Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)

2.4 m/s
8.4 m/s
10.8 m/s
15.3 m/s

Answers

Answer 1

Answer:

[tex]8.4\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Let [tex]x[/tex] denote the displacement of the ball. It is given that [tex]x = 12\; {\rm ft}[/tex]. Apply unit conversion and ensure that the displacement [tex]x\![/tex] of the ball is measured in meters:

[tex]\begin{aligned}x &= (12\; {\rm ft})\, \frac{(0.30\; {\rm m})}{(1\; {\rm ft})} = 3.6\; {\rm m}\end{aligned}[/tex].

Let [tex]u[/tex] denote the initial velocity of the ball, and let [tex]v[/tex] denote the velocity of the ball right before it hits the ground. Note that since the question states the ball was "dropped", assume that the ball was initially at rest with initial velocity [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].

Under the assumptions, the acceleration [tex]a[/tex] of the ball will be constantly [tex]a = g = 9.8\; {\rm m\cdot s^{-2}[/tex].

Rearrange the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find the final velocity [tex]v[/tex] of the ball right before landing:

[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &= \sqrt{(0\; {\rm m\cdot s^{-1}})^{2} + 2\, (9.8\; {\rm m\cdot s^{-1}})\, (3.6\; {\rm m})} \\ &= 8.4\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer 2

Answer: B (8.4 m/s)

Explanation:


Related Questions

the perfect order measures how effectively logistics serves the customer while the landed cost measures how efficiently logistics provides that service.a) trueb) false

Answers

AutoLFADS is a framework for large-scale neural network training for generalised estimate of single-trial population dynamics.

What is neural network ?

Deep learning techniques are based on neural networks, sometimes referred to as artificial neural networks (ANNs) or simulated neural networks (SNNs), which are a subset of machine learning. Their structure and nomenclature are modelled after the human brain, mirroring the communication between organic neurons.

Deep neural network of population dynamics models require considerable hyperparameter adjustment for each dataset to function at the cutting edge. Without using behavioural or task information, AutoLFADS is a model-tuning framework that generates high-performing autoencoding models automatically using data from a range of brain regions and activities.

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A transformer has 132 kV across its primary coil and 33 kV across the secondary coil. It produces a current of 8 A in its secondary coil. How much current flows through the primary coil? [4 marks]

Answers

Answer:

32 A.

Explanation:

In a transformer, the ratio of the voltage across the primary coil to the voltage across the secondary coil is equal to the ratio of the current in the primary coil to the current in the secondary coil. This relationship is known as the transformer equation, and it can be written as follows:

Vp / Vs = Is / Ip

Where Vp is the voltage across the primary coil, Vs is the voltage across the secondary coil, Is is the current in the secondary coil, and Ip is the current in the primary coil.

Given the values provided in the problem, we can substitute them into the transformer equation to find the current in the primary coil:

Ip = (Vp * Is) / Vs

= (132 kV * 8 A) / 33 kV

= 32 A

Therefore, the current in the primary coil of the transformer is 32 A.

a current of 0.30 a is passed through a lamp for 2.0 minutes. the energy dissipated by this lamp during the 2.0 minutes is 216 j. what is the potential (in v) of the power supply?

Answers

The potential (in v) of the power supply 6 Volt.

Calculation :

The energy formula ,

H = VIT

Where,

H=energy dissipated the lamp

I=current through the lamp

V = potential difference

t=time during which the current flows through the lamp

From the question we have,

Current(I)=0.3A

The Energy supplied(E)=216 j

Time during which the current flows(t)=2 minutes=2×60 seconds

H = VIT  

216 j = v * 0.3 A *120 sec

V = 216/( 0.3 A *120 sec)

V = 6 Volt

Scientists define energy as the ability to do work. Modern civilization is possible because humans have learned to convert energy from one form to another and use it for work.

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how many kilometers would you have to go above the surface of the earth for our weight to decrease to half of what it was at the surface

Answers

We have to go above 2651 kilometres of the earth for our weight to decrease to half of what it was at the surface.

Geophysics applies the concepts of physics to take a look at the Earth. It deals with things like the motion of the Earth's crust and the temperatures of its interior.

Geology is the scientific study of the earth, including the elements that make it up, the processes that shape it, how it was formed, and information about the planet's past and how it has been ever since. Today, geology also encompasses an examination of distant worlds.

We define the earth via thinking about other people as our enemies, outsiders and foreigners; by dividing our earth into international locations and by developing enmity in opposition to any other institution of people.

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question a 2 kg mass and a 5 kg mass are connected by a massless cord suspended over a massless and frictionless pulley. if the acceleration due to gravity is g, what will be the acceleration of the masses after they are released from rest?

Answers

The masses accelerate at a speed of 3g13 m/s2 after being released from rest.

m1 is 2 kilogram, while m2 is 5 kg. Because the pulley has no friction and no mass, the tension force on both sides of the string will be the same. [An inactive state for the Law of Motion] Both masses accelerate at the same rate. Let's say it is a m/s2.

Remember that 2 kg of mass is flowing downhill while 5 kg of mass is traveling upward, or the opposite is true.

Let's begin writing equations.

m1 g - T = m1a for a mass of 2 kg.

2g - T = 2a ————— (I) For a mass of 5 kg, T - m2g = m2a

T - 5g = 5a ————— (II) By combining the two equations, we arrive at 2g - 5g = 13a a = 3g13 m/s2.

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A proton is traveling to the right at 2.0x 10^7m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speeds of each after the collision? What is the direction of the proton after the collision? (up/down,left/right) What is the direction of the carbon atom after the collision? (up/down,left/right)

Answers

The speeds and the direction of each after the collision:

The proton = -1.692 x 10⁷m/s to the left.

The carbon = 0.3076 x 10⁷ m/s to the right.

The law of conservation of momentum

If there are two object each with masses m₁ and m₂ move with speed v₁ and v₂, then the two objects collide, so after colliding the speed of each object becomes v₁’ and v₂'.

Since there are no external forces acting on the system, then the momentum of the system is conserved, meaning the momentum before and after the same collision.

The equation is:

(m₁)(v₁) + (m₂)(v₂) = (m₁v₁)' + (m₂v₂)'

We have,

velocity of the proton = 2.0 x 10⁷ m/s ⇒ v₁

Mass of the proton = m₁

The velocity of a carbon atom = 0 ⇒ v₂  

The mass of the carbon atom = 12 m₁

So, the speeds of proton after the collision:

(m₁)(v₁) + (m₂)(v₂) = (m₁v₁)' + (m₂v₂)'

v₁’ = (m₁ - m₂) / (m₁ + m₂) (v₁)

= (m₁ - 12m₁) / (m₁ + 12m₁) (2.0 x 10⁷)

= (-11 m₁) / (13m₁) (2.0 x 10⁷)

= -1.692 x 10⁷m/s ⇒ it's negative, so to the left.

Now, the speed of the carbon atom after collision:

V₂’ = (2m₁) / (m₁ + m₂)v₁

= 2/13 (2 x 10⁷)

=  0.3076 x 10⁷ m/s ⇒ it's positive, so to the right.

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the highest frequency ultrasound devices have frequencies around 20 mhz. what is the wavelength of these sound waves inside the body?

Answers

The highest frequency ultrasound devices have frequencies around 20 mhz. 5 m lambda to 15 m lambda is the wavelength of these sound waves inside the body.

Formula for sound wave = sound wavelength (λ) = sound velocity (v) / Sound frequency (F).

After calculating sound wavelength by the above formula we get the wavelength in between 5 to 15.

A periodic wave's wavelength is its spatial period, or the length over which its shape repeats. It is a property of both traveling waves and standing waves, as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two neighboring crests, troughs, or zero crossings.  The spatial frequency is the reciprocal of wavelength. The Greek letter lambda is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids. Wavelength and frequency are inversely related, with shorter wavelengths for higher frequencies and longer wavelengths for lower frequencies, assuming a sinusoidal wave travelling at a constant wave speed. The medium (such as a vacuum, air, or body of water) through which a wave travels determines its wavelength. Waves can be anything from music to light to water to periodic electrical signals in a conductor.

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Describe how the nature vs nurture debate can be applied to the study of phobias.

Answers

The nature vs. nurture debate has been around since Ancient Greece when Plato and Aristotle disputed the origins of human behaviour. However, the development of genetics has gained a renewed interest in the scientific community and the eugenics community

a line in the spectrum of atomic mercury has a wavelength of 255 nm. when mercury emits a photon of light at this wavelength, the frequency of this light is

Answers

The frequency of light is 1.18 × 10^15Hz

Wavelength = 255m

The frequency of light is the number of cycles of the light wave that pass a given point in a specified amount of time. Light waves are electromagnetic waves that are characterized by their frequency, which is measured in hertz (Hz).

Calculating the frequency -

f = c/ λ

where c is the speed of light and λ is the wavelength

Thus,

c = 3 × 10^18 m/s

λ = 255 × 1/ 10^9

= 255 × 10^7

Substituting the values in the formula -

= 3 × 10^18 / 255 × 10^7

= 1.18 × 10^15

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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the first hill, where a horizontal spring that is initially compressed 0.25 m will push a small car forward.

A hill with a compressed spring at the top. There is a car sitting against the compressed spring. The portion of the hill is vertically colored orange, the middle section is vertically colored yellow and the last section colored green.
Each group of students will choose a car and a spring to push the car and then build a track. The assignment is to make the car go 5.0 m/s when it reaches the bottom of the first hill. Four groups of students choose springs and build tracks as described in the table.

A 4 column table with 4 rows. The first column is labeled group with entries A, B, C, D. The second column is labeled car mass in kilograms with entries .75, .60, .55, .84. The third is labeled spring constant in newtons per meter with entries 65, 35, 40, 32. The last column is labeled hill height in meters with entries 1.2, .90, 1.1 , .95.
Which group’s roller coaster will most likely make the car travel closest to 5.0 m/s when it is at the bottom of the first hill?

A
B
C
D

Answers

The group that will make the car travel closest to 5 m/s when it is at the bottom of the first will is of:

Group C.

How to obtain the velocity for each car?

The velocity that each car will assume at the bottom of the hill is given by the equation presented as follows:

v = square root (2gh + kx²/m).

The parameters are given as follows:

g = 9.8 m/s is the acceleration relative to the gravity.h is the height of the hill.k is the spring constant.x = 0.25 represents the initial compression of the spring.m is the mass of the car.

The parameters are given by the table in this problem, hence the velocities of each car are given as follows:

Group A: v = square root(2 x 9.8 x 1.2 + (65 x 0.25²)/0.75) = 5.38 m/s.Group B: v = square root(2 x 9.8 x 0.9 + (35 x 0.25²)/0.6) = 4.61 m/s.Group C: v = square root(2 x 9.8 x 1.1 + (40 x 0.25²)/0.55) = 5.05 m/s. -> closest to 5 m/s.Group D: v = square root(2 x 9.8 x 0.95 + (32x 0.25²)/0.84) = 4.58 m/s.

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Answer: group C

Explanation:

You have lenses with the following focal lengths:f= 25mm, 50mm, 100mm, and 200mm. - n what arrangement would you use these lenses to get the highest-power telescope? Fo=200, fe=25

Answers

Answer:To use these lenses to get the highest power telescopeFo=

Explanation:

The magnifier. Then we need only one lens. The angular exaggeration of the magnifier when the object is put veritably near to the focal point is given by the formula

M =

f

25 cm

where ff is the focal point. thus, the lower the focal length the larger the exaggeration therefore we should elect the lens with the shortest focal length which is the one with

f = 25 mm.

Step 2

2 of 4

{ The microscope.} The microscope. Then we need two lenses. The angular exaggeration of the microscope is given by

L ⋅ 25 cm

thus, lower the product of the focal lengths of the two lenses the larger the exaggeration. thus we should elect the two lenses with the lowest focal lengths i.e. boxed{f_0 = 25

= 25 mm and f

e

= 50 mm.

Step 3

3 of 4

{ The telescope.} The telescope. Then we need two lenses. The angular exaggeration of the telescope is given by

M = - frac{f_0}{f_e}

M = −

f

e

f

0

i.e. it's directly commensurable to the focal length of the objective lens and equally commensurable to the focal length of the eye piece lens. thus we should elect the objective lens with the loftiest possible focal length and the eyepiece lens with the lowest possible focal length

Pellets of mass 2.0 g are fired in parallel paths with speeds of 120 m/s through a hole 1.5 mm in diameter.How far from the hole must you be to detect a 1.0-cm-diameter spread in the beam of pellets?

Answers

The distance from the hole must be 2.244 × 10⁻²⁷ m to detect a 1.0 cm diameter spread in the beam of pellets.

Given that,

Mass of the pellet m = 2 g

Speed of the pellet v = 120 m/s

Diameter of the hole d = 1.5 mm = 1.5 × 10⁻³ m

Distance between the pellet and the hole D = 1 cm = 1 × 10⁻² m

Let us find the wavelength,

λ = h/mv = (6.626 × 10⁻³⁶)/(2 × 120) = 2.76 × 10⁻³³ m

So, y = (1.22 λD)/d = (1.22 × 2.76 × 10⁻³³ × 1 × 10⁻²)/(1.5 × 10⁻³) = 2.244 × 10⁻²⁷ m

Thus, the distance from the hole must be 2.244 × 10⁻²⁷ m to detect a 1.0 cm diameter spread in the beam of pellets.

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A figure skater is spinning slowly with arms outstretched. Shebrings her arms in close to her body and her moment of inertiadecreases by 1/2. Her angular speed increases by a factor ofA. 2B. 1C. 4D. square root of 2E. 1/2

Answers

She draws her arms close to her body, which causes her moment of inertia to fall by half. She moves at an angle that is faster by a factor of A. 2B. 1C. 4D. L=Iω

L=Iω

L is a constant that is inversely proportional to the rotational inertia and the angular speed due to the conservation of angular momentum.

Whenever "I" falls, rises proportionately (inversely related) A - Unless you meant increased, which you didn't, Part B is asking the same thing. B would follow from that. If not, Part C proceeds as in Part A: That key equation indicates that they are inversely proportional. In terms of, we solve for ω

I=L/ω

The moment of inertia of a rigid body, also known as the mass moment of inertia, angular mass, second moment of mass, or more precisely, rotational inertia, is a property that establishes the torque required for the desired angular acceleration about a rotational axis, much like mass establishes the force required for the desired acceleration. Depending on the axis selected and the distribution of the body's mass, it may take more torque to modify the body's rate of rotation for bigger moments.

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a body of mass 2.64 kg is pushed straight upward by a 32.9 n vertical force. what is its acceleration (in m/s2)?

Answers

A body of mass 2.64 kg is pushed straight upward by a 32.9 N vertical force, then its acceleration is 2.67m/s^2

Given the mass of body (m) = 2.64kg

Force acting on the body (F) = 32.9N

We know that F = ma where a is the acceleration on the body.

Here when body is thrown upwards acceleration acts upwards along with the force but in the opposite direction the gravitational acceleration acts downwards.

Then Fg force due to gravity is = mg where g is the gravitational acceleration = 9.8m/s^2

Fnet = F - Fg

ma = F- mg

a = 32.9 - (2.64x9.8)/2.64 = 2.66

Hence the acceleration acting on the body is 2.67m/s^2

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Weighing a Bacterium Scientists are using tiny, nanoscale cantilevers 4 micrometers long and 500 nanometers wide-essentially miniature diving boards-as a sensitive way to measure mass. The cantilevers oscillate up and down with a frequency that depends on the mass placed near the tip, and a laser beam is used to measure the frequency. A single E. coli bacterium was measured to have a mass of 710 femtograms = 7.10×10−16 kg with this device, as the cantilever oscillated with a frequency of 17.1 MHz .
Treating the cantilever as an ideal, massless spring, find its effective force constant.

Answers

If we consider the cantilever to be a perfect, massless spring, then K = 5.952N/m is its effective force constant.

How does science define force?

The meaning of the term "force" is obvious. At about this point, it's perfectly permissible to describe a force as shoving or pulling. An object doesn't "have a force in it" or "contain a force." A force is applied to one thing by another.

Why is force such a big deal?

A force can be a push or a pull, and it affects how we go about our daily lives because we can't open or close items, raise our legs or arms or accomplish a lot more without it.

Briefing

Numerous E coli bacteria, m = 7.10 × 10⁻¹⁶

frequency =17.1 MHz

frequency of vibration, F = 1/2π√K/M

17.1×10⁶= 1/2π√K/ 7.10 × 10⁻¹⁶

K/ 7.10 × 10⁻¹⁶ = 7.10 ×10¹⁵

K = 5.952N/m

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which arrangement of the sun, the moon, and the earth results in the highest high tides, and the lowest low tides on earth

Answers

Due to the alignment of the earth, moon, and sun, the moon and sun's gravitational pulls work in the same direction, producing the highest high tides and the lowest low tides.

What path does physics follow?

Location in relation to another object is referred to as direction. Relative terminology, such as up, down, out, left, right, forward, backward, or sideways, can be used to express direction by comparing the position of one item to another.

How do you define direction?

Direction can refer to the way something moves, the way you must go to get somewhere, the way something is beginning to take shape, or the direction you are facing.

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need help with the answer

Answers

1st law because Newton’s first law states that an object in motion (your body) remains in motion until another force (seatbelt) acts on it

what is the wavelength of an electromagnetic wave that has a frequency of 1 hertz? a. less than 1 m b. 1 m c. more than 1 m

Answers

The wavelength of an electromagnetic wave that has a frequency of 1 hertz  more than 1 m.

In a vacuum, electromagnetic waves move at a constant speed of 3.00 x 108 ms-1. Both the magnetic and electric fields have no effect on them. However, they can exhibit diffraction or interference. Any medium, including air, a solid, or a vacuum, can be traversed by an electromagnetic wave. It is not dependent on a medium to spread or move from one location to another. The opposite is true for mechanical waves, which require a medium to propagate (like sound or water waves). Transverse waves are electromagnetic waves. Thus, they are evaluated based on their amplitude (height) and wavelength (distance between the highest and lowest points of two successive waves).

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when ultraviolet light with wavelength of 370.0 nm falls on certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.40 ev. find the work function (binding energy) of the metal (in ev). (10 pts.)

Answers

The work function of the metal is 2.25 eV.

The work function, or binding energy, of a metal is the minimum energy required to remove an electron from the surface of the metal. In this case, the maximum kinetic energy of the emitted photoelectrons is 1.40 eV, which means that the energy required to remove the electron from the metal was at least 1.40 eV.

To find the work function of the metal, we can use the following equation:

Work function (eV) = Photon energy (eV) - Maximum kinetic energy of photoelectrons (eV)

Plugging in the values given in the problem, we get:

Work function (eV) = Photon energy (eV) - 1.40 eV

The energy of a photon can be calculated using the equation:

Photon energy (eV) = hc / wavelength (m)

Where h is the Planck constant (6.626 x [tex]10^{-34}[/tex] J × s), c is the speed of light (2.998 x [tex]10^{8}[/tex] m/s), and wavelength is the wavelength of the photon in meters.

Plugging in the values and converting the wavelength to meters, we get:

Photon energy (eV) = (6.626 x [tex]10^{-34}[/tex] J × s) × (2.998 x [tex]10^{8}[/tex] m/s) / (3.70 x [tex]10^{-7}[/tex] m)

= 3.65 eV

Substituting this value back into the first equation, we get:

Work function (eV) = 3.65 eV - 1.40 eV

= 2.25 eV

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an object of mass m moves at a constant speed v around a circular path of radius r. the net force applied on the object is f. if the acceleration of the object is halved, what happened to the mass?

Answers

A mass m object travels along a circular path with a radius r at a constant speed v. The object is subjected to a net force of f. The mass stays unchanged if the object's acceleration is cut in half.

An item moving in a circle at a consistent speed is known as uniform circular motion. An object constantly changes its direction as it moves in a circle. The object is always traveling tangent to the circle.

Given that an object experiences centripetal force as it moves in a circular motion.

So, Force (f) is = mv2/r, where m is an object's mass, v is its speed, and r is its path's radius.

In this case, centripetal acceleration is equal to (ac) = v2/r.

Given that when acceleration is halved:

ac' = ac/2 = v^2/2r

There is no effect on mass as it will remain constant throughout the circular motion of the object but net force will be effected.

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5 V battery with metal wires attached to each end.
The figure shows two wires connected to a battery’s positive and negative terminals. These wires stop short of completing a circuit. Four points are marked on the wires. Point 1 is at the end of the wire connected to the positive terminal. Point 2 is at the negative terminal. Point 3 is at the positive terminal. Point 4 is at the end of the wire connected to the negative terminal.
What are the potential differences ΔV12=V2−V1, ΔV23=V3−V2, ΔV34=V4−V3, and ΔV41=V1−V4?
Enter your answers numerically separated by commas

Answers

The potential differences is ΔV12=V2−V1=0V

ΔV23=V3−V2=5V

ΔV34=V4−V3= 0V

and ΔV41=V1−V4=-5V

From the figure, there is no connection between the points 1 and 4 . Hence, the potential at the points 4 and 3 is same and is equal to zero.

[tex]\Delta V_{34} & =V_4-V_3 \\& =0[/tex]

Similarly, the potential at the points 1 and 2 is same and is equal to zero.

[tex]\Delta V_{12} & =V_2-V_1 \\& =0[/tex]

The potential difference between the points 4 and 1 is equal to potential difference between the points 3 and 2.

Given that the potential difference across the points 3 and 2 is,

[tex]\Delta V_{23} & =V_3-V_2 \\& =5 \mathrm{~V}[/tex]

The potential difference between the points 4 and 1 is also 5V.

[tex]\Delta V_{41} & =-\Delta V_{23} \quad\left(\because \Delta V_{23}=\Delta V_{14}\right) \\& =-5 \mathrm{~V}[/tex]

Therefore, the required answer is 0,5V, 0,-5V.

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Note:- The correct question could be,

5 V battery with metal wires attached to each end.

The figure shows two wires connected to a battery’s positive and negative terminals. These wires stop short of completing a circuit. Four points are marked on the wires. Point 1 is at the end of the wire connected to the positive terminal. Point 2 is at the negative terminal. Point 3 is at the positive terminal. Point 4 is at the end of the wire connected to the negative terminal.

What are the potential differences ΔV12=V2−V1, ΔV23=V3−V2, ΔV34=V4−V3, and ΔV41=V1−V4?

Enter your answers numerically separated by commas

the north celestial pole is 35∘ above your northern horizon. this tells you that you are at:

Answers

You are at  latitude 35∘ N , if the north celestial pole is 35∘ above your northern horizon . The north celestial pole is a point where all stars can be seen from the northern hemisphere.

All of the stars visible from the Northern Hemisphere spin about the North Celestial Pole in the sky. North Street Polaris stays virtually immobile in the sky and all the stars in the northern sky appear to revolve around it because it is almost directly aligned with the Earth's axis of rotation "above" the North Pole, the north celestial pole. This spot in the sky is virtually exactly where ar, also known as Polaris, is located.

The reasonably brilliant Pole Star is located extremely close to the North Celestial Pole, making it simple to detect it in the night sky. But there are no observable stars near the South Celestial Pole. All of the stars in the sky are centred on the North Celestial Pole.

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A ray of light travelling in air strikes the surface of Jell-O at an angle of 37.1 degrees. If the light travels at 2.92 x 10^8 m/s through the Jell-O a
a) what is the index of refraction for Jell-O
b) What is the angle of refraction


Someone please help me fast

Answers

a) The index of refraction for Jell-O is  1.027.

b) The angle of refraction is 38.28°.

What is the index of  refraction?

The index of refraction measures how a light beam bends when it travels through different media. The refractive index n is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

if i is the angle of incidence of a ray in vacuum (angle between the incoming ray and the perpendicular to the surface of a medium, known as the normal), and r is the angle of refraction (angle between the ray in the medium and the normal). Then:

[tex]n = \frac{sini}{sinr}[/tex]

a)   the index of refraction for Jell-O is = speed of light in air/speed of the light in the medium

=   3.00x 10^8 m/s /2.92 x 10^8 m/s

= 1.027

b) the angle of refraction is = sin⁻¹(1.027sin37.1°)

= 38.28°

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the moon has a mass of 1×1022 kg, and the gravitational field strength at a distance r from the planet is 0.001 n/kg. what is the gravitational force exerted on the moon while it is in orbit around the planet?
a. 0 N
b. 1 x 10^19 N
c. 1 x 10^22 N
d. 1 x 10^25 N

Answers

The gravitational force exerted on the moon of mass 1×10²² kg  is 1×10¹⁹ N. And the right option is b. 1×10¹⁹ N

What is gravitational force?

Gravitational force is the most prevalent force in the universe, pulling together on any two objects with mass in the universe.

To calculate the gravitational force exerted on the moon, we use the formula below.

Formula:

F = mE.................... Equation 1

Where:

F = Gravitational forcem = Mass of the moonE = Gravitational field strength

From the question,

Given:

m = 1×10²² kgE = 0.001 N/kg

Substitute these values into equation 1

F = ( 1×10²²×0.001)F = 1×10¹⁹ N

Hence, the gravitational force exerted on the moon is 1×10¹⁹ N.

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Which occurs when white light separates into a spectrum of colors when it passes through a glass prism?

Answers

A spectrum of seven colors ( red, orange, yellow, green, blue, indigo, and violet) is created when white light travels through a glass prism and is dispersed.

When white light travels through a prism, dispersion is the breaking of the light into its individual colors. The seven colors that make up the spectrum are produced. White light experiences varying degrees of color bending when it passes through a prism.

The longest wavelength and least bending color are red, yet violet bends the most. A red light would therefore be at the apex of the created spectrum, violet would be at the bottom, and the remaining hues or colors would be in between.

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Two positive point charges q are placed on the x-axis, one at x = a and one at x = -a.a. Derive an expression for the electric field at points on the x-axis, where ?a < x < a.

Answers

Th electric field at a point between 2 points is given by E =   [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

Formula for electric field at a point on x-axis is given by:

E = [tex]\frac{kq}{r^2}[/tex]  ----(i)

where k is constant, q is magnitude of charge and r is the distance from charge where electric field is to be found.

                                         

   Q (-a,0)____________________(x,0)_____________________Q(a,0)

so we have to fing electric field at point (x,0) due to the both charges shown in the figure.

due to point charge at pont (-a,0) the electric field is in the positive x direction and due to charge at point (a,0) the electric field is in negative x direction.

let [tex]E_1[/tex] be the electric field due to charge at point (-a,0)

let [tex]E_2[/tex] be the electric field due to charge at point (a,0)

[tex]E_1[/tex] = [tex]\frac{kQ}{(a+x)^2}[/tex] (i^)

where i^ is unit vector along x- axis

[tex]E_2[/tex] = [tex]\frac{kQ}{(a-x)^2}[/tex] (-i^)

[tex]E_1+E_2 = \frac{kQ}{(x+a)^2}[/tex] i^  [tex]+ \frac{kQ}{(a-x)^2}[/tex] (-i^)

=>   [tex]\frac{-4kQax}{(a^2-x^2)^2}[/tex] i^

so [tex]E_1 + E_2 =[/tex] [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

so for any point a < x < a .

Th electric field is given by E =   [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

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what is the frequency of a photon that is emitted when the dipole moment of a proton flips in a magnetic field of 2.1 t? the energy of a photon is related to its frequency by the equation e

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42 Hz is the frequency of a photon that is emitted when the dipole moment of a proton flips in a magnetic field of 2.1 t, the energy of a photon is related to its frequency by the equation e.

The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field. The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. Paramagnetism, diamagnetism, and antiferromagnetism are three additional magnetic effects that a nonuniform magnetic field can have on "nonmagnetic" materials, albeit these forces are often so minute that they can only be detected by laboratory equipment. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things. Electric charges in motion and the intrinsic magnetic moments of elementary particles, which are connected to their spin, a fundamental quantum feature, create magnetic fields. As parts of the electromagnetic force, one of the four fundamental forces of nature, magnetic and electric fields are interdependent.

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What type of separation methods are described in the recipe? decantation and screening distillation and screening decantation and centrifugation distillation and filtration.

Answers

Type of separation methods described in the recipe  are: A. decantation and screening.

What is decantation and screening?

Process for the separation of mixtures of immiscible liquids or of a liquid and a solid mixture such as a suspension is called decantation. In this process, the precipitated liquid is separated from solid. According to the given instruction for the recipe, fat which is in liquid state is separated from the meat.

Practice of taking granulated ore material and separating it into multiple grades by particle size is called screening. In the process of screening, more liquid is separated by placing mixture on the screen. In this, gravity plays an important role for the process of separation.

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Note: The question given on the portal is incomplete. Here is the complete question.

Question: A recipe gives the instructions below

After browning the meat pour off fat from the pan to further reduce fat use a strainer.

what type of separation methods are described in the recipe

A decantation and screening

B distillation and screening

C decantation and centrifugation

D distillation and filtration

a 0.950-kg ball is dropped from rest at a point 3.30 m above the floor. the ball rebounds straight upward to a height of 1.90 m. taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?

Answers

The impulse of is the product of the average force and the duration it is excreted. The impulse of the net force exerted to the ball during the impact with the floor was 4.28 N-s in magnitude and direction.

The ball weighs 0.500 kg, according to the information provided.

The floor is 1.2 meters high.

The force of impulse

The impulse of is the product of the average force and the duration it is excreted.

One instant before the collision, the velocity is 4.852 meters per second.

One instant after the collision with the floor, the velocity was 3.71 meters per second squared.

The magnitude and direction of the ball's net force impulse during the collision with the floor.

4.28N-s.

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if two cars are both travelling at 50 km/h and they collide head on, the effect is similar to a car colliding with a wall at what velocity ?

Answers

The effect is similar to a car colliding with a wall at a velocity  of 50 km/h.

What is the principle of conservation of linear momentum?

The principle of conservation of linear momentum states that the sum of the initial momentum is equal to sum of the final momentum, provided that the system is Isolated.

Mathematically, the law of conservation of linear momentum is given as;

Pi = Pf

where;

Pi is the sum of the initial momentumPf is the sum of the final momentum

m₁u₁ = m₂u₂

where;

m₁ is the initial massu₁ is the initial velocitym₂ is the final massu₂ is the initial velocity

Based on Newton's third law of motion, action and reaction are equal and opposite.

If the car collides with a stationary wall, the wall will exert equal and opposite reaction to the car. If a car hits a wall with a velocity of 50 km/h, the wall will move at a velocity of 50 km/h provided it has equal mass with the car.

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