A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket motor is ignited and the rocket burns fuel at a rate of 95 kg/s. The exit speed of the exhaust gas relative to the rocket is 2900 m/s. Neglecting drag and friction forces, determine the acceleration and the velocity of the car at time t = 10 s. Plot the acceleration and velocity from time t0 to t = 10 s.

Answer: \(5.77\ rps\)

Explanation:

Given

Initial angular velocity is \(\omega_i=8\ rps\)

rate of reduction \(\alpha=0.9 rev/s^2\)

after 17 revolution i.e. \(\theta =17\ rev\)

using \(\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta\)

Insert the values

\(\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps\)

Therefore, the magnetic force on an electron is 1.175 x 10⁻¹⁴N in the positive y-direction.

The force experienced by a moving charge in a magnetic field is given by the Lorentz force law. Since the charge on a proton is positive and that on an electron is negative, the direction of the magnetic force experienced by each is different.In this question, we need to find the magnitude and direction of the magnetic force on a proton and an electron as they travel from the Sun to the Earth.

Let's first calculate the magnetic force on a proton:

F = qvBsinθ

where q = charge of the particle

v = velocity of the particle

B = magnetic field strength

θ = angle between the velocity of the particle and the magnetic field = 90° (since the proton is moving perpendicular to the magnetic field)

Therefore,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= 1.175 x 10⁻¹⁴ N

Direction of magnetic force on a proton:The direction of the magnetic force on a proton can be found using the right-hand rule. According to this rule, if we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the negative y-direction.

Therefore, the magnetic force on a proton is 1.175 x 10^-14 N in the negative y-direction.

- Now, let's calculate the magnetic force on an electron:

Again using the Lorentz force law,

F = qvBsinθ

= (1.6 x 10⁻¹⁹) x (3.99 x 10⁵) x (2.93 x 10⁻⁸) x sin 90°

= -1.175 x 10⁻¹⁴ N (the negative sign indicates that the direction of the magnetic force is opposite to that of the proton)

Direction of magnetic force on an electron:Again using the right-hand rule, we can find the direction of the magnetic force on an electron. If we point the thumb of our right hand in the direction of the particle's velocity (in the positive x-direction) and the fingers in the direction of the magnetic field (in the positive z-direction), then the magnetic force will be perpendicular to both and will be in the positive y-direction.

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Since no work is done, the power exerted is zero. Therefore, the man exerts no power on the wall.

In physics, force is defined as any action that can change the motion of an object or cause an object to accelerate. Force is a vector quantity, meaning that it has both magnitude (size or strength) and direction. The unit of force in the International System of Units (SI) is the Newton (N), which is defined as the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared (1 N = 1 kg × 1 m/s^2). Force can be measured using a variety of instruments, such as spring scales, strain gauges, or force plates. Some common types of forces include gravitational force, electromagnetic force, frictional force, and normal force. The study of forces and their effects on the motion of objects is known as mechanics and is a fundamental concept in physics.

Here,

a. The man does not do any work on the wall because the wall does not move. Work is only done when there is a displacement in the direction of the force applied.

b. Since no work is done, no energy is used or transferred.

c. The power exerted by the man can be calculated using the formula:

Power = Work / Time

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Walking, running, and breathing are examples of motion. For rolling of ball, the street is a frame of reference and an insect on the ball is an inappropriate frame of reference.

A frame of reference, also known as a reference frame, is a perspective used in physics to assess whether an object is moving.  A frame of reference is an environment or object that is thought to be stationary. The Earth itself, despite its motion, serves as the most frequently used frame of reference.

Some examples of motion are:

Motion is a part of many of our daily activities, including walking, running, closing doors, etc.

Another illustration of motion is the passage of air into and out of our lungs.

The vehicles that transport passengers between the point of pickup and the destination have motion.

A ball is rolling down the street that it is moving.

For a ball rolling down the street, the street is an appropriate frame of reference as it is rest as compared to the ball.

For the same example, the insect sitting on the rolling ball is a part of the rolling ball.

Therefore, the insect is an inappropriate frame of reference for the motion of the ball rolling down the street.

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Answer:

Explanation: An astronaut on Jupiter will feel much heavier than on Earth due to the higher force of gravity. If the astronaut weighs 100 kg on Earth, he would weigh approximately 300 kg on Jupiter. This increased force of gravity can make it difficult to move around and perform physical tasks.

Answer: put them in the microwave

Explanation:

The work input to pump is 207.8 J

Heat is the degree of hotness or coldness of a body.

Work is the product of force and distance int he direction of the force.

To find the work input to the pump, we use the formula for the efficiency of the Carnot engine, we have

W/Q = 1 - (T/T') where

Making W subject of the formula, we have

W = Q[1 - (T/T')]

Given that

Substituting the values of the variables into the equation, we have

W = Q[1 - (T/T')]

W = 2.00 kJ[1 - (263 K/293.5 K)]

W = 2.00 kJ[1 - (263/293.5)]

W = 2.00 kJ[1 - 0.8961]

W = 2.00 kJ[0.1039]

W = 0.2078 kJ

W = 207.8 J

So, the work input is 207.8 J

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Differentiation

Derivative Rule [Basic Power Rule]:

\(\displaystyle\begin{aligned}f(x) & = cx^n \\f'(x) & = c \cdot nx^{n - 1} \\\end{aligned}\)

Derivative Rule [Chain Rule]:
\(\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)\)

Let's define what the problem gives us:

We know from our trigonometric derivatives that the derivative of \(\displaystyle \sin x\) is equal to \(\displaystyle \cos x\). However, since we have some arbitrary constant \(\displaystyle a\) multiplying \(\displaystyle x\) inside our \(\displaystyle \sin x\) function, we will have to apply the derivative rule of Chain Rule:

\(\displaystyle\begin{aligned}y & = \sin ax \\y' & = \boxed{ \cos (ax) (ax)' } \\\end{aligned}\)

To further simply the derivative, we now apply the derivative rule of Basic Power Rule and simplify:

\(\displaystyle\begin{aligned}y & = \sin ax \\y' & = \cos (ax) (ax)' \\& = (ax)' \cos ax \\& = 1 \cdot ax^{1-1} \cos ax \\& = ax^0 \cos ax \\& = \boxed{ a \cos ax } \\\end{aligned}\)

∴ the derivative of the function \(\displaystyle y = \sin ax\) is equal to \(\displaystyle \boxed{ y' = a \cos ax }\).

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Topic: Calculus

Unit: Differentiation

The force needed to accelerate a mass of 40 kg from velocity v₁ = (4î - 5) + 3k)m/s to v = (8î + 3) - 5k)m/s in 10 seconds is 12.4 N.

Start by calculating the change in velocity (Δv) experienced by the object. This can be done by subtracting the initial velocity v₁ from the final velocity v.

Δv = v - v₁ = ((8î + 3) - 5k) - ((4î - 5) + 3k)

= 8î + 3 - 5k - 4î + 5 - 3k

= 4î - 8k + 8

Next, calculate the acceleration (a) using the formula:

a = Δv / t

where t is the time interval, given as 10 seconds.

a = (4î - 8k + 8) / 10

= (0.4î - 0.8k + 0.8) m/s²

The force (F) required to accelerate the object can be found using Newton's second law:

F = m * a

where m is the mass, given as 40 kg.

F = (40 kg) * (0.4î - 0.8k + 0.8) m/s²

= (16î - 32k + 32) N

Simplify the expression to obtain the final answer:

F = 16î - 32k + 32 N

≈ 12.4 N

Therefore, the force needed to accelerate a mass of 40 kg from velocity v₁ = (4î - 5) + 3k)m/s to v = (8î + 3) - 5k)m/s in 10 seconds is approximately 12.4 N.

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To solve this problem, we can use the principles of Newton's laws of motion and the conservation of energy.

At the moment of release, the second block will start to accelerate downwards due to gravity, and the first block will start to move to the right due to the tension in the rope. Since the surface is frictionless, there is no horizontal force acting on the first block once it starts moving.

Using the free-body diagrams for the two blocks, we can write the following equations of motion:

For the second block:

m2g - T = m2a

where g is the acceleration due to gravity, T is the tension in the rope, a is the acceleration of the second block, and m2 is the mass of the second block.

For the first block:

T = m1a

where m1 is the mass of the first block and a is its acceleration.

Since the two blocks are connected by a rope, they must have the same acceleration, so we can set the two equations for acceleration equal to each other:

m2g - T = m1a

T = m1a

m2g - m1a = T = m1a

Solving for a, we get:

a = (m2/m1 + m2)g

We can also use the conservation of energy to find the final velocity of the first block after 1.2 seconds. At the moment of release, the total mechanical energy of the system is given by:

E = m1gh

where h is the initial height of the second block. As the blocks move, the potential energy of the second block is converted into the kinetic energy of both blocks. At the end of the 1.2 seconds, all of the potential energy will be converted into kinetic energy, so we can write:

E = (1/2)m1v^2 + (1/2)m2v^2

where v is the final velocity of the first block.

Solving for v, we get:

v = sqrt(2gh(m1+m2)/m1)

Plugging in the given values, we get:

a = (2/5)g ≈ 3.92 m/s^2

v = sqrt(2gh(m1+m2)/m1) ≈ 2.36 m/s

Therefore, the displacement of the velocity of the first block 1.2 s after the release of the blocks is approximate:

vt + (1/2)at^2 = 2.361.2 + (1/2)3.92(1.2)^2 ≈ 5.52 m/s

So the answer is not given in the options.

Answer:

Force

Explanation:

Displacement of the airplane (s) = 2000 m

Initial velocity of the airplane (u) = 0 m/s (Starts from rest)

Final velocity of the airplane = 56.4 m/s

Equation used to solve the problem:

\( \boxed{ \bf{ {v}^{2} = {u}^{2} + 2as}}\)

By substituting values in the equation, we get:

\( \rm \longrightarrow {56.4}^{2} = {0}^{2} + 2 \times a \times 2000 \\ \\ \rm \longrightarrow 3180.96 = 0 + 4000a \\ \\ \rm \longrightarrow 4000a = 3180.96 \\ \\ \rm \longrightarrow \dfrac{4000a}{4000} = \dfrac{3180.96}{4000} \\ \\ \rm \longrightarrow a = 0.80 \: m {s}^{ - 2} \)

\( \therefore \) Minimum uniform acceleration necessary for the plane to take flight (a) = 0.80 m/s²

A 66.1-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.59 m, and ends with a speed of 8.51 m/s. The nonconservative work done on the boy is approximately -42.7 kilojoules.

To find the nonconservative work done on the boy, we need to consider the change in the boy's mechanical energy during the process. Mechanical energy is the sum of the boy's kinetic energy (KE) and gravitational potential energy (PE).

The initial mechanical energy of the boy is given by the sum of his kinetic energy and potential energy when he catches the wave:

E_initial = KE_initial + PE_initial

The final mechanical energy of the boy is given by the sum of his kinetic energy and potential energy after he drops through the height:

E_final = KE_final + PE_final

The nonconservative work done on the boy is equal to the change in mechanical energy:

Work_nonconservative = E_final - E_initial

Let's calculate each term:

KE_initial = (1/2) * m * v_initial^2

= (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = m * g * h_initial

= 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * m * v_final^2

= (1/2) * 66.1 kg * (8.51 m/s)^2

PE_final = m * g * h_final

= 66.1 kg * 9.8 m/s^2 * 0

Since the boy ends at ground level, the final potential energy is zero.

Substituting the values into the equation for nonconservative work:

Work_nonconservative = (KE_final + PE_final) - (KE_initial + PE_initial)

Simplifying:

Work_nonconservative = KE_final - KE_initial - PE_initial

Calculating the values:

KE_initial = (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * 66.1 kg * (8.51 m/s)^2

Substituting the values:

Work_nonconservative = [(1/2) * 66.1 kg * (8.51 m/s)^2] - [(1/2) * 66.1 kg * (1.60 m/s)^2 - 66.1 kg * 9.8 m/s^2 * 1.59 m]

Calculating the result:

Work_nonconservative ≈ -42.7 kJ

Therefore, the nonconservative work done on the boy is approximately -42.7 kilojoules. The negative sign indicates that work is done on the boy, meaning that energy is transferred away from the boy during the process.

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Answer:

It is Conductivity because it is the measure of the ease.

Answer:

The value is \(w= 2*10^{-3} \ rad/s\)

Explanation:

From the question we are told that

  The diameter of the storm is  \(d = 10.0 \ km = 10 000 \ m\)

   The wind speed is  \(v = 10.0 \ m/s\)

Generally the radius is mathematically represented as

       \(r = \frac{d}{2}\)

=>  \(r = \frac{10000}{2}\)

=>    \(r = 5000 \ m\)

Generally the angular velocity is mathematically represented as  

        \(w= \frac{v}{r}\)

=>     \(w= \frac{10}{5000}\)

=>     \(w= 2*10^{-3} \ rad/s\)

The magnitude of the force P provided that the stress in the part AB is two times that of BC part is 0.8 kN.

The magnitude of the force P provided that the stress in the part AB is two times that of BC part is calculated as follows;

Take moment about the joint to determine the magnitude of the force along part BC.

120 kN x 750 mm  =  F x 1000 mm

F = ( 120 kN x 750 mm ) / ( 1000 mm )

F = 90 kN

Stress is given as force divided by area. The following equation can be used to determine the magnitude of force P.

Stress in AB = 2 times stress in BC

P/A₁  = 2F/A₂

where;

A₁ =  πd²/4  =  π(50 x 10⁻³)²/4

A₁ = 1.96 x 10⁻⁵ m²

A₂ = πd²/4  =  π(75 x 10⁻³)²/4

A₂ = 4.42 x 10⁻³ m²

P/A₁ = 2F/A₂

P = (2F x A₁) / (A₂)

P = (2 x 90 kN x 1.96 x 10⁻⁵ m² ) / ( 4.42 x 10⁻³ m² )

P = (2 x 90,000 N x 1.96 x 10⁻⁵ m² ) / ( 4.42 x 10⁻³ m² )

P = 798.2 N

P = 0.798 kN

P ≈ 0.8 kN

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Answer:

Electric charge is a fundamental property of matter and causes all static electrical phenomenon

Answer:

\(\theta_c = 36.78^o\)

Explanation:

The relationship between the refractive index and the critical angle is given as follows:

\(\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )\)

where,

η = refractive index = 1.67

θc = critical angle =?

Therefore,

\(\theta_c = Sin^{-1}(\frac{1}{1.67} )\)

\(\theta_c = 36.78^o\)

Answer:

which sport are you referring to?

Explanation:

Answer:

Because as we climb higher the amount of oxygen in the air decreases and this makes the air thinner and dryer. And at higher altitudes the blood pressure inside of our body is more than the atmospheric pressure which forces the blood to come out from openings like the nose.

Explanation:

Answer:

ALT TO FRIEND

HI FIREND

COPY BOB IN ALT ACC

CANT GET BAN

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Explanation:

Answer:

T = 1733.16 s = 28.88 min

Explanation:

The orbital velocity of a satellite about Earth is given as follows:

\(v = \sqrt{\frac{GM}{R}}\)

where,

v = orbital speed = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

M = Mass of Earth = 5.982 x 10²⁴ kg

R = Orbit Radius = 3.117 x 10⁶ m

Therefore,

\(v = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(5.982\ x\ 10^{24}\ kg)}{(3.117\ x\ 10^{6}\ m)}}\\\\v = 11.3\ x\ 10^{3}\ m/s\)

but the velocity is given as:

\(v = \frac{distance}{time}\)

for distance = circumference = 2πR

time = time period = T = ?

Therefore,

\(11.3\ x\ 10^{3}\ m/s = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{T}\\\\T = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{11.3\ x\ 10^{3}\ m/s}\\\\\)

T = 1733.16 s = 28.88 min

Answer:

a has a faster period

Explanation:

Period only depends on the length, irrelevant to the mass.

a has a shorter length, so has a small period, or faster.

Answer:

The focal length of the mirror is 10.4 cm.

Explanation:

So substitute the values of object distance and image distance in the mirror equation,

1/f = 1/(-18cm) + 1/(-25.0cm)

1/f = -25cm/(-18cm x -25cm) - 18cm/(-18cm x 25cm)

1/f = ( -25cm - 18cm)/(18cm x 25cm)

1/f = -43.0/450.0

f = -10.4651 cm.

The focal length of the mirror is approximately -10.4 cm.

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While certain types of glass containers may be suitable for short-term storage of vinegar, it is generally recommended to use non-reactive materials, such as plastic or stainless steel, for long-term storage to avoid any potential chemical reactions or corrosion.

Vinegar is an acidic liquid that contains acetic acid. When vinegar comes into contact with certain types of glass, particularly those made of lead or other reactive materials, a chemical reaction can occur. This reaction can lead to the leaching of potentially harmful substances into the vinegar.

Glass containers made from specific types of glass, such as soda-lime glass, are generally safe for storing vinegar.

However, it is important to note that prolonged storage or exposure to vinegar can still cause the glass to corrode over time. This can result in the deterioration of the glass container, potentially leading to breakage or the release of glass fragments.

To avoid any potential issues, it is recommended to use containers made of non-reactive materials, such as food-grade plastic or stainless steel, for long-term storage of vinegar. These materials do not react with the acidic nature of vinegar and do not pose a risk of leaching harmful substances.

Additionally, it is important to store vinegar in a cool, dark place to maintain its quality and prevent spoilage. Exposure to light and heat can degrade the quality of vinegar over time.

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When materials vibrate, waves are created that travel through the substance, and this energy is what we hear as sound. Earthquakes are earth vibrations that cause the (potential) energy held within rocks to be released (as a result of their pressure-generating relative positions). Seismic waves are produced by earthquakes.

How do sound waves and earthquakes compare?

The waves lose energy as they move through the air with sound or through the ground with shaking during an earthquake. Therefore, a band can be heard louder close to the stage than farther away, and an earthquake can be felt more strongly close to the fault than farther away.

In actuality, sound in the air cannot match how quickly earthquake waves move. In rock, the compressional or "P" wave of an earthquake moves at the In actuality, sound in the air cannot match how quickly earthquake waves move. The speed of a P wave is typically 10,000 mph. The speed of sound through air is roughly 750 mph.

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Answer:

1 km

Explanation:

displacement =velocity ×time

displacement =40m/s ×25s

displacement =1000m equivalent to 1km