PLZZZZ HELPPPPPPPPPppppp​

Answer:

gas, liquid, solid

sound cannot travel in space

Answer:

1. gas , liquid , gas

2.sound cannot travel in space

Answer:

a) the body  is changing direction,

b)the body must go to the left and the acceleration to the right

c) the movement has not started.

d) all points of the motion

Explanation:

In this exercise you are asked to find in which position you have the following characteristics of the movement

a) The velocity is zero and the acceleration is negative

This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.

In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases

b) The velocity is zero, but the acceleration is positive

This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right

c) Both are zero

This only occurs where the body is stopped and the movement has not started.

d) both the velocity and the relation are nonzero.

This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration

Answer:

All answer are explained below in the explanation section.

Explanation:

1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.

As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.

Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.

2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.

Tire pressure at temperature 18.6 degree C is ~ 35 psi.

Temperature at air pressure of 37 psi is ~26.1 degree C

3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.

3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32

3. c) It is already done in part a of this question.

4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.

5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]

Plugging in the values, we get:

Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:

% Error = 21.33%

6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi

6.  b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.

6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.

7.  a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.

7.  b.) Estimation of temperature from best fit value of line is = 26.64 degree C

7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.

Answer:

the change in kinetic energy of the system is 0.577 J

Explanation:

Given;

mass of the bullet, m₁ = 6.25 g = 0.00625 kg

initial velocity of the bullet, u₁ = 365 m/s

mass of the crate, m₂ = 4.5 kg

initial velocity of the crate, u₂ = 0

distance moved by the system after collision, d = 0.15 m

Determine the final velocity of the system after collision;

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.00625 x 365 + 4.5 x 0 = v(0.00625 + 4.5)

2.2813 + 0 = v(4.5063)

2.2813 =  v(4.5063)

v = 2.2813 / 4.5063

v = 0.506 m/s

The change in kinetic energy of the system after collision is calculated as;

ΔK.E = ¹/₂ (m₁ + m₂)v²

ΔK.E =  ¹/₂ (4.506) x 0.506²

ΔK.E = 0.577 J

Therefore, the change in kinetic energy of the system is 0.577 J

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

Answer:

a) v_f = 0.898 m / s, b)   v₂ = -6.286 m / s

Explanation:

a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s

Initial instant. Before you start to pass gas

        p₀ = (M + Δm) v

M is the mass of the astronaut M  = 80Kg and Δm the masses of the gases

Final moment. When you expel the gases

        p_f = M (v + Δv) + Δm (v-v_e)

where v_e is the gas velocity v_e = 100 m / s

momentum is conserved

        p₀ = p_f

        M v + Δm v = Mv + M Δv + Δm v -Δm ve

          0 = M Δv - Δm v_e

if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM

         M dv = -v_e dM

         ∫ dv = - v_e ∫ dM / M

We solve, between the lower limits v₀ = v with M = M₀   and the upper  limit v = v_f for M = M_f

         v_f - v₀ = - v_e (ln M_f - Ln M₀)

         v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])

         v_f = v₀ + v_e ln (\frac{M_o}{M_f})

let's calculate

         v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)

          v_f = -1.3 +2.20

          v_f = 0.898 m / s

b) launch the jetpack to increase its speed up to the speed of the platform

  initial instant. Before launching the tanks

        p₀ = (M + m') v_f

final instnte. After launching the tanks

       p_f = M v₁ + m' v₂

indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it

       p₀ = p_f

       (M+ m) v_f = M v₁ + m v₂2

       v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]

        v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f

let's calculate

        v₂ = 80/10 (0.898 - 0) + 0.898

        v₂ = -7.1874 + 0.898

        v₂ = -6.286 m / s

Answer:

The direction of the force on the vertical wire is towards the East or right.

Explanation:

Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.

Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.

Answer:

2 m

Explanation:

speed=frequency×wavelength

wavelength=speed/frequency

wavelength=1000/500

=2 m

Answer:

Workdone = -2730 J

Explanation:

Formula for workdone is;

W = Force × Displacement

Now, according to Newton's 3rd law of motion, to every action, there is an equal and opposite reaction.

In the question given, we are told that a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus. This means that the force exerted by the rope on Magnus acts in an opposite direction to that which Magnus does to the rope.

Therefore, the force will be in the negative direction.

So;

Workdone = -2100 N × 1.3 m

Workdone = -2730 J

Answer:

a) electric field at point P must be zero

b) harged must be positive

c) force ais in the direction of the electric field

d)  force is in the opposite direction to the electric field

e)  force is in the opposite direction to the field

Explanation:

After reading your exercise, it is unfortunate that the diagram did not come out, but we are going to answer the questions in general.

a) force on a charge (+) is zero

this implies that the electric field at point P must be zero

        F = q E

b) the magnitude of the charge on the left is on the right

this indicates that the charged must be positive since the lines must exit the charge

c) force on load directed towards (direction not indicated)

since the charge is positive the force at point L is in the direction of the electric field at this point

d) force on test load (-) does not indicate direction

The force on a negative charge is in the opposite direction to the electric field at point J

e) Force on a test load (-) at point N

the force is in the opposite direction to the field at point N

Answer:

5 m/s

Explanation:

50=1/2*4v^2

4*1/2=2

25*2=50

so...

square rood of 25 is 5

answer 5 m/s

sorry if that didn't make since

The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule.  The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.

To compute the Molecular Mass of [tex]$O_{3}[/tex]

Atomic mass of oxygen(O) = 16

As [tex]$O_{3}[/tex] contains 3 atoms,

The molecular mass of [tex]$O_{3}[/tex]

= (16 x 3) = 48g/mol

Hence the mass of 9.11 moles O3

= 9.11 mol x 48g/mol

= 437.28g

= 0.43728kg

Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

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Answer:

      I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

we see the intensity decreases with the inverse of the distance squared

Explanation:

Intensity is defined as power per unit area,

           I = P / A

in this case we have that the sound is emitted in a spherical form therefore the area is

           A = 4 pi r2

therefore the intensity is

          I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

as we see the intensity decreases with the inverse of the distance squared

Answer:

-0.54m/s²

Explanation:

According to Newton's second law of motion

F = ma

Force = mass * acceleration

Given

Mass m = 4490kg

Take the sum of forces

Sum of force along the east = 890+1210 = 2100N

Sum of forces along the west = -4520N

Net force = -4520+2100

Net force = -2420N

Acceleration = Net force/Mass

Acceleration = -2420/4490

Acceleration = -0.54m/s²

Hence the boat acceleration is -0.54m/s²

Answer:

[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

[tex]13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}[/tex]

[tex]current(I) \times 10.92= 19292[/tex]

[tex]current(I) = \dfrac{19292}{10.92}[/tex]

[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]

Explanation:

Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero. Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero.

Answer:

a)  W = 1.63 10⁻²⁸ J,  b)  W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d)  W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

          W = ΔU = U_∞ -U

the potential energy for point charges is

           U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]

in this case we only have two particles

           U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]

the distance is

           r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]

           r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)

           r₁₂ = √2= 1.4142 m

we substitute

           W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

            W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142

            W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

            U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]

in this case all charges are electrons

             q₁ = q₂ = q₃ = q

             W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]

the distances are

            r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0

            r₁₃ = 3

            r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2

            r₂₃ = √13

            r₂₃ = 3.606 m

let's look for the job

            W = U

let's calculate

            W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]

            W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

             W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]

all charges are equal q₁ = q₂ = q₃ = q₄ = q

             W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]

let's look for the distances

             r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]

             r₁₄ = 5 m

             r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]

             r₂₄ = √13 = 3.606 m

             r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]

            r₃₄ = 4 m

we calculate

           W = 9 10⁹ (1.6 10⁻¹⁹)²  [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]

           W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

            W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]

in this case the charges have the same values ​​but charge 5 is positive and the others negative, so the products of the charges give a negative value

            W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]

we look for distances

            r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2

            r₁₅ = √ 2 = 1.4142 m

            r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]

            r₂₅ = √2 = 1.4142 m

            r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]

            r₃₅ = √5 = 2.236 m

            r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]

            r₄₅ = √13 = 3.606 m

we calculate

           W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]

            W = - 4.93 10⁻²⁸ J

Answer:B

Explanation:

C and O

Atoms of C and O pair of elements will form covalent bonds in a compound.Therefore the correct option is B.

The chemical compound is a combination of two or more  either similar or dissimilar chemical elements

for example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.

These chemical compounds are formed because of different types of bonds between the constituents elements ,the chemical bonds are mainly ionic bonds, covalent bonds,s, and hydrogen bonds.

Ionic bonds are formed due to the transfer of electrons between two bond forming pairs differentiated by their electronegativity.

Covalent bonds are formed by the sharing of electrons.Generally organic compound are formed as the reason of covalent bonds.

The carbon and oxygen atoms share their valence electrons to form a covalent bond, therefore the correct option is B.

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Answer:

Explanation:

Displacement is minimum distance between initial and final point .

Distance is total length of path covered in a journey .

a )

Tyler covered a longer distance in the journey because total length of path covered by him is longer due to curved path .

b )

Both have same displacement , because minimum distance between initial and final point in both the case is same .

c )

average speed = distance / time

as time is same for both the case ,

average speed ∝ distance

As distance covered by Tyler is more , his average  speed is more .

d )

average velocity = displacement / time

As both displacement and time are same in both the case , average velocity in both the case is same .

e )

They start at the same time and place at the bottom of the mountain and finish at the same time , both have tie and nobody won the race , in spite of speed of Tyler being greater  .

Answer:

Option 2 both take millions of years to form

Explanation:

Both  coal and ores take millions of years to form.

Ore is a naturally occurring rock or silt that has precious minerals in it that may be extracted, processed, and sold for a profit. These minerals are usually metals. Mining is the process of removing ore from the soil. The valuable metals or minerals are then removed by treating or refining the ore, frequently through smelting.

The concentration of the desired ingredient in an ore is referred to as its grade. To decide if a rock has a high enough grade to be worth mining and is therefore regarded as an ore, the value of the metals or minerals it contains must be evaluated against the expense of extraction.

Typically, oxides, sulphides, silicates, or native metals like copper or gold are the minerals of interest. To separate the valuable components from the waste rock, ore must be treated. Numerous geological processes collectively known as ore genesis are responsible for the formation of ore deposits.

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Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

Answer:

[tex]\boxed {\boxed {\sf 3.75 \ seconds }}[/tex]

Explanation:

Average acceleration is found by dividing the change in acceleration by the time.

[tex]a=\frac{ v_f-v_i}{t}[/tex]

The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:

[tex]a= -2.4 \ m/s^2 \\v_f= 0 \ m/s \\v_i= 9 \ m/s[/tex]

Substitute the values into the formula.

[tex]-2.4 \ m/s^2=\frac{0 \ m/s - 9 \ m/s}{t }[/tex]

Solve the numerator.

[tex]-2.4 \ m/s^2 = \frac{-9 \ m/s}{t}[/tex]

We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.

[tex]\frac{-2.4 \ m/s^2}{1} = \frac{-9 \ m/s}{t}[/tex]

[tex]-9 \ m/s *1= -2.4 \ m/s^2 *t[/tex]

[tex]-9 \ m/s=-2.4 m/s^2*t[/tex]

t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4

[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =\frac{ -2.4 \ m/s^2*t}{-2.4 \ m/s^2}[/tex]

[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =t[/tex]

[tex]3.75 \ s=t[/tex]

It takes 3.75 seconds.

Answer:

Explanation:

Total time taken = 0.85 s .

Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85

5.43 / 343 + time taken by coin to fall by 5.43 m = .85

time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s

Let the initial velocity of throw of coin = u

displacement of coin s  = 5.43 m

time take to fall t =  .834 s

s = ut + 1/2 gt²

5.43 = u x .834 + .5 x 9.8 x .834²

5.43 = u x .834 + 3.41

u x .834 = 2.02

u = 2.42 m /s .

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

Explanation:

Answer:

0.0119502868 kilocalorie

Explanation:

Answer:

increases by 50

Explanation: