PHYSICS COLLEGE

Nina and Jon are practicing an ice skating routine. Nina is standing still. Jon, who is twice as heavy as Nina, skates toward her, pushing Nina away with force f. Assuming the system is closed, which statement is correct about this system? a. Nina experiences a force equal to f/2. b. Nina experiences a force equal to f^2. c. Nina experiences a force equal to 2f. d. Nina experiences a force equal to f.

Answers

Answer 1

Answer:

Answer:

Explanation:

• Nina experiences a force equal to f.

Answer 2

Answer:

Answer:

Nina experiences a force equal to f

Explanation:

got to get that 2nd answer slot correct too before an abusive expert verifier with an alt comes in and purposely verifies the wrong answer

Related Questions

If anyone can help with this, before 8 am eastern time i’ll give you 20$. Thank you
A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spectrometer. What is the mass of this ion?
If you travel 2 km north, then travel 5 km south, what is your displacement?
Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?
When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = (1)/(2)mv^(2)$

So, $v = \sqrt{(2E)/(m)}$ .... (1)

Now,

$r = (mv)/(Bq)$

Substitute the value of v from equation (1), we get

$r = (√(2mE))/(Bq)$

Let the radius of the alpha particle is r2.

For proton

So, $r_(1) = \frac{\sqrt{2m_(1)E}}{Bq_(1)}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_(2) = \frac{\sqrt{2m_(2)E}}{Bq_(2)}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle $=6.64* 10^(-27) kg$

mass of proton $=1.67* 10^(-27) kg$

charge on alpha particle is twice of proton

radius of Proton is given by

$r=(mv)/(|q|B)$

and Kinetic energy $K=(P^2)/(2m)$

where P=momentum

$P=√(2Km)$

$r=(√(2km))/(qB)$ ---1

Radius for Alpha particle is

$r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}$ -----2

Divide 1 & 2 we get

$(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}$

$r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5$

$r_(alpha)=0.997* 16$

$r_(alpha)=15.95 cm$

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.

Answers

Answer: Length axis f= 114.3 Hz, Width axis f=228.67 Hz

Explanation:

We are given that,

Length of tub= 1.5 m

Width of tub= 0.75 m

Sound speed= 343 m/s

Now, we are also given shower is closed.

So, frequency is given as:

f= $m* (v)/(2L)$

For length axis

Put v= 343 m/s, m=1 and L=1.5 m

f= 1 * $(343)/(2*1.5)$

f= 114.3 Hz

For next resonant frequency, m=2

f= 2* $(343)/(2*1.5)$

f= 228.67

For width axis

Put v= 343 m/s, m=1 and L= 0.75 m

f= 1* $(343)/(2*0.75)$

f= 228.67 Hz

For next frequency, m=2

f= 2* $(343)/(2*0.75)$

f= 457.34 Hz

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Express your answer in micrometers(not in nanometers).

Answers

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=(y)/(D)\n\Rightarrow \theta=tan^(-1){(y)/(D)}\n\Rightarrow \theta=tan^(-1){(4.84* 10^(-3))/(3)}\n\Rightarrow \theta=0.09243\ ^(\circ)$

$sin\theta=(\lambda)/(d)\n\Rightarrow d=(\lambda)/(sin\theta)\n\Rightarrow d=(598* 10^(-9))/(sin0.09243)\n\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=(\lambda')/(2)\n\Rightarrow \lambda'=2dsin\theta\n\Rightarrow \lambda'=2* 0.00037066* sin0.09243\n\Rightarrow \lambda'=1.196* 10^(-6)\n\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

Final answer:

The wavelength of light that will produce the first-order dark fringe at the same point on the screen is the same as the original wavelength of the light, which is 598 nm (0.598 μm).

Explanation:

To find the wavelength of light that will produce the first-order dark fringe at the same point on the screen, we can use the equation dsinθ = nλ, where d is the separation between the slits, θ is the angle of the fringe, n is the order of the fringe, and λ is the wavelength of the light.

In this case, the first-order bright fringe is located at a distance of 4.84 mm from the center of the central bright fringe. Since this is a first-order fringe, n = 1.

Plugging in the values, we have (0.120 mm)(sinθ) = (1)(λ). Rearranging the equation gives sinθ = λ/0.120 mm.

Since the first-order dark fringe is located at the same point as the first-order bright fringe, the angle of the first-order dark fringe can be calculated by taking the sine inverse of λ/0.120 mm.

Finally, to find the wavelength of light that will produce the first-order dark fringe at this point, we can rearrange the equation to solve for λ: λ = (0.120 mm)(sinθ).

Now, substitute the known values into the equation to calculate the wavelength of light:

λ = (0.120 mm)(sinθ) = (0.120 mm)(sin sin^-1(λ/0.120 mm)) = λ.

The wavelength of light that will produce the first-order dark fringe at this point on the screen is the same as the original wavelength of light, which is 598 nm. Converting this value to micrometers, we get 0.598 μm.

Learn more about interference pattern here:

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An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

What it the Zeeman effect and what does it tell us about the Sun?

Answers

Answer:

Explanation:

Zeeman Effect -

In the presence of static magnetic field , the process of splitting of the spectral line into several components , is called Zeeman effect .

On the sun , there is a strong magnetic field , and hence , can show Zeeman effect .

And due to this ,

The sunspots can be studied by using this effect .

Sunspots are the darker region on the surface of Sun .

Lab: Weather Patterns

Answers

A weather pattern is defined as a period of time when the weather remains consistent. In the lab, a lot of observation about weather is obtained

What is the definition of a weather pattern?

A weather pattern is defined as a period of time when the weatherremains consistent. Weather changes are crucial to humanexistence.

because they influence our everyday activities and provide moisture for crops.

The rain does not always end within the day, and gloomy days might last just as long as sunny days. Tornadoes and hurricanes, for example, may inflict tremendous damage.

In the lab the following observation about weather is obtained;

1. We will find the graphs and statistics that indicate signs of climate change and engage with an interactivegraphic.

2. You'll also look at and debate maps of global temperature and precipitation patterns that are changing.

3. This lab will teach you about Earth's biomes and the close relationship that exists between them and the climates that serve to define them.

To learn more about the weatherpattern refer to the link;

brainly.com/question/2497685

Final answer:

The question pertains to meteorology, climatology, and atmospheric science. These are disciplines that study weather and climate, respectively, and their effects on the planet. Atmospheric Science is a broad field that includes both and employs physics principles.

Explanation:

The question refers to the subjects of meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric impacts on the Earth's weather. It involves the prediction of weather in the short term based on thousands of measurements of variables such as air pressure and temperature.

Climatology, on the other hand, is the study of climate, which involves analyzing averaged weather conditions over longer time periods using atmospheric data. Unlike meteorologists, climatologists focus on patterns and effects that occur over longer timescales of decades, centuries, and millennia.

Atmospheric Science is a broad field that encompasses both meteorology and climatology, as well as other disciplines that study the atmosphere. This discipline is typically based heavily on physics and involves the study of weather and climate patterns, predictions of developments in weather and climatic events, and the analysis of the effects of these events on the planet and its inhabitants.

Learn more about Weather Patterns here:

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