Plot the velocity vs. time and the position vs. time for a car that travels at 20 m/s for 20 seconds, then accelerates in 10 seconds to 30 m/s, travels at this speed for 20 seconds, and the brakes and comes to rest in 10 s.

Answer:

The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Explanation:

Given that,

Average pressure [tex]P=1.01\times10^{5}\ Pa[/tex]

Radius of earth [tex]R_{E}=6.38\times10^{6}\ m[/tex]

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

[tex]P=\dfrac{F}{A}[/tex]  

[tex]F=PA[/tex]

[tex]F=P\times 4\pi\times R_{E}^2[/tex]

Where, P = pressure

A = area

Put the value into the formula

[tex]F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2[/tex]

[tex]F=516.6\times10^{17}\ N[/tex]

Hence, The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Answer:

5.164 x 10^19 N

Explanation:

P = 1.01 x 10^5 Pa

R = 6.38 x 10^6 m

Area = 4 π R²

[tex]A= 4\times 3.14\times6.38\times10^{6}\times6.38\times10^{6}[/tex]

A = 5.112 x 10^14 m^2

Pressure is defined as the force exerted per unit area.

The formula for the pressure is

P = F / A

Where, F is the force, A be the area

here force is the weight of atmosphere.

F = P x A  

F = 1.01 x 10^5 x 5.112 x 10^14

F = 5.164 x 10^19 N

Answer:

[tex]t=4.06s[/tex]

Explanation:

From the exercise we know

[tex]y_{o}=1.2m\\v_{o}=20m/s\\y=1.6m\\g=-9.8m/s^2[/tex]

To find how long does it takes the marker to reach the highest point we need to use the equation of position:

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]1.6m=1.2m+(20m/s)t-\frac{1}{2}(9.8m/s^2)t^2[/tex]

[tex]0=-0.4+20t-4.9t^2[/tex]

Now, we need to use the quadratic formula:

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-4.9\\b=20\\c=-0.4[/tex]

Solving for t

[tex]t=0.020s[/tex] or [tex]t=4.06s[/tex]

So, the answer is t=4.06s because the other option is almost 0 and doesn't make any sense for the motion of the marker

Answer:

a) Linear equation

Explanation:

Definition of acceleration

[tex]a=\frac{dv}{dt}\\[/tex]

if a=constant and we integrate the last equation

[tex]v(t)=v_{o}+a*t[/tex]

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

Answer:

2.551 m/s

Explanation:

Given:

Boulder is dropped from the cliff

thus,

Initial speed of the boulder, u = 0 m/s

Final speed to be obtained, v = 90 km/h =[tex]90\times\frac{5}{18}[/tex] =25 m/s

also, in the case of free fall the acceleration of the boulder will be equal to the acceleration due to the gravity i.e g = 9.8 m/s²

Now, from the Newton's equation of motion

v = u + at

where, a is the acceleration = g = 9.8 m/s²

t is the time

on substituting the respective values, we get

25 = 0 + 9.8 × t

or

t = [tex]\frac{\textup{25}}{\textup{9.8}}[/tex]

or

t = 2.551 m/s

Answer:

Part 1) Number of electrons in 1 liter of water equals[tex]N=3.346\times 10^{26}[/tex]

Part 2) Net charge of all the electrons equals [tex]Charge=53.61\times 10^{6}[/tex]

Explanation:

Since we know that the density of water is 1 kilogram per liter thus we infer that mass of 1 liter of water is 1 kilogram hence we need to find electron's in 1 kg of water.

Now since it is given that molar mass of water is 18.0 grams this means that 1 mole of water contains 18 grams of water.

Hence by ratio and proportion number of moles in 1 kg water equals

[tex]n=\frac{1000}{18}[/tex]

Now by definition of mole we know that 1 mole of any substance is Avagadro Number of particles.

Hence the no of molecules in 'n' moles of water equals

[tex]n'=\frac{1000}{18}\cdot N_a\\\\n'=\frac{1000}{18}\cdot 6.023\times 10^{23}\\[\tex][tex]\\n'=3.346\times 10^{25}[/tex]

Now since it is given that each molecule has 10 electron's thus the total number of electrons in n' molecules equals

[tex]N=10\times 3.346\times 10^{25}\\\\N=3.346\times 10^{26}[/tex]

Part 2)

We know that charge of 1 electron equals [tex]1.602\times 10^{-19}C[/tex] the the charge of electrons in 'N' quantity equals

[tex]Charge=1.6022\times 10^{-19}\times 3.346\times 10^{26}\\\\Charge=52.61\times 10^{6}Columbs[/tex]  

The statement that heuristics allow a priori theoretical concepts is true. Heuristics are mental shortcuts that can definitely be informed by a priori theoretical concepts, while empiricism emphasizes the role of experience in the formation of ideas.

Regarding the question on the statements about heuristics and empiricism, the best choice would be 'A'. Heuristics allow a priori theoretical concepts'. Heuristics are mental shortcuts or rules of thumb that simplify decisions, particularly under conditions of uncertainty. They are not necessarily based on theoretical concepts but rather on practical, personal experience or common sense. However, they can certainly be informed by a priori theoretical concepts in the sense that our theoretical understanding can shape the rules of thumb we use.

On the other hand, empiricism is a philosophical system that emphasizes the role of experience, especially sensory perception, in the formation of ideas, while discounting a priori reasoning, intuitiveness, and innate ideas. Hence, options B, C, and D are incorrect. Option E is also incorrect as heuristics can definitely be based on a trial and error approach.

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Answer:

Answered

Explanation:

impacts of the physics of matter on aviation operations.

Thrust, drag and lift.

Thrust:

It is the force developed by airplane engines that cause it to pull forward. With the help of huge propellers of course attached to the wings.

Drag:

It is the resistive force on the plane caused by the friction between air and plane. Its magnitude depends upon surface area, speed and viscosity of the air.

Lift:

The drag produced is utilized such that one of its component acts opposite to the weight. This causes the plane to take flight and stay in air. Lift can be deduced using Bernoulli's principle.

Bernoulli's principle is equivalent to law of conservation of energy. Meaning it tries to keep the energy of a system constant. In doing so, it produces low pressure zone above the wing. Which causes a net upward force, lift.

Answer:

[tex]10942249.24 m^{-1}[/tex]

Explanation:

Rydberg's formula is used to describe the wavelengths of the spectral lines of chemical elements similar to hydrogen, that is, with only one electron being affected by the effective nuclear charge. In this formula we can find the rydberg constant, knowing the wavelength emitted in the transcision between two energy states, we can have a value of the constant.

[tex]\frac{1}{\lambda}=Z^2R(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})[/tex]

Where [tex]\lambda[/tex] it is the wavelength of the light emitted, R is the Rydberg constant, Z is the atomic number  of the element and [tex]n_{1} n_{2}[/tex] are the states where [tex]n_{1}<n_{2}[/tex].

In this case we have Z=1 for hydrogen, solving for R:

[tex]R=\frac{1}{\lambda}*(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})^{-1}\\R=\frac{1}{658.9*10^{-9}m}*(\frac{1}{2^2}-\frac{1}{3^2}})^{-1}\\R=1.52*10^6m^{-1}*(\frac{36}{5})=1.09*10^7 m^{-1}=10942249.24m^{-1}[/tex]

This value is quite close to the theoretical value of the constant [tex]R=10967758.34 m^{-1}[/tex]

Answer:

Power factor of the AC series circuit is [tex]cos\phi=0.5[/tex]

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

[tex]cos\phi=\dfrac{R}{Z}[/tex]

[tex]cos\phi=\dfrac{30}{60}[/tex]

[tex]cos\phi=\dfrac{1}{2}[/tex]

[tex]cos\phi=0.5[/tex]

So, the power factor of the ac series circuit is [tex]cos\phi=0.5[/tex]. Hence, this is the required solution.

The power factor of an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms is 0.5, indicating that half of the power is effectively used.

The power factor in an AC circuit is a measure of how much of the power is being effectively used to do work. It's calculated as the ratio of the resistance R to the impedance Z. Given an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms, the power factor is the resistance divided by the impedance.

The calculation is straightforward: Power factor = R / Z = 30 Ohms / 60 Ohms = 0.5.

This means that the power factor for this particular AC circuit is 0.5, which indicates that only half of the total power is being used effectively while the other half is reactive power, which does not perform any work but is necessary for the functioning of the circuit's reactive components.

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The minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

To stop the automobile before it hits the barrier, the minimum coefficient of kinetic friction needed can be found using the equation F = μk Mcg. The force of friction is equal to the product of the coefficient of kinetic friction (μk), the mass of the car (Mc), and the acceleration due to gravity (g). Since the car is stopping, its acceleration is negative, equal in magnitude to the square of its velocity divided by twice the stopping distance.

Using the given information, we can solve for the coefficient of kinetic friction:

F = μk Mcg

μk Mcg = (Mc)((-v^2)/(2d))

Substituting the values, we get:

μk = ((-v^2)/(2d))/g

where v = 20 m/s and d = 50 m.

Substituting the values in the equation, we get:

μk = ((-20^2)/(2*50))/9.8 ≈ -0.2041 ≈ 0.2041

Therefore, the minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

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Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m[/tex]

Second rocket

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m[/tex]

When this will collide the total distance they would have covered would be 6000 m.

[tex]6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0[/tex]

[tex]t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72[/tex]

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

[tex]s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m[/tex]

The second rocket would have gone 3469.788 m when they collide

Final answer:

To find the distance covered by the second rocket at the point of collision, we analyze both rockets' motion based on their initial velocities and accelerations. By using the equations of motion, we can solve for the time of collision and calculate the distance each rocket has traveled, specifically looking for the second rocket's distance.

Explanation:

To find how far the second rocket has gone when they collide, we must analyze the motion of both rockets taking into account their initial velocities and accelerations. The first rocket has an initial velocity of 150 m/s and accelerates at 5 m/s2, while the second rocket starts with the same velocity but accelerates at 15 m/s2. Both rockets start 6000 meters apart. To find the point of collision, we use the equation of motion s = ut + (1/2)at2 for both, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time until collision.

Let's denote the distance covered by the first rocket as s1 and the second as s2, with the total distance being s1 + s2 = 6000 m. By finding the time of collision and plugging it back into the equation of motion for the second rocket, we can find s2, the distance covered by the second rocket. The actual calculation involves setting up equations based on the motion of both rockets and solving for the time t, after which s2 can be calculated.

Answer:

Frequency will be 4.30 kHz

Explanation:

We have inductance [tex]L=37\mu H=37\times 10^{-6}H[/tex]

And capacitance [tex]C=37\mu H=37\times 10^{-6}F[/tex]

Inductive reactance is given by [tex]X_L=\omega L[/tex]

And capacitive reactance [tex]X_C=\frac{1}{\omega C}[/tex]

As in question it is given that inductive reactance and capacitive reactance is same

So [tex]X_L=X_C[/tex]

[tex]\omega L=\frac{1}{\omega C}[/tex]

[tex]\omega ^2=\sqrt{\frac{1}{LC}}=\frac{1}{\sqrt{37\times 10^{-6}\times 37\times 10^{-6}}}=27027.027rad/sec[/tex]

We know that angular frequency [tex]\omega =2\pi f[/tex]

[tex]2\times 3.14\times f=27027.027[/tex]

f = 4303.66 Hz =4.30 kHz

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = [tex]P= \frac{RT}{\sqrt{2}N_A\sigma\lambda }[/tex]

putting values we get

=[tex] \frac{8.314\times400}{\sqrt{2}\times6.022\times10^{23}\times0.28\times10^{-18}\times5\times10^{-6} }[/tex]

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Answer:

213 ft/s

Explanation:

Given:

Δx = 330 ft

v₀ = 0 ft/s

t = 3.1 s

Find: Δv

x = x₀ + ½ (v + v₀)t

330 ft = ½ (v + 0 ft/s) (3.1 s)

v = 213 ft/s

The change in velocity is:

Δv = 213 ft/s − 0 ft/s

Δv = 213 ft/s

Answer:

Explanation:

Given

mass of each arrow=0.1 kg

velocity of arrow=30 m/s

One arrow is fired u=due to east and another towards south

Momentum of first arrow

[tex]P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}[/tex]

[tex]P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}[/tex]

Total momentum P

[tex]P=P_1+P_2[/tex]

[tex]P=3\hat{i}-3\hat{j}[/tex]

magnitude [tex]|P|=\sqrt{3^2+3^2}=3\sqrt{2}[/tex]

direction

[tex]tan\theta =\frac{-3}{3}[/tex]

[tex]\theta =45^{\circ}[/tex] clockwise w.r.t to east

The magnitude of the total momentum of the two-arrow system is approximately 4.24 kg·m/s, and the direction is 45° south of due east.

The question involves calculating the total momentum of a system comprising two arrows fired in perpendicular directions. Momentum is a vector quantity, meaning it has both magnitude and direction. Since the two arrows have the same speed but are fired in perpendicular directions, their momenta are also perpendicular to each other. The momentum of each arrow can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.

The momentum of the first arrow (due east) is peast = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s east, and the momentum of the second arrow (due south) is psouth = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s south. To find the total momentum of the system, we need to find the resultant of these two momentum vectors.

The magnitude of the total momentum is found using the Pythagorean theorem: ptotal = √(peast² + psouth²) = √(3.0² + 3.0²) kg·m/s = √(9 + 9) kg·m/s = √18 kg·m/s ≈ 4.24 kg·m/s.

The direction of the total momentum with respect to due east can be found using trigonometry, specifically tan-1 (psouth/peast), which gives us 45° south of east, since the magnitudes are equal.

Answer:

Explanation:

We know that the differential work made by the gas  its defined as:

[tex]dW =  P \ dv[/tex]

We can solve this by integration:

[tex]\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv[/tex]

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

[tex]P \ V = \ n \ R \ T[/tex]

[tex]P = \frac{\ n \ R \ T}{V}[/tex]

This give us

[tex] \int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv [/tex]

As n, R and T are constants

[tex] \int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv [/tex]

[tex] \Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1} [/tex]

[tex] \Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 ) [/tex]

[tex] \Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 ) [/tex]

[tex] \Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})[/tex]

But the volume is:

[tex]V = \frac{\ n \ R \ T}{P}[/tex]

[tex] \Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )[/tex]

[tex] \Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})[/tex]

Now, lets use the value from the problem.

The temperature its:

[tex]T = 27 \° C = 300.15 \ K[/tex]

The ideal gas constant:

[tex]R = 8.314 \frac{m^3 \ Pa}{K \ mol}[/tex]

So:

[tex] \Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm}) [/tex]

[tex] \Delta W = 7475.69 joules[/tex]

We know that, for an ideal gas, the energy is:

[tex]E= c_v n R T[/tex]

where [tex]c_v[/tex] its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

[tex]\Delta E = \Delta Q - \Delta W[/tex]

where [tex]\Delta W[/tex] is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

[tex]\Delta E = 0[/tex]

[tex]\Delta Q = \Delta W[/tex]

Answer:

[tex]t=6.19s[/tex]

Explanation:

With a initial velocity is zero, kinematics equation is:

[tex]y=1/2*g*t^2[/tex]

[tex]t=\sqrt{2y/g}=\sqrt{2*188/9.81}=6.19s[/tex]

Answer:

[tex]\frac{dv_{r}}{dt}=g-k_{r}v_{r}^2[/tex]

Explanation:

Second Newton's Law:

[tex]F=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt} \\[/tex]    (1)

m and v are the instantaneous mass and instantaneous velocity

The only force is the weight:

[tex]F=mg[/tex]          (2)

On the other hand we know:

[tex]\frac{dm}{dt}=k*m*v[/tex]        (3)

We replace (2) and (3) in (1), and we solve for dv/dt :

[tex]\frac{dv}{dt}=g-kv^2[/tex]

Answer:

This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:

[tex]a=\frac{F}{m}[/tex]

Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.

The situation is impossible because the gravitational force between equally charged particles is much weaker than the electric force, making it impossible for the two forces to have the same magnitude, resulting in non-zero net force and movement.

The scenario described by the student is impossible because the gravitational force and electric force between two particles with identical mass and charge are not equal in magnitude. According to physics, specifically Coulomb's Law and Newton's Law of Universal Gravitation, the gravitational force is significantly weaker than electric force when the particles have the same magnitude of charge as that of their mass in standard units.

The electric force (Coulomb's Law) is known to be much stronger than gravitational force. For example, the electric repulsion between two electrons is approximately 1042 times stronger than their gravitational attraction. Therefore, if two dust particles are floating in empty space and are at rest with respect to each other, with identical charges and mass, the electric forces would cause them to repel each other with much greater force than the gravitational forces would attract them. Hence, they cannot remain at a constant distance with zero net force on each other.

Answer:

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

Answer:

Explanation:

The rock will decelerate at the rate of g or 9.8 m s⁻²while on its way up because of gravitational attraction towards the earth.  . In other words , it will accelerate with the magnitude - 9.8 m s⁻² while on its way up.

On its way down , it will accelerate due to gravitational attraction. In this case acceleration will be positive and equal to 9.8 m s⁻².

At the highest point in its trajectory , rock  becomes stationary or comes to  rest momentarily. So at the peak position , velocity is zero.

At the highest point , as the gravitational force continues to act on the body undiminished, its acceleration will remain the same ie 9.8 m s⁻² at the highest point as well. It will act in the downward direction.

The distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

The distance traveled by the child on the toboggan can be calculated using the equations of motion for uniformly accelerated motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is given by:

[tex]d = ut + (1/2)at^2[/tex]

In this case, the child starts at rest [tex](u = 0)[/tex] and has an acceleration of a[tex]= 1.8 m/s^2[/tex].

Calculate the distances for the given times:

(a) For t = 1.0 s:

[tex]d = (0) \times (1.0 ) + (1/2) \times (1.8 ) \times (1.0 )^2\\d = 0 + 0.9 \\d = 0.9\ m[/tex]

(b) For t = 2.0 s:

[tex]d = (0) \times (2.0 ) + (1/2) \times (1.8 ) \times (2.0 )^2\\d = 0 + 3.6\\d = 3.6\ m[/tex]

(c) For t = 3.0 s:

[tex]d = (0) \times (3.0 ) + (1/2) \times (1.8 ) \times (3.0)^2\\d = 0 + 8.1\\d = 8.1\ m[/tex]

So, the distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

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Final answer:

Using the kinematic equation for motion with constant acceleration, the distance traveled by the child on the toboggan is 0.9 m after 1.0 s, 3.6 m after 2.0 s, and 8.1 m after 3.0 s.

Explanation:

To solve for the distance traveled by the child on the toboggan in each case, we'll use the kinematic equation for motion with constant acceleration: d = v_i * t + \frac{1}{2} * a * t^2, where d is the distance traveled, v_i is the initial velocity, t is the time, and a is the acceleration.

Since the child starts at rest, the initial velocity v_i is 0 m/s. The acceleration a is given as 1.8 m/s^2.

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, [tex]d=24.44\times 2=48.88\ m[/tex]

(a) Acceleration, [tex]a=-4\ m/s^2[/tex]

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{-(24.44)^2}{2\times -4}[/tex]

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, [tex]a=-8\ m/s^2[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{-(24.44)^2}{2\times -8}[/tex]

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

Answer:

T = 450 C

Explanation:

given,

pressure of rigid tank containing water = 200 kPa

water start condensing at 100 °C

vapor saturation pressure at 100 °C = 101421 Pa (from saturation table)

specific volume  = 1.6718 m³/kg

From super heated steam table

steam table is two set of table which is of energy transfer properties of water and steam.

at v = 1.6718 m³/kg and pressure = 200 kPa

temperature is equal to T = 450 C

Answer:7 cm

Explanation:

Given

one sphere has a radius(r) of 4.85 cm

Also another sphere has a mass 3 times the previous one

since the density remains same therefore

second sphere has a volume 3 times of former.

Let R be the radius of bigger Sphere

[tex]\frac{4\pi R^3}{3}=3\times \frac{4\pi r^3}{3}[/tex]

[tex]R=\left ( 3\right )^\frac{1}{3}r=1.44\times 4.85=6.99 \approx 7 cm[/tex]

Answer:

Refractive index of slab = 1.22

Explanation:

Using Snell's law as:

[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]

Where,  

[tex]{\theta_i}[/tex]  is the angle of incidence  ( 27.0° )

[tex]{\theta_r}[/tex] is the angle of refraction  ( 21.8° )

[tex]{n_r}[/tex] is the refractive index of the refraction medium  (slab, n=?)

[tex]{n_i}[/tex] is the refractive index of the incidence medium (vacuum, n=1)

Hence,  

[tex]1\times {sin27.0^0}={n_r}\times{sin21.8^0}[/tex]

Refractive index of slab = 1.22

The index of refraction of the transparent material is found using Snell's law and is approximately 1.333.

To find the index of refraction of the transparent material, we can use Snell's law, which states the relationship between the angles of incidence and refraction at the boundary between two media with different refractive indices. Snell's law is given by the equation:

n1sin(θ1) = n2sin(θ2)

Where n1 is the refractive index of the first medium (vacuum in this case, which is 1), θ1 is the angle of incidence, n2 is the refractive index of the second medium (the transparent material), and θ2 is the angle of refraction.

Here, we have θ1 = 27.0° and θ2 = 21.8°. Plugging in the known values:

1 * sin(27.0°) = n2 * sin(21.8°)

n2 = sin(27.0°) / sin(21.8°)

Using a calculator, we find:

n2 ≈ 1.333

Thus, the index of refraction of the transparent material is approximately 1.333.

Answer:

(e) The particles move apart with a velocity that increases for a while and then becomes constant.

Explanation:

Each particle feels a repulsive (because they have same sign of charge) electric force from the each other:

[tex]F=\frac{kq_{1}q{2}}{d^{2}}[/tex]

and

[tex]F=ma\\[/tex]

So each particle feels a repulsive force proportional to the quadratic inverse of the distance.that means that the charges begin to move away, and the further away they are from each other, the force (and therefore the acceleration) decreases, at a rate inversely proportional to the square of the distance. Theoretically this acceleration will never be zero, but in practice it will at some point reach a value very close to zero. Then the speed will grow for a while and when the acceleration has reached almost zero, the speed will practically remain constant.

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant [tex]\gamma =1.4[/tex]

specific heat is given as [tex]=\frac{\gamma R}{\gamma -1}[/tex]

gas constant =287 J⋅kg−1⋅K−1

[tex]Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}[/tex]

specific heat at constant volume

[tex]Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}[/tex]

change in internal energy [tex]= Cv(T_2 -T_1)[/tex]

                            [tex]  \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}[/tex]

change in enthalapy [tex]\Delta H = Cp(T_2 -T_1)[/tex]

                                 [tex] \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}[/tex]

change in entropy

[tex]\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})[/tex]

[tex]\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})[/tex]

[tex]\Delta S = 382.79 J kg^{-1} K^{-1}[/tex]

Answer:

1.86 75 X 10⁻⁵ C.

Explanation:

Capacitance of parallel plate capacitor is given by

C = K∈₀ A /d

C is capacitance , A is plate area , d is distance between plate and ∈₀ is permittivity of air which is 8.85 x 10⁻¹²  and K is dielectric constant

Using the data given in the question

C = [tex]\frac{6.7\times 8.85\times10^{-12}\times 1.4}{.04\times10^{-3}}[/tex]

C = .2075 X 10⁻⁵ F

Charge Q = CV

.2075 X 10⁻⁵ X 9 = 1.86 75 X 10⁻⁵ C.

Final answer:

To achieve an overall magnification of 200 with the given measurements, the eyepiece lens would need to have a focal length of approximately 5.68 cm.

Explanation:

To calculate the eyepiece focal length that will give an overall angular magnification of 200 in a microscope, we can use the formula of magnification for a compound microscope. Given that the distance between the objective and eyepiece lenses is 19 cm and the objective lens has a focal length of 5.5 mm, we can set up the equation as follows:

M = (D/fo) × (25/fe)

Where:

In the given problem, M = 200 and fo = 5.5 mm (which we should convert to centimeters to keep consistent units, thus fo = 0.55 cm). We are solving for fe.

Using the formula, we substitute the values:

200 = (25/0.55) × (25/fe)

Solving for fe, we find:

fe = (25/0.55) × (25/200)

fe = (45.45) × (0.125)

fe = 5.68 cm

Therefore, the eyepiece lens would need to have a focal length of approximately 5.68 cm to achieve an overall magnification of 200.