year 1800Year 1800 percent = 0.028% Mass of air = 500 KgMass of CO₂ =?

Mass = percent × mass of air

Mass of CO₂ = 0.028% × 500

Mass of CO₂ = 0.14 Kg

How to determine the mass of CO₂ in 500 Kg in year 2013Year 1800 percent = 0.040% Mass of air = 500 KgMass of CO₂ =?

Mass = percent × mass of air

Mass of CO₂ = 0.040% × 500

Mass of CO₂ = 0.2 Kg

How to determine the differenceMass of CO₂ in year 1800 = 0.14 KgMass of CO₂ in year 2013 = 0.2 KgDifference =?

Difference = mass in 2013 - mass in 1800

Difference = 0.2 - 0.14

Difference = 0.06 Kg

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While standing at the top of an 100 m high observatory, you accidentally dropped your phone through the grates (it is now falling straight down to the ground). a. What is the velocity of the phone after 4.0 s? b. How far does the phone fall during this time? C. Will your phone hit the ground? If not, how long more before it hits the ground?​

Answers

Using the idea of motion under gravity, it will take 0.5 seconds more before the phone hits the ground.

What is the distance covered?

The distance covered is obtained form the equations of kinematics under gravity.

Now;

v = u + gt

Recall that it was dropped from a height hence u = 0 m/s

v = gt

v = 9.8 m/s^2 * 4 s

v = 39.2 m/s

Now;

h = ut + 1/2gt^2

but u = 0 m/s

h = 1/2gt^2

h = 1/2 * 9.8 * (4)^2

h = 78.4 m

The phone  will not hit the ground within this time

For the phone to hit the ground;

h = 1/2gt^2

if h = 100 m

100 = 1/2 * 9.8 * t^2

2 * 100/9.8 =  t^2

t = √2 * 100/9.8

t = 4.5 seconds

It will take about 0.5 seconds more before the phone will hit the ground.

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A 500 N weight is hung at the middle of a rope attached to two buildings at the same level. If the breaks in the tension exceed 1800 N, what is the minimum angle the rope can make with the horizontal? Group of answer choices 4o 8o 11o 18o

Answers

Answer:

The correct choice is 8o.

Explanation:

The given weight is hanging at the mid point of the rope and the buildings are at the same level, Obtain the given equation by equating the vertical components of force,

2Tsin(φ)=W

where φ is the angle made by rope with the horizontal.

Given T=1800 N and W=500 N, the value of φ is calculated as follows,

2*1800*sin(φ)=500

sin(φ)=0.1388

φ=7.90

φ≈8o

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An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h.Compare it's centripetal acceleration with the acceleration due to gravity.​

Answers

Answer:

900 km/hr = 9E5 m / 3600 sec = 250 m/sec

ah = v^2 / R = 250^2 m/s / 1000 m = 62.5 m/s^2   horizontal acceleration

av = 9.8 m/s^2      vertical acceleration (due to gravity)

ah / av = 62.5 / 9.8 = 6.38   centripetal acceleration is greater

A block of mass 60kg measures 6cm by 5cm by 4cm. Calculate i) The maximum pressure it can exert. The minimum pressure. ii) A block of mass 60kg measures 6cm by 5cm by 4cm . Calculate i ) The maximum pressure it can exert . The minimum pressure . ii )​

Answers

Answer:

4.9 is the answer

Explanation:

pressure is force divided by area

force = mass x acceleration due to gravity

= 60kg x 9.8 m/s

area = length x width x height

=6 x 5 x 4

= 120

pressure = 588/120

= 4.9 pascal

The correct answer for the value of (i) Maximum pressure is [tex]24.5 \dfrac{N}{cm^2}[/tex](ii)Minimum pressure is [tex]19.6 \dfrac{N}{cm^2}[/tex].

Given:

Weight of the block:  [tex]60[/tex] kg

Dimensions of the box:  [tex]6 *4* 5[/tex]

Value of gravitational constant:  [tex]9.8[/tex] m/s²

Now,

[tex]Pressure = \dfrac{Force}{Area}[/tex]

[tex]Force = mass*accelaration[/tex]

[tex]= 60 * 9.8[/tex]

[tex]= 588 N[/tex]

Minimum pressure is when area is maximum:

[tex]= \dfrac{588}{6*5}\\\\= 19.6 \dfrac{N}{cm^2}[/tex]

Maximum pressure when area is minimum:

[tex]= \dfrac{588}{6*4} \\\\= 24.5 \dfrac{N}{cm^2}[/tex]

Maximum and minimum pressure are   [tex]24.5 \dfrac{N}{cm^2}[/tex]  and  [tex]19.6 \dfrac{N}{cm^2}[/tex] respectively.

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Calculate the frequency of the 3rd harmonic of a 0.5 meter string with a linear density of 0.03 kg/m at a tension of 50 newtons.

Answers

Frequency of third harmonic is 122.48 HZ

Theory:

Frequency of third harmonic with length L , linear density μ , Tension

T is given as

                  fₙ= n v/2L

                   fₙ = n/2L × √T/√μ

      plugging in the values here-

                   fₙ =  3/ 2×0.5 × √50 / √0.03

                   fₙ = 122.48  Hz

          Thus, the frequency of third harmonic is 122.48 HZ

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Which statement is true of a piece of ice at 0°C that is put into a freezer at
-18°C?

Answers

The statement which is true about a piece of ice at 0°C which is put into a freezer at -18°C is it having the temperature of the freezer.

What is Temperature?

This is referred to the degree of hotness or coldness of a body and the unit is Celsius or Kelvin.

The ice at 0°C will experience a change in temperature of the freezer when put in it in this scenario.

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If the radio waves transmitted by a radio station have a frequency of 79.5 MHz, what is the wavelength of the waves, in meters

Answers

If the radio waves transmitted by a radio station have a frequency of 79.5 MHz, the wavelength of the waves, in meters is 3.77m.

What are radio waves?The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz. They may be reflected and refracted to change direction, and they do not harm the human body if they are absorbed by it. They are perfect for communicating due to their characteristics.

What are Frequency and wavelength?When describing the temporal rate of change seen in oscillatory and periodic phenomena like mechanical vibrations, radio waves, and light, frequency is a crucial parameter utilized in science and engineering.The length of a waveform signal that is propagating in space or over a wire is measured by the separation between identical points (adjacent crests) in the adjacent cycles. Frequency, or the number of wave cycles per second, is inversely related to wavelength.

Given: Velocity of light,  c = 3.00 x 10⁸ m/s  m/s

Frequency, f = 7.95 x 10⁷/s

For wavelength, the required formula is

[tex]λ = c/ f \\λ = 3 * 10 ^{8} / 7.95 * 10^{7} \\λ = 3.77[/tex]

Hence, the required wavelength is 3.77 meters.

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You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The frequency of the other is:

Answers

The frequency of the other wave is 613 Hz or 607 Hz.

The difference between the frequencies of two waves is called the beat frequency.

Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.

Which has a higher frequency is not mentioned. Therefore, there are two possibilities.

Δf = | 610 - 613 | = 3

or

Δf = | 610 - 607 | = 3

Therefore, the frequency of the other wave is 613 Hz or 607 Hz.

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Which equation can be used to solve for the magnitude of the velocity (v) at which Frances slides backwards after giving the curling stone a push in Test 3?

A
0 = –(45 kg • v) + (20 kg • 3 m/s)

B
0 = –[(45 kg + 20 kg) • v]

C
60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s)

D
195 kg • m/s2 = (45 kg + 20 kg) • v

Answers

A

A 0 = –(45 kg • v) + (20 kg • 3 m/s)

Not moving, means momentum is zero. Thus, momentum after = momentum before (& that momentum is zero)

V = 4/3m/s

(-45 x 4/3) + (20x3) = 0

Solving to find v :

0. = –(45 kg • v) + (20 kg • 3 m/s)

= - (45 kg • v) + 60

-60

-60 = - 45 x v

÷ — 45

4/3 = v

(4/3 = 1.33333333333)

Hope this helps!

The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).

Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.

What is velocity?

A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.

A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").

Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.

Therefore, Thus, option C is correct.

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4. An applied force of 6.2 N acts on a 2.1-kg object, pushing it horizontally across a surface
where the coefficient of kinetic friction is 0.15.
a. Draw a free body diagram for the body
b. Determine the frictional force acting.
c. Determine the net force acting on the body
d. Determine the object's acceleration.
Step by step, thank you
[2C]
[2A]
[2A]
[2A]

Answers

The frictional force on the body is  0.25 N the net force on the body is  5.95 N while the acceleration produced is 2.83 m/s^2

What is the net force?

The net force is used to describe the resultant force that acts on the object.

The frictional force that acts on the object s obtained from;

Ff = 0.15 * 2.1-kg * 9.8 m/s^2

Ff = 0.25 N

The net force that acts on the body = F - Ff where F is the applied the force

=  6.2 N - 0.25 N

Net force = 5.95 N

Now;

Net force = ma

a= Net force /m

a = 5.95 N/2.1-kg

a = 2.83 m/s^2

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A speed skater moving across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadily, then continues on at 6.0 m/s. What is her acceleration on the rough patch

Answers

Her acceleration on the rough patch is 2.8m/s².

To find the answer, we need to know about the Newton's equation of motion.

What are the Newton's equation of motion?

These are as follows

V= U +atS = Ut+1/2 at²V²-U²= 2aS

V= Final velocity

U = initial velocity

t= time

S= distance

What will be acceleration if initial velocity, final velocity and distance are 8 m/s , 6 m/s and 5m respectively?Here U= 8 m/s, V = 6 m/s and S = 5mSo, 6²-8²= 2a× 5=10a

36-64 = 10a

-28= 10 a

a= -28/10= -2.8 m/s²

Negative sign indicates the decrease of acceleration.

Thus, we can conclude that the acceleration on the patch is 2.8 m/s².

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Without using afood flask, suggest another ways would you keep food warm a till to the hospital

Answers

Another way to keep food warm till you arrive at the hospital is by putting the food in a foil and wrapping it up with a towel to retain the heat.

What is Food preservation?

Food preservation is defined as the process by which a cooked food or process food is kept in such a way that microorganisms cannot affect their taste and texture.

To prevent a cooked food from getting cold, you can put the food in an aluminium foil and wrap it with a towel to prevent heat loss.

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A vehicle starts from rest and accelerates uniformly for 12 seconds to a
velocity of 10ms1. It then runs at a constant velocity and is finally came to rest
in 66m with constant . The total distance covered by the vehicle is
580m. Find the value of acceleration and the time taken.

Answers

The value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

Acceleration of the vehicle

The acceleration of the vehicle before coming to rest is calculated as follows;

v² = u² + 2as

where;

v is the final velocityu is the initial velocitya is the accelerations is the distance traveled before stopping

the car came to rest with constant velocity attained after 12 seconds.

the initial velocity of the car before 12 seconds is zero.

v² = 0 + 2as

a = v²/2s

a = (10²)/(2 x 66)

a = 0.76 m/s²

Time of motion of the vehicle

d = ut + ¹/₂at²

where;

d is the total distance traveledt is the time of motiona is accelerationu is initial velocity of the vehicle

580 = 0 + ¹/₂(0.76)t²

580 = 0.38t²

t² = 580/0.38

t² = 1,526.3

t = √1,526.3

t = 39 seconds

Thus, the value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

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If f is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?.

Answers

The net force acting on test charge is resultant of two force i.e.,  √2 F

The magnitude of force on test charge due to one charge Q is:

                F = k qQ / r²

where k = 9 × 10^9

Similarly the force due to second charge will be:

         F = k qQ / r²

Now both the charges of magnitude Q is present and it is pointed at right angle to each other then the force  will be resultant i.e.,

                    F = √ F² + F² = F√2

 Hence, net force acting on test charge is √2 F.

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The density of a certain type of plastic is 0.77 g/cm3. If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick, what is its mass

Answers

Mass is 77,000 g

Given

The density of a certain type of plastic = 0.77 [tex]$\mathrm{g} / \mathrm{cm}^{3}$[/tex]

Length = 10.0 m = 1000 cm

Width = 1 m = 100 cm

Thickness = 1 cm

Formula

Volume = length x width x thickness

Mass = Density x Volume

Explanation

Density

Density is how much "stuff" is contained in a specific quantity of space is determined by its density. For instance, a block of the harder, lighter element gold (Au) will be denser than a block of the heavier element lead (Pb) (Au). Styrofoam blocks are less dense than bricks. Mass per unit volume serves as its definition.

Volume V = 1000 x 100 x 1

             V = 100,000 [tex]cm^{2}[/tex]

Mass M= 0.77 [tex]$\mathrm{g} / \mathrm{cm}^{3}$[/tex] x 100,000 [tex]cm^{2}[/tex]

M = 77,000 g

So, Mass is 77,000 g

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A mass is placed on a smooth inclined plane with an angle of 37e to the horizontal. If theinclined plane is 5.0-m long, how long does it take for the mass to reach the bottom of theinclined plane after it is released from rest?

Answers

The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.

Height of the inclined plane

The height of the inclined plane is calculated as follows;

h = L(sin 37)

where;

h is height of the planeL is length of the plane

h = 5 x sin(37)

h = 3.01 m

Time of motion of the mass

t = √(2h/g)

t = √(2 x 3.01 / 9.8)

t = 0.78 s

Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.

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Throw two balls from the same height at the same time at an initial speed of 20 m/s. One is thrown vertically down, while the other is thrown vertically up. We want to find the time difference between their landing.

Answers

The time difference between their landing is 2.04 seconds.

Time of difference of the two balls

The ball thrown vertical upwards will take double of the time taken by the ball thrown vertically downwards.

Time difference, = 2t - t = t

t = √(2h/g)

where;

h is the height of fallg is acceleration due to gravity

Apply the principle of conservation of energy;

¹/₂mv² = mgh

h = v²/2g

where;

v is speed of the ball

h = (20²)/(2 x 9.8)

h = 20.41 m

Time of motion

t = √(2 x 20.41 / 9.8)

t = 2.04 s

Thus, the time difference between their landing is 2.04 seconds.

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At which points on the roller coaster is the car not moving?

A
A and E

B
B and C

C
C and D

D
D and E

Answers

The maximum potential energy of the car or the when car is not moving at points A and E; option A.

What is potential energy?

Potential energy is the energy of a body due to its state or position.

In the rollercoaster motion shown, the maximum potential energy occurs when the car is no longer moving.

At different points other than at maximum potential energy, the energy is a combination of potential and kinetic energy.

The maximum potential energy occurs at A and E, at which point the car is not moving.

In conclusion, at maximum potential energy, the car is not moving.

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. A man of mass 80 kg stands in a lift, what is the reaction from the floor of the lift if the lift;A. moves upwards with steady speed. moves upwards with acceleration 0.5m/sC. moves downward with acceleration 0.4m/s(g - 9.8m/s')​

Answers

The reaction of the floor is as follows:

R = 784 NR = 816 N

What is the reaction from the floor for a man in a lift?

The reaction of the floor on the man is equal to the weight of the man and the upward force.

Reaction, R = mg + ma

A)  When the lift moves upwards with steady speed.

R = 80 * 9.8 + * 80 * 0

R = 784 N

B) When the lift moves upwards with acceleration 0.4m/s

R = 80 * 9.8 + * 80 * 0.4

R = 816 N

In conclusion, the reaction of the floor changes as the body accelerates upwards.

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a ray of light incident on a mirror, at an angle of 45°. Another mirror is placed at an angle of 45° to the first ones as shown. Sketch the patch of the ray until it emerges from the two mirrors (2mks) 45° www.www 45° ​

Answers

Answer:

If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)

The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite  direction to the  incident ray.

The eyes of a farsighted person are Blank______ than normal; therefore light rays tend to focus Blank______ the retina.

Answers

The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.

What is Farsightedness:

It is also known as hyperopia, is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.

Farsightedness is caused by a cornea that isn't curved enough or It causes by an eyeball that's too short.

These two problems prevent light from focusing directly on the retina.

so, the light rays focus behind the retina.

Hence,

The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.

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Why should a scientist not be concerned when a fellow scientist responds to their conclusions with skepticism?

Answers

ABCDEDJGHIKLMNOP

just wanted to see if bold work

Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the: Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle: Mass of sand Mass of water that filled the space above the sa Volume of the sand: Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the : Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle : Mass of sand Mass of water that filled the space above the sand
Volume of the sand :​

Answers

The values of the density, volume and masses are:

mass of water that completely filled the bottle = 23.2 gvolume of water that completely filled the bottle = 23.2 cm³volume of the density bottle = 23.2 cm³mass of sand = 24.4 gmass of water that filled the space above the sand = 14.8 gvolume of sand = 8.4 cm³density of sand = 2.9 g/cm³What is density?

Density is defined as the ratio of mass and volume of  a substance.

Density = Mass/volume

The given data is as follows:

Mass of empty density bottle = 43.2 g

Mass of density bottle full of water = 66.4 g

Mass of density bottle with some sand = 67.5 g

Mass of density bottle with the sand

filled up with water = 82.3 g:

(a) mass of water that completely filled the bottle = 66.4 - 43.2

mass of water that completely filled the bottle = 23.2 g

(b) volume of water that completely filled the bottle = mass/density

density of water = 1 g/cm³

volume of water that completely filled the bottle = 23.3 g/1 g/cm³

volume of water that completely filled the bottle = 23.2 cm³

(c) volume of the density bottle = 23.2 cm³

(d) mass of sand = 67.5 g - 43.2 g = 24.4 g

(e) mass of water that filled the space above the sand = 82.3 - 67.5 = 14.8 g

(f) volume of sand = volume of bottle - volume of water

volume of water = 14.8 cm³

volume of sand = 23.2 cm³ - 14.8 cm³

volume of sand = 8.4 cm³

(g) density of sand = 24.4 g/8.4 cm³ = 2.9 g/cm³

Therefore, the density of sand is obtained by taking the ratio of the mass of the sand and the volume if the sand.

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A brightfield microscope equipped with a ____________ ___________ stop is able to block light from the illuminator creating bright objects on a dark background.

Answers

Answer: A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.

Explanation: To find the correct statement about Brightfield microscope, we need to know more about the Brightfield microscope.

what is Brightfield microscope?The Brightfield microscopic technique is generally used with the compound microscopes and, these are one, among the widely used optical illumination techniques. The Brightfield microscope or Compound light microscope is an optical instrument, which uses light rays and produces dark images with a bright background.How Bright field microscope works?A brightfield microscope equipped with a opaque light stop, will produce bright objects on a dark background by blocking the light illuminated from the light source. This opaque light stop, blocks the light rays that arising from the illuminator to the objective lenses and, only allows the reflected or transmitted light of the specimen. As a result, we get bright objects with a dark background.

Thus, we can conclude that, A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.

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A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory?

Answers

The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s

What is Trajectory ?

When any object is thrown from horizontal at an angle θ except 90°, then the path followed by it is called trajectory, the object is called projectile and its motion is called projectile motion.

Here, we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v = 0.

Given :

Horizontal speed= 30m/sVertical Speed= 30 m/s

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy = vy0 + ayt

Plug in our givens

0 = 30  - 10t

solve for t

t = 3 seconds

Hence, The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s

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If an electron has an uncertainty in its velocity of 1.40 m/s, what is the uncertainty (in meters) in its position

Answers

Uncertainty in position:

If the electron has an uncertainty in its velocity of 1.40 m/s then the uncertainty in its position is [tex]4.14\times10^{-4} \text{ m}[/tex].

Heisenberg's uncertainty principle to calculate the required:

Step-1:

We have to apply Heisenberg's uncertainty principle to calculate the uncertainty in the position of an electron. According to the principle:

[tex]$\Delta x \Delta p \geq \frac{h}{4 \pi}$[/tex]

Here,

[tex]$\Delta x$[/tex] is the uncertainty on the position measurement

[tex]$\Delta p$[/tex] is the uncertainty on the momentum measurement

h is the Planck constant.

It is known that the momentum of a particle is calculated as, the product of the mass of the particle and its velocity.

Therefore,

p=mv

Thus the Heisenberg principle becomes:

[tex]$m \Delta x \Delta v \geq \frac{h}{4 \pi}$[/tex]

Here [tex]$\Delta v$[/tex] is the uncertainty in the velocity measurement.

It is given that, [tex]$\Delta v$[/tex]=1.40 m/s. The particle is electron here, thus the mass of the particle m=[tex]9.1 \cdot 10^{-31} \mathrm{~kg}[/tex] and the value of the Plank's constant is, h=[tex]6.62 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}[/tex]

Step-2:

Substituting the values into the equation to get the value of the position uncertainty.

[tex]\Delta x \geq \frac{h}{4 \pi\times m\times \Delta v}\\\geq\frac{6.62\times10^{-34}}{4\pi \times 9.1\times10^{-31}\times1.40} \text{ m}\\\geq 4.14\times10^{-4} \text{ m}[/tex]

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In which of the following scenarios is the total momentum of the system conserved?

Answers

Answer:

The total momentum of a system is conserved only when the system is closed.

Explanation:

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.510 and is submerged in water, which has an index of refraction of 1.333. What is the critical angle (in degrees) for the light ray to remain inside the fiber

Answers

The answer is 73.13°.

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

The critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°

This means i=C and r = 90°

The Snell's law formula will become

n1/n2 = sinC/sin90°

n2/n1 = 1/sinC

Where n1 is the refractive index of the less dense medium = 1.473

n2 is the refractive index of the denser medium = 1.540

Substituting the values in the formula,

1.540/1.473 = 1/sinC

1.045 = 1/sinC

SinC = 1/1.045

SinC = 0.957

C = sin^-1(0.957)

C = 73.13°

The refractive index of glass is 1.5. It way that the velocity of mild in glass is 1.5 times slower than the rate of mild in a vacuum, the speed of light in glass isn't always unbiased of the shade of mild.

Refractive index is likewise the same as the velocity of light c of a given wavelength in an empty area divided with the aid of its speed v in a substance or n = c/v.

The index of refraction of fabric is a ratio that compares the velocity of light in a vacuum ( c=3.00x108ms ) to the velocity of mild in that precise medium. Because the index of refraction increases, the amount that the material bends the mild increases.

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The critical angle (in degrees) for the light ray to remain inside the fiber is 73.13°.

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

The critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°.

This means i=C and r = 90°

The Snell's law formula will become

n1/n2 = sinC/sin90°

n2/n1 = 1/sinC

Where n1 is the refractive index of the less dense medium = 1.473

n2 is the refractive index of the denser medium = 1.540

Substituting the values in the formula,

1.540/1.473 = 1/sinC

1.045 = 1/sinC

SinC = 1/1.045

SinC = 0.957

C = sin^-1(0.957)

C = 73.13°

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If you place a 3.5-kg object on a spring which has a spring constant 144-N/m, and stretch the spring so the object starts oscillating. How much time does it take to complete one cycle

Answers

Answer:

M = 3.5 kg         mass of object

F = - K x     where K is the spring constant

K = 144 N/m

ω = (K/M)^1/2     for simple harmonic motiom

ω = (144 kg/s^2 / 3.5 kg)^1/2 = 6.41 / sec

f =  ω / (2 * π)  = 1 / P

P = (2 Π / ω) = 2 * 3.14 / 6.41 sec

P = .98 sec

Check: f = 1 / P = 1.02 /sec

ω = 2 * π * f = 2 * 3.14 * 1.02 = 6.41 sec

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