Why are the meters squared in the formula to calculate acceleration?

Answer:

Vrel= 0.75c

Explanation:

See attached file

Answer: i think it is d. none of them.

Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.

Answer:

292 nm

Explanation:

The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J

at maximum wavelength, the energy of the photons is equal to its work function

Energy of the electron = hf

but hf = hc/λ

where h is the planck's constant = 6.63 × 10-34 m^2 kg/s

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength that this occurs, which is the maximum wavelength

Equating, we have

hc/λ =  ∅

substituting, we have

(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19

(1.989 x 10^-25)/(6.81 x 10^-19) = λ

λ = 292.07 x 10^-9 = 292 nm

Answer:

d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Explanation:

Let us take the momentum of a photon unit as u

we know that the rate of change of momentum is proportional to the force exerted.

For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that

F = (u - 0)/t = u/t

for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,

F = u

For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to

F = (u - (-u))/t = 2u/t

just as the we did above, it becomes

F = 2u.

From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Answer:

The moment of inertia is  [tex]I= 312.09 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The  mass of the platform is   m =  137 kg

     The radius is  r  =  1.53 m

    The mass of the person is  [tex]m_p = 68.7 \ kg[/tex]

    The distance of the person from the center is  [tex]d_c =1.19 \ m[/tex]

    The mass of the dog is  [tex]m_d = 25.9 \ kg[/tex]

     The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

Generally the moment of inertia of the system is mathematically represented as

      [tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

          [tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

           [tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

           [tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also  [tex]I_2[/tex]  is the moment of inertia of the person about the axis which is mathematically represented as

          [tex]I_2 = m_p * d_c^2[/tex]

substituting values

          [tex]I_2 = 68.7 * 1.19^2[/tex]

          [tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also  [tex]I_3[/tex]  is the moment of inertia of the dog about the axis which is mathematically represented as

          [tex]I_3 = m_d * d_d^2[/tex]

substituting values

          [tex]I_3 = 25.9 * 1.45^2[/tex]

          [tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus  

        [tex]I= 160.35 + 97.29 + 54.45[/tex]

        [tex]I= 312.09 \ kg \cdot m^2[/tex]

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Answer:

The new volume will be 0.918 x 10^5 L

Explanation:

initial volume  [tex]V_{1}[/tex] = 1.21 x 10^5 L

Initial temperature [tex]T_{1}[/tex] = 265 K

Final volume [tex]V_{2}[/tex] = ?

Final temperature [tex]T_{2}[/tex] = 201 K

Th gas is an ideal gas.

For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant

substituting value, we have

(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201

[tex]V_{2}[/tex] =  24321000/265 = 91777.4 L

= 0.918 x 10^5 L

Answer:

30.93 cm

Explanation:

Given that:

A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses

The power of the old pair of lens p = 2.25 diopters

The focal point length = 1/p

The focal point length =  1/2.25

The focal point length = 0.444 m

The focal point length = 44.4 cm

The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens

The near point of the person from the glass = 83 cm

Let consider s' to be the image on the same sides of the lens,

∴ s' = -83 cm

We known that:

the focal length of a mirror image 1/f =1/u +1/v

Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.

Then:

1/f = 1/s + 1/s'

1/s = 1/f - 1/s'

1/s = (s' -f)/fs'

s = fs'/(s'-f)

s =( 44.4× -83)/(-83 - 44.4)

s = - 3685.2 / - 127.4

s = 28.93 cm

Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm

= 30.93 cm

Answer:

D Is 430m

Explanation:

See attached file

Answer:

Explanation:

Charge on an electron (q) = 1.6 * 10 ^ -19 C

Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec

We know that, Force exerted on moving particle moving through a magnetic field :

[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]

1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B

B =  0.08573 T

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

Answer:

d = 6.38 x 10⁻³ m = 6.38 mm

Explanation:

Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:

Fringe Spacing = Δx = 1 cm/34

Δx = 0.0294 cm = 2.94 x 10⁻⁴ m

But, the formula for fringe spacing in a double slit experiment is:

Δx = λL/d

where,

λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m

L = Distance between screen and slits = 3.1 m

d = slit separation = ?

Therefore,

2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d

d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)

d = 6.38 x 10⁻³ m = 6.38 mm

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]

Thus, the effective dose allowable yearly = 16 times

Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish  = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]

(B). the depth of the fish in the mirror.

[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]

Now find the depth of reflection of the fish in the bottom of the tank.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]

Answer:

A = 6.8 km²

Explanation:

A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.

B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;

F_rad = 2IA/c

I is given by the formula;

I = P/(4πr²)

Thus;

F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²

Where;

A is the area of the sail

r is the distance of the sail from the sun

c is the speed of light = 3 × 10^(8) m/s

P is total power output of the sun = 3.90 × 10^(26) W

Now,F_rad = F_g

Where F_g is gravitational force.

Thus;

PA/2cπr² = G•m•M_sun/r²

r² will cancel out to givw;

PA/2cπ = G•m•M_sun

Making A the subject, we have;

A = (2•c•π•G•m•M_sun)/P

Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg

G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²

Thus;

A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))

A = 6.8 × 10^(6) m² = 6.8 km²

Answer:

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

[tex]B = \frac{N\mu_o I}{2R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Answer:

The work done by the system is 550 J

Explanation:

Given;

heat added to the system, Q = 550 J

Apply the first law of thermodynamics;

ΔU = Q - W

Where;

ΔU is change in internal energy

Q is the heat added to the system

W is the work done by the system

During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.

0 = Q - W

W = Q

W = 550 J

Therefore, the work done by the system is 550 J

Answer:

1988.05 rad/s^2

Explanation:

The angular speed of the optical disk ω = 998.0 rad/s

the time taken to come to rest t = 0.502 s

The magnitude of the average angular acceleration ∝ = ω/t

∝ = 998.0/0.502 = 1988.05 rad/s^2

Answer:

Explanation:

Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/

α = 26.56°

Laser angle with Normal = 90 - 26.56 = 63.44 °

Assuming a red laser, refractive index in water is 1.331.

Angle of refraction in water is given by:

Ref Ind = Sin i / Sin r

1.331 = Sin 63.44 / Sin r

Sin r = 0.8945 / 1.331 = 0.6721

Angle r = 42.22°

For the path in water:

Tan 42.22 = x / 3.2

x = 2.9m where x is the lateral displacement of the laser ince it hits the water

So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool

Answer:

objetive: a converging lens for large diameter lenses

eyepiece you must select a lens with a small focal length and the diameter is not important

The selected lenses should decrease chromatic aberration.

Explanation:

A telescope is an instrument that collects light from very distant objects, therefore very weak.

Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.

For the eyepiece you must select a lens with a small focal length and the diameter is not important

the telescope magnification is

                 m = f_objective / F_ocular

The selected lenses should decrease chromatic aberration.

In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is                 m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Explanation:

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Hope this Helps!! Sorry its late