When driving on a highway, sudden strong cross wind gusts: Select one: a. Do not affect a car as much as a strong head wind b. Affect large cars more than small cars c. Always cause severe dust problems d. Can move a car sideways into another lane

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

[tex]\qquad \sf  \dashrightarrow \:E = \dfrac{hc}{\lambda} [/tex]

here,

____________________________________

[tex]\qquad \sf  \dashrightarrow \: E = \cfrac{1.99 \times 10 {}^{ - 25} }{7 \times 10 {}^{ - 12} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: E= \cfrac{1.99}{7} \times 10 {}^{ (- 25 + 12)} { } [/tex]

[tex]\qquad \sf  \dashrightarrow \:E= 0.284 \times 10 { }^{ - 13} [/tex]

[tex]\qquad \sf  \dashrightarrow \: E= 2.84 \times 10 { }^{ - 14\:\; } J\: \: [/tex]

so, correct choice is A

The buoyant force acting on the block is 58.8 N.

The dimensions of the block are given as 0.10 m, 0.20 m and 0.30 m.

The volume of the block will be = 0.10 × 0.20 × 0.30 m³

                                                     = 0.006 m³

We know that,

W = mg                                 .....(1)

where,

W = weight of the block

m = mass of the block

g = acceleration due to gravity = 9.8 m/s²

We also know that mass is proportional to density and volume

m = ρ.V                                  .....(2)

Substituting the equation (2) in (1)

W = ρ.V.g                               .....(3)

Density of water, ρ = 1000 kg/m³

Force is the net weight acting on the block

F = W = ρ.V.g

F = 1000 × 0.006 × 9.8

F = 58.8 N

Therefore, the buoyant force acting on the block is 58.8 N.

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The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

            p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

  = 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

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0.07  is the magnitude of the magnetic field at a distance of 2d from the wire

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The hottest celebrity is this one.

The majority of stars fall under a select few spectral categories. The hottest to the coldest stars are listed in the spectral types of the Henry Draper and Bright Star Catalogues (see stellar classification). The letters O, B, A, F, G, K, and M are used to indicate these categories in decreasing order of temperature.

Using spectral types, we can categorise stars based on their colour or the spectral lines that are visible in their light. The spectral types are OBAFGKM, with the hottest stars being O and the coldest being M. To humans, cooler stars appear redder and hotter stars appear bluer.

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The frequency heard by an observer in the moving car as he approaches the police car is 558.6 Hz.

The frequency heard by the observer is calculated by applying the principle of Doppler effect.

f₀ = f'[(v/v - vs)]

where;

f₀ = 500[(343 / 343 - 36)]

f₀ = 500(343/307)

f₀ = 558.6 Hz

Thus, the frequency heard by an observer in the moving car as he approaches the police car is 558.6 Hz.

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Hi there!

First, let's find the period of the pendulum. This can be found by solving for the amount of time it takes for the pendulum to make ONE complete swing.

[tex]T = \frac{\text{Total time}}{\text{Number of complete swings} }\\\\T = \frac{145}{110} = 1.318 s[/tex]

Now, let's use the equation for the period of a simple pendulum:
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]

T = Period (1.318 s)
L = length of string (0.55 m)

g = acceleration due to gravity on planet (? m/s²)

Let's solve for 'g' doing some quick rearranging of the equation:
[tex]T^2 = 4 \pi^2 (\frac{L}{g})\\\\g = \frac{4\pi^2 L}{T^2}\\\\[/tex]

Solving for 'g' by plugging in values:
[tex]g = \frac{4\pi^2 (0.55)}{(1.318)^2}\\ \\= \boxed{12.496 \frac{m}{s^2}}[/tex]

Answer:

The final temperature is [tex]29.6^oC[/tex].

Explanation:

When the two masses come in contact, one releases heat and the other absorbs it. The former can be modelled with the equation [tex]HeatLost = (Mass1 (kg))(c1)(T1-T)[/tex], and the latter by [tex]HeatGained=(Mass2(kg))(c2)(T-T2)[/tex]

m1=0.15 kg

m2=0.06 kg

T1 = 70 degrees Celsius = 343 K

T2 =  20 degrees Celsius = 293 K

T= Final temperature

c1 = Specific heat capacity of copper

c2 = Specific heat capacity of water

Because there is no heat lost into the surroundings, the heat removed from one substance is the same as the heat gained in the other. Therefore:

[tex](Mass1)(c1)(T1-T)=(Mass2)(c2)(T-T2)[/tex]

[tex](0.150)(400)(343-T)=(0.06)(4184)(T-293)[/tex]

[tex](60)(343-T)=(251.04)(T-293)[/tex]

[tex]20580-60T=251.04T-73554.72[/tex]

[tex]-311.04T=-94134.72[/tex]

[tex]T=302.6 K[/tex]

[tex]T=29.6^oC[/tex]

Hope this helps! (My apologies if the answer is wrong, it has been a while since I've done this)

Sales Promotion, Income advertising publications to short-term and transient incentives to buy or result in trials of new items.

Sales advertising is an advertising approach in which an enterprise uses a transient campaign or provide to increase interest or demand in its service or product. There are numerous motives why a commercial enterprise may additionally choose to apply an income promotion (or 'promo'), however, the primary purpose is to reinforce income.

sales promotion is one of the elements of the promotional mix. The number one element within the promotional blend is advertising and marketing, private selling, direct marketing, and exposure/public relations.

Examples encompass contests, coupons, freebies, loss leaders, a factor of purchase shows, charges, prizes, product samples, and rebates. sales promotions may be directed at both the patron, sales group of workers, or distribution channel individuals (such as outlets).

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The energy of a photon of ultraviolet radiation is E = 2.91 * 10^-18 J

What is a photon?

A photon is a particle of light defined as a discrete bundle (or quantum) of electromagnetic energy.

The energy of a photon is obtained by multiplying Planck's constant by the frequency of electromagnetic radiation:

E = h * f

where,

h is Planck's constant

h= 6.63×10⁻³⁴ Js

f= 4.4×10^15 Hz

Substituting the values in the equation for Energy:

E = (6.63×10⁻³⁴) * ( 4.4×10^15)

E = 2.91 * 10^-18 J

Hence,

The energy of a photon of ultraviolet radiation is E = 2.91 * 10^-18 J

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Answer: the first choice!

Explanation: did it the question on edg!:)

Answer:

See below

Explanation:

Power = amps * volts  = .27 * 12 = 3.24 watts

V= IR

R = V/I = 12/.27 = 44.44 ohms

The net gain in potential energy by the crate is 120 J.

Potential energy is the energy that an object has stored in it as a result of its position or condition. Potential energy exists in a stretched spring, a book held above your head, and a bicycle on top of a hill. The joule, which is represented by the letter "J," is the standard unit for calculating potential energy.

We use the idea of potential energy, which is given by the equation,

PE = mgh

where,

m = mass of the crate

g = gravitational acceleration

h = height

Given,

mg = 40 N

L = 5 m

angle of inclination, ∅ = 37°

h = L sin∅

h = 5 (sin 37°)

h = 5 (3/5)

h = 3 m

Put the values in the above formula,

PE = mgh

PE = 40 (3)

PE = 120 J

Therefore, the net gain in potential energy by the crate is 120 J.

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C. The work done by the 5500.0 N force on the elevator is 275,000 J.

The work done by the 5500.0 N force on the elevator is calculated as follows;

W = Fd

where;

W = 5500 x 50

W = 275,000 J

Thus, the work done by the 5500.0 N force on the elevator is 275,000 J.

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The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.

Given:

mass of block 1, m₁ = 40 gm = 40×10⁻³ kg

mass of block 2, m₂ = 20 gm = 20×10⁻³ kg

Applied force, F = 0.03 N

Calculation:

Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:

F = ma

where F is the applied force

            m is the total mass of the system

            a is the acceleration of block 2 (as it is pulled by horizontal force)

From the above equation we get:

0.03 N = (m₁+m₂) a

⇒ a = (0.03 N) / (m₁+m₂)

⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)

⇒ a = 0.5 m/s²

Now, from the free-body diagram of block 2 as shown in figure 3, we get:

Balancing the forces along the horizontal:

∑Fₓ = 0

∴ T = m₂ a

where T is tension in the string

           a is the acceleration of block 2

Applying values in the above equation we get:

T = (20×10⁻³ kg) × (0.5 m/s²)

  = 0.01 N

Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.

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The sound that Tina produces while singing is loud and high-pitched.

How is a sound produced?

Vibrations that travel through the air or another medium and can be heard by an observer are called sounds. Sound is produced by the following cycle:

Therefore, when Tina sings larger vibrations are created, and the tension within her vocal cords increases producing a very high-pitched loud sound.

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Answer:

31.43 Ω

Explanation:

Using Ohm's Law:

[tex]\displaystyle{V=IR}[/tex]

Where V = voltage (V) , I = electric current (A) and R = resistance (ohm)

We know that a computer uses 3.5 ampere at 110 volt, we want to know the resistance. Therefore substitute V = 110 and A = 3.5

[tex]\displaystyle{110=3.5R}\\\\\displaystyle{\dfrac{110}{3.5}=R}\\\\\displaystyle{R=31.43}[/tex]

Therefore, the resistance will be around 31.43 Ohm

The period remains the same if the amplitude of the motion is increased to 2a

What is Time period?

The time period is the time taken by a complete cycle of the wave to pass a point.

The equation of the period for spring is:

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where,

Time period = T  

mass of system = M

k is the spring constant

Amplitude of the motion is a

From the equation we can say that,

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

Now if we increase the amplitude of the motion to 2a, the period of time remains the same.

Because the time period is not dependent on amplitude, it is independent of a.

Hence,

The period remains the same if the amplitude of the motion is increased to 2a.

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Answer:

2.72×10^-7

Explanation:

velocity = frequency × wavelength

2.05×10^8=7.55×10^14 wavelength

wavelength = 2.05×10^8/7.55×10^14

wavelength = 2.72×10^-7

Answer:

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

The orbital speed of the satellite is 7.35*10^3 m/s.

What is orbital speed?

The speed of the satellite in its orbit is termed the orbital speed.

The orbital speed is given by the formula,

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

where G is the universal gravitational constant, M is the mass of the planet and r is the distance of the satellite from the center of the planet.

Here the distance of the satellite from the center of the planet is the sum of the planet's radius and the height attained by the satellite above the ground. So

r=6370 + 1000

r=7370 km

Given the mass of the planet is 5.97*10^24 kg and the value of the gravitational constant is 6.67*10^(-11) N m^2 kg^(-2), substitute these values in the formula of the orbital speed.

Note: 1 km=1000 m

[tex]v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370 \text{ km}}} \\ v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370\times 1000 \text{ m}}} \\ v= 7.35\times 10^3 \text{ m/s}[/tex]

Hence the orbital velocity of the satellite is 7.35*10^3 m/s.

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

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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a) Density is mass of a unit volume of a material substance

b) Algebraic expression relating Density = m / v

c) Density is an intensive property, which means the density does not change as the amount of the substance present changes

a) Density is mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is the measurement of how tightly a material is packed together.

The more closely packed particles are, the more dense the object. The amount of mass in a particular space or volume. The more particles (or mass) in a given space, the more dense it is.

b) Algebraic expression relating Density = mass / volume or m/v

where d is density, M is mass, and V is volume.

c) Density (ρ) is the amount of mass (m) per unit volume (V) of a substance. Density is an intensive property, which means the density does not change as the amount of the substance present changes.

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It looks like you're given

[tex]x(t) = -11\,\mathrm m + \left(1\dfrac{\rm m}{\rm s}\right) t - \left(5\dfrac{\rm m}{\mathrm s^3}\right) t^3[/tex]

[tex]y(t) = 19\,\mathrm m + \left(3\dfrac{\rm m}{\rm s}\right) t - \left(9\dfrac{\rm m}{\mathrm s^2}\right) t^2[/tex]

The particle's position vector at time [tex]t[/tex] is given by

[tex]\vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

Differentiate [tex]\vec r[/tex] twice to recover the velocity and acceleration vectors.

[tex]\vec a(t) = \dfrac{d\vec v(t)}{dt} = \dfrac{d^2\vec r}{dt^2} = x''(t)\,\vec\imath + y''(t)\,\vec\jmath \\\\ \implies \vec v(t) = \left(1\dfrac{\rm m}{\rm s} - \left(15\dfrac{\rm m}{\mathrm s^3}\right) t^2\right)\,\vec\imath + \left(3\dfrac{\rm m}{\rm s} - \left(18\dfrac{\rm m}{\mathrm s^2}\right) t\right) \,\vec\jmath \\\\ \implies \vec a(t) = -\left(30\dfrac{\rm m}{\mathrm s^3}\right)t \, \vec\imath - \left(18\dfrac{\rm m}{\mathrm s^2}\right) \,\vec\jmath[/tex]

At [tex]t=1.4\,\rm s[/tex], the particle has acceleration

[tex]\vec a(1.4\,\mathrm s) = \left(-42\,\vec\imath - 18\,\vec\jmath) \dfrac{\rm m}{\mathrm s^2}[/tex]

with magnitude

[tex]\|\vec a(1.4\,\mathrm s)\| = \sqrt{\left(-42\dfrac{\rm m}{\mathrm s^2}\right)^2 + \left(-18\dfrac{\rm m}{\mathrm s^2}\right)^2} \approx 45.695\dfrac{\rm m}{\mathrm s^2}[/tex]

and making an angle [tex]\theta[/tex] relative to the positive [tex]x[/tex]-axis such that

[tex]\tan(\theta) = \dfrac{-18}{-42} = \dfrac37[/tex]

Since both components of the acceleration vector have negative sign, the acceleration points into the third quadrant, so that we add a multiple of 180° after taking the inverse tangent of both sides, namely

[tex]\theta = \tan^{-1}\left(\dfrac37\right) - 180^\circ \approx -156.801^\circ[/tex]

Now, (a) the magnitude of the net force acting on the particle is, by Newton's second law,

[tex]F = (0.41\,\mathrm{kg})\|\vec a(1.4\,\mathrm s)\| \approx \boxed{18.735\,\mathrm N}[/tex]

and (b) makes the same angle as the acceleration vector, [tex]\theta \approx \boxed{-156.801^\circ}[/tex].

At this moment, its velocity vector is

[tex]\vec v(1.4\,\mathrm s) = \left(-28.4\vec\imath - 22.2\,\vec\jmath\right) \dfrac{\rm m}{\rm s}[/tex]

which (c) makes an angle [tex]\theta[/tex] such that

[tex]\tan(\theta) = \dfrac{-22.2}{-28.4} = \dfrac{111}{142}[/tex]

This vector also points in the third quadrant, so

[tex]\theta = \tan^{-1}\left(\dfrac{111}{142}\right) -180^\circ \approx \boxed{-142.986^\circ}[/tex]

Status refers to a person's prestige, social honor, or reputation in society. Weber stated that political strength become now not rooted entirely in capital fees, but additionally in a single's individual popularity. Poets or saints, for example, can own a massive impact on society, often with little monetary worth.

The German sociologist Max Weber formulated a three-component theory of stratification that defines a status group (additionally status class and status estate) as a group of humans within a society who can be differentiated by means of non-financial characteristics along with honor, prestige, ethnicity, race, and religion.

Social stratification entails society as a system of hierarchical categories. Max Weber defined stratification because the division of a society into distinct communities, that have varying assignments of “status honor” or status.

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Graviton

Graviton, postulated quantum that is thought to be the carrier of the gravitational field. It is analogous to the well-established photon of the electromagnetic field.

String theory states that everything in our Universe is made up of tiny vibrating strings. These strings are one dimensional objects and are identical to one another.

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When they meet the 40-kg boy will have moved a distance of 6 m.

Displacement of the 40 kg boy

The displacement of the 40 kg boy is calculated from the principle of center mass.

X(40 kg) = (60 x 10 m    +  40 x 0)/(40 kg + 60 kg)

X(40 kg) = (600)/(100) = 6 m

X(60 kg) = (60 x 0    +  40 x 10 m)/(40 kg + 60 kg)

X(60 kg) = (400)/(100) = 4 m

Thus, when they meet the 40-kg boy will have moved a distance of 6 m.

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The beat frequency will be 8.1 Hz  when the horns are sounded on a day when the speed of sound is 340 m/s

frequency = speed / wavelength

f1 = 340 / 6 = 56.67 Hz

f2 = 340 / 7 =  48.57 Hz

Beat frequency  = fb = | f2 - f1 |

                                   = | 48.57  - 56.67 |

                                   = 8.1 Hz

The beat frequency will be 8.1 Hz  when the horns are sounded on a day when the speed of sound is 340 m/s

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The decreasing order of the momenta are b>a=d=e>c.

Solution: The momentum of all the options are mentioned below according to the formula:

a. p = 20 × 1 = 20gm/s

b. p = 20 × 2 = 40gm/s

c. p =10 × 1 = 10gm/s

d. p = 10 × 2 = 20 gm/s

e. p =200 × 0.1 = 20gm/s

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Question:

I hope you are trying to ask the question below:

Rank in order, from largest to smallest, the momenta a to e of the objects

a. 20 g & 1 m/s

b. 20 g & 2 m/s

c. 10 g & 1 m/s

d. 10 g & 2 m/s

e. 200 g & 0.1 m/s

Debris collecting around pilings is most likely to create a boating hazard around river bridges.

The option (c) is the correct option.

Boating Hazards:

Dams, submerged items, freezing water, rapidly changing weather, sunstroke, and current are just a few of the dangers that boaters may encounter. It's not always easy to see these risks. These risks must be recognized by boaters, and they must always be prepared to prevent hazards.

Operator negligence is the most frequent reason for boating accidents, according to US Coast Guard (USCG) recreational boating statistics from 2019. Inattentiveness on the part of the operator can result in crashes, people falling overboard, and slip-and-fall incidents on board, all of which can result in life-threatening injuries.

The greatest places to find more about any potential local risks are marinas and local boaters. Check any nearby marine charts as well to learn about potential dangers and how to avoid them.

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