Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?​

Answer:

Explanation:

(i) μs = F/N = mgsinθ/mgcosθ = tanθ

  tanθ = 0.3

  θ = 16.7°

(ii) a = F/m

    a = (mgsinθ - (μk)mgcosθ) / m

    a = g(sinθ - (μk)cosθ)

    a = 9.8(sin16.7 - (0.2)cos16.7)

    a = 0.94 m/s²

(iii) s = ½at²

     t = √(2s/a)

     t = √(2(1)/0.94)

     t = 1.5 s

Answer:

(i) 0.6s (ii) 8.42m

Explanation:

U² = 6.02² + 7²

U = 9.23

angle of projection

tanø = 6.02/7

ø = 40.7

Time of fligt

t = Usinø/g

t = 9.23 sin 40.7/10

t = 0.6

H range = U²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

The time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.

Velocity ​​is defined as the directional motion of an object which is indicated by the rate of change of position as observed from a particular frame of reference. It is measured by a particular standard of time. It is a vector quantity as it has both magnitude and direction.

It can be expressed as:

v= d/t

Where. v is the velocity in m/s

d is the displacement measured in meter 'm'

t is the time measured in seconds 's'

For above given information,

v = 6.02 +7.0j, so the initial velocity will be u

u² = 6.02² + 7²

u = 9.23m/s

Angle of projection, tanø = 6.02/7

ø = 40.7

Time taken=  t = u sinø/g

t = 9.23 sin 40.7/10

t = 0.6s

Horizontal range  = u²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

Thus, the time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.

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The increase in tension on the steel wire is 8,484.75 N.

The given parameters;

The area of the steel wire is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]

The extension of the steel wire is calculated as follows;

[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]

The increase in tension on the steel wire is calculated as follows;

[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]

Thus, the increase in tension on the steel wire is 8,484.75 N.

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Answer:

I think it's c but don't know for sure

Answer:

P1 = 1.3 (500 + 60) = 728 kg-m      total momentum to right at start

P2 = (v2 - 10) 60 + 500 v2

total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart

728 = 560 v2 - 600

v2 = 1328 / 560 = 2.37 m/s    new speed of cart

Check:

After:    p2 for cart = 500 * 2.37 = 1186

p1 for man = (2.37 - 10) * 60 = -458

P2 = p1 + p2 = 728       total momentum unchanged

Answer:

Explanation:

Given,

As we know,

Kinetic Energy,

Therefore ,

Kinetic energy of the object is,

[tex] = \frac{1}{2} \times 12 \: kg \times 10 \: m {s}^{ - 1} \times 10 \: m {s}^{ - 1} [/tex]

= (6 × 10 × 10) J [As 'J' stands for 'joule']

= 600 J (Ans)

Answer:

iIn a nuclear plant, the heat source is from the nuclear reaction whereas in a thermal power plant it is from the combustion of coal. The difference is in the inlet steam parameters to the turbine in a nuclear plant. Thermal power plants use steam at superheated conditions. ... The nuclear plant uses a 'wet steam turbine'.

Explanation:

The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.

Power plant is an industrial structure that generates electricity. The majority of power plants are linked to the electrical grid.

Nuclear power bare a form of thermal power plant. You have a reactor where fission takes place and heat is generated, a heat exchanger that transports this heat to where it is needed.

Thermal power plant equipment converts this heat into electric energy, usually via a steam turbine.

The reactor, heat exchanger, and thermal conversion technology all have different designs and technologies, but the overall architecture is quite similar to other types of thermal power plants.

The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.

Hence, the two types of power plants differ in  difference is in the turbine's input steam characteristics.

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Consult the attached free body diagram.

If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces

• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0

• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0

since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).

We only really need the first equation. Simplifying it, we get

F [archer] - T cos(θ) - T cos(θ) = 0

F [archer] - 2T cos(θ) = 0

F [archer] = 2T cos(θ)

cos(θ) = F [archer] / (2T)

We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say

T = 0.842 F [archer]

and from this we have

cos(θ) = F [archer] / (2 • 0.842 F [archer])

cos(θ) = 1/1.684

cos(θ) ≈ 0.593

Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.

Answer:

The answer is D

Explanation:

The chloroplast absorbs sunlight for energy not the cell wall

Answer:

Increases with increase in mass

Explanation:

gravity is proportional to mass and inversely proportional to the square of the distance between them

F = GMm/d²

In Physics, energy can be defined as the ability and capacity to do work by an object or physical body.

In Physics, energy can be defined as the ability to do work. Thus, energy must be possessed or transferred to a physical object (body) before it can be used in doing a work or heating a system.

Generally, there are two (2) main types of energy and these are;

For example, you require a sufficient amount of energy to move a crate of egg across a given distance.

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Answer:. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another. ... All forms of energy are associated with motion.

Explanation:. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another. ... All forms of energy are associated with motion.

The Rayleigh criterion allows finding the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

The Rayleigh criterion is used to find the separation of two points, it is based on the fact that the diffraction maximuum pattern of the first object coincides with the first minimum of the second object.

By entering in the diffraction ratio for slits you will find.

           sin θ  = [tex]\frac{\lambda}{a}[/tex]  

In general in diffraction experiments the angles are very small,

           [tex]tan \theta = \frac{y}{x} = \frac{sin \theta}{cos \theta} \\sin \theta = \frac{y}{x}[/tex]

For the case of circular apertures, when solving in polar coordinates, a constant appears.

        [tex]\frac{y}{x} = 1.22 \frac{\lambda}{D}[/tex]

       [tex]D = 1.22 \frac{\lambda \ x}{y}[/tex]

Where λ is the wavelength of light and D is the diameter of the aperture.

They indicate that the separation between the star and the planet is 1 AU and the distance from the system to the Earth is 3 parce.

Let's reduce the parce to astronomical units

       x = 3 pc (  [tex]\frac{206264 AU}{1 pc}[/tex] )

       x = 6.18 10⁵ AU

Let's calculate

          D = [tex]D = 1.22 \ \frac{550 \ 10^{-9 } \ 6.18 \ 10^5 }{1}[/tex]  

          D = 0.415  m

In conclusion, using the Rayleigh criterion we can find the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

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Answer:

B.-80.35 J

i dont know the calculation

Answer:

1

Explanation:

A=mgh=0.2*10*0.5=1 J

[tex]\boxed{\sf PE=mgh}[/tex]

[tex]\boxed{\sf KE=\dfrac{1}{2}mv^2}[/tex]

[tex]\boxed{\sf ME=KE+PE}[/tex]

#1

[tex]\\ \sf\longmapsto PE=60(0)(10)=0J[/tex]

[tex]\\ \sf\longmapsto KE=\dfrac{1}{2}(60)(8)^2=30(64)=1920J[/tex]

[tex]\\ \sf\longmapsto ME=1920+0=1920J[/tex]

#2

[tex]\\ \sf\longmapsto PE=60(10)(1)=600J[/tex]

[tex]\\ \sf\longmapsto KE=600J[/tex]

[tex]\\ \sf\longmapsto ME=1200J[/tex]

Now

[tex]\\ \sf\longmapsto \dfrac{1}{2}mv^2=600\implies 30v^2=600\implies v^2=20\implies v=4.2m/s[/tex]

Answer:

If the mass of geranium is 75 and the atomic number is 32 then it must have

N = 75 -32 = 43 neutrons

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass =  x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained  by glass + Heat gained by water

Heat gained by glass = x ×  0.18 × 20

Heat gained by water = 74  ×  1.0 × 20

Hence;

2000 =  (x ×  0.18 × 20) + ( 74  ×  1.0 × 20)

2000 - 1480 =  (x ×  0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.​

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Answer:

Acceleration due to gravity on a certain planet = 4 m/s²

Explanation:

According to the question,

Weight = 30 N

Mass = 7.5 kg

Let acceleration due to gravity be 'a'

Formula:

Weight = Mass × Acceleration due to gravity

30 = 7.5 × a

a = 30/7.5

a = 4 m/s²

100000 Pascal

Explanation:

pressure= force/area

Max pressure= force/min area

so f=5

min area= 5×10^-5

5÷5*10^-5 = 100000pascal

Hi there!

On a level road:

∑F = Ff (Force due to friction)

The net force is the centripetal force, so:

mv²/r = Ff

Rewrite the force due to friction:

mv²/r = μmg

Cancel out the mass:

v²/r = μg

Solve for v:

v = √rμg

v = √(25)(9.81)(0.8) = 14.01 m/s

Answer:

Vertical acceleration 9.8 m/s² downward

Horizontal acceleration 0.0 m/s²

assuming no air resistance.

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

Age of the universe; [tex]t = \ ?[/tex]

We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:

[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]

Now,

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

We know that;

[tex]1\ light\ years = 9.46*10^{15}m[/tex]

so

[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]

Therefore;

[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]

Now, we input this Hubble's constant value into our equation;

[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;

       x = 0.026 m

Given parameters

To find

The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. With maximum compression.

        Em₀ = U = ½ k x²

Final point. Where the kinetic and potential energy are equal.

        [tex]Em_f = K +U[/tex]  

Since the mechanical energy is constant at this point K = U, therefore we can write the energy.

        [tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]

Energy is conserved.

        [tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf

        ½ k x² = 2 (½ k xf²)

        [tex]x_f = \frac{x_o}{2}[/tex]  

let's calculate.

        [tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]  

In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;

       x = 0.026 m

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Answer:

A i think its right im postivite

Answer: A. Faults

Explanation: ;-; its easy because faults are cracks and they can push layers past each other making tectonic plates rub against each other then creating a earthquake

Answer:

Water Vapour, Carbon dioxide, methane,nitrous oxide,ozone.

Answer:

Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.

Answer:

a = 6.22 m//s²

Explanation:

F = ma

a = F/m

a = 573.0 / 92.1

a = 6.221498...

Answer:

[tex]\boxed {\boxed {\sf 6.22 \ m/s^2}}[/tex]

Explanation:

We are asked to find the acceleration due to gravity on another planet.

Weight is the measure of the force of gravity. Therefore, we can use the following version of the force formula:

[tex]F_g=mg[/tex]

In this formula, [tex]F_g[/tex] is the weight, m is the mass, and g is the acceleration due to gravity.

The object weights 573.0 Newtons (or 573.0 kg*m/s²) on the planet. The object has a mass of 92.1 kilograms.

Substitute these values into the formula.

[tex]573.0 \ kg*m/s^2 = 92.1 \ kg * g[/tex]

We are solving for g, so we must isolate the variable. It is being multiplied by 92.1 kilograms. The inverse of multiplication is division, so divide both sides of the equation by 92.1 kg.

[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}= \frac{92.1 \ kg*a}{92.1 \ kg}[/tex]

[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}=a[/tex]

The units of kilograms cancel.

[tex]6.22149837 \ m/s^2=a[/tex]

The original measurements of weight and mass have 4 and 3 significant figures. Our answer must have the least number of sig figs, or 3. For the number we found, that is the hundredth place. The 1 in the thousandths place tells us to leave the 2 in the hundredth place.

[tex]6.22 \ m/s^2=a[/tex]

The acceleration due ot gravity on planet Xyleneer is approximately 6.22 meters per second squared.

Answer:

sorry for you

The ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.

The relationship between the rotational frequency [tex](\(\omega\))[/tex] and moment of inertia (I) is given by the equation:

[tex]\[I_1\omega_1 = I_2\omega_2\][/tex]

where [tex]\(I_1\)[/tex]and [tex]\(I_2\)[/tex] are the moments of inertia in the two cases, and [tex]\(\omega_1\) and \(\omega_2\)[/tex] are the corresponding rotational frequencies.

Let's denote the moment of inertia in the first case (arms extended) as [tex]\(I_1\)[/tex] and in the second case (arms drawn in) as [tex]\(I_2\)[/tex]. The given rotational frequencies are [tex]\(\omega_1 = 0.25 \, \text{rev/s}\) and \(\omega_2 = 0.80 \, \text{rev/s}\)[/tex].

Using the equation [tex]\(I_1\omega_1 = I_2\omega_2\)[/tex], we can rearrange it to solve for the ratio of moments of inertia:

[tex]\[\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}\][/tex]

Substituting the given values, we have:

[tex]\[\frac{I_1}{I_2} = \frac{0.80 \, \text{rev/s}}{0.25 \, \text{rev/s}}\][/tex]

Simplifying the expression, we get:

[tex]\[\frac{I_1}{I_2} = 3.2\][/tex]

Therefore, the ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.

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