TIME REMAINING 54:08 If a cup of coffee is at 90°C and a person with a body temperature of 36°C touches it, how will heat flow between them? from the hand to the cup from the cup to the hand from the cup to the surrounding air from the hand to surrounding air

The bandwidth of the periodic composite signal is drawn as a range between 100 Hz and 2100 Hz , The data rates for sending a digital signal using one harmonic, three harmonics, and five harmonics on a TV channel with a 6 MHz bandwidth would be 6 MHz, 18 MHz, and 30 MHz .

For the first question

Draw the bandwidth of a periodic composite signal, we need to consider the highest frequency component present in the signal.

We have two sine waves one with a frequency of 100 Hz and the other unspecified. Since the bandwidth is given as 2000 Hz, we can assume that the second sine wave has a frequency of 2100 Hz (2000 Hz above the first sine wave frequency).

Draw the bandwidth, we can create a graph with frequency on the x-axis and amplitude on the y-axis.

We plot the amplitude values for the two sine waves at their respective frequencies (100 Hz and 2100 Hz). The bandwidth will be the range between these two frequencies on the x-axis.

For the second question

The data rate for a digital signal transmitted using one harmonic, three harmonics, and five harmonics can be calculated by multiplying the channel bandwidth by the number of harmonics used. Since the bandwidth is given as 6 MHz, the data rates would be as follows:

One harmonic: 6 MHz

Three harmonics: 18 MHz

Five harmonics: 30 MHz

The data rate increases with the number of harmonics used because each harmonic contributes additional information to the signal, allowing for a higher data transmission rate.

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Answer:

60 seconds

Explanation:

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The sun will become a red giant but it will not be a problem for humanity because the Earth will become inhabitable before that.

As per the research conducted it is said that the sun will eventually become a red giant and it engulf our planet Earth. The Earth will become extremely giant. This in turn will not become a problem for humanity, flora and fauna as the Earth will be completely inhabitable. All the forms of life will completely perish and no life form will exist. The planet will be engulfed by the red giant Sun. This is the reason it is not a problem for humanity.

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Answer:

A toaster because you plug it up for electric energy and heat to toast it

Without calibration, the data would no longer be accurate since it would deviate from real values. Therefore, calibration is occasionally required.

It converts an applied mechanical force—such as load, weight, tension, compression, or pressure—into another physical variable—in this case, an electrical output signal—that can be measured, converted, and standardized. The electrical signal changes in direct proportion to the force acting on the sensor.

A load cell or weight sensor is often referred to as a force sensor. They are employed to gauge load, strain, and compression. A lot of them contain internal strain gauges that are attached to the metal structure and respond even to the smallest compression changing the resistance and reporting on the outcomes.

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Answer:

encyclopedia most reliable I think

The correct statement about the half-lives of radioactive elements is: The half-lives of radioactive elements vary from fractions of a second to billions of years. Option D

The half-life of a radioactive element is the time it takes for half of a sample of the element to decay into another, more stable form.

Radioactive elements have varying half-lives, depending on their specific isotopes and nuclear properties. These half-lives can range from extremely short durations, such as fractions of a second, to incredibly long periods, like billions of years.

The variation in half-lives reflects the diversity in stability among different isotopes of radioactive elements, which is influenced by factors like the ratio of protons and neutrons in the nucleus.

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When Weight of 7294.3 N can be supported by boom if force 7800 is applied on the cable and there is no resultant rotation at point C.

Torque is defined as force times perpendicular distance from axis of rotation to the point at which force is applied. Torque is denoted by τ and its SI unit is Nm. i.e Torque is given by, τ = F×d sinθ. Torque is vector quantity.

Torque is analogous to force in rotational motion. like there angular velocity ω is analogous  linear velocity v. means Torque is a force in rotational motion.

Looking at the answered figure I have resolved forces, there will be no resultant rotation when opposite torques due to opposite horizontal forces gets balanced.

We get,

τ₁ = τ₂

F×d Cos40⁰ = W×d Sin55°

F Cos40° = W Sin55°

W = F×(Cos40°÷Sin55°)

W = F×0.93516

W = 7800 ×0.93516

W = 7294.3 N

Hence Weight that can be supported is 7294.3 N.

b) When we assume that there is no cable, Then

Torque τ = W×d Sin35°

τ = 7294.3×10 Sin35°

τ = 41838.3 Nm

bloom will rotate with 41838.3 Nm torque.

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Answer:

a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is 89.289 joules.

e) 89.289 joules must be done to stop the bullet-block system.

f) The bullet-block system will travel 13.007 meters before stopping.

Explanation:

a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by  the initial momentum of the bullet:

\(p = m\cdot v_{o}\) (1)

Where:

\(p\) - Net momentum, in kilogram-meters per second.

\(m\) - Mass of the bullet, in kilograms.

\(v_{o}\) - Initial speed of the bullet, in meters per second.

If we know that \(m = 0.5\,kg\) and \(v_{o} = 50\,\frac{m}{s}\), then the net momentum of the bullet-block system before the collision is:

\(p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)\)

\(p = 25\,\frac{kg\cdot m}{s}\)

The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The total energy of the bullet before the collision is its initial translational kinetic energy (\(K\)), in joules:

\(K = \frac{1}{2}\cdot m \cdot v_{o}^{2}\) (2)

\(K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}\)

\(K = 625\,J\)

The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:

\(v_{f} = \frac{m\cdot v_{o}}{m+M}\) (3)

Where:

\(v_{f}\) - Final speed, in meters per second.

\(M\) - Mass of the block, in kilograms.

If we know that \(m = 0.5\,kg\), \(v_{o} = 50\,\frac{m}{s}\) and \(M = 3\,kg\), then the final speed of the bullet-block system is:

\(v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)\)

\(v_{f} = 7.143\,\frac{m}{s}\)

The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system (\(K\)), in joules, is:

\(K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2}\) (4)

\(K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}\)

\(K = 89.289\,J\)

The total energy of the bullet-block system after the collision is 89.289 joules.

e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.

f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:

\(W = \mu_{k}\cdot (m+M)\cdot g\cdot s\) (5)

Where:

\(W\) - Work done by friction, in joules.

\(g\) - Gravitational acceleration, in meters per square second.

\(s\) - Travelled distance, in meters.

\(\mu_{k}\) - Kinetic coefficient of friction, no unit.

If we know that \(m = 0.5\,kg\), \(M = 3\,kg\), \(\mu_{k} = 0.2\), \(g = 9.807\,\frac{m}{s^{2}}\) and \(W = 89.289\,J\), then the travelled distance of the bullet-block system is:

\(s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}\)

\(s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\)

\(s = 13.007\,m\)

The bullet-block system will travel 13.007 meters before stopping.

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F\(_{axial}\)

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

Dimensionless quantities commonly used in aerodynamics include the coefficient of lift, coefficient of moment, and coefficient of drag. Hence option d is correct.

Dimensionless quantities are important in aerodynamics as they provide standardized values that can be compared across different systems. The coefficient of lift (CL) is a dimensionless quantity that relates the lift generated by an airfoil to its shape and the angle of attack. It is calculated by dividing the lift force by the product of the dynamic pressure of the flow and the reference area of the airfoil.

The coefficient of moment (CM) is another dimensionless quantity that relates the pitching moment of an airfoil to its shape and the reference length. It is calculated by dividing the pitching moment by the product of the dynamic pressure of the flow, the reference area, and the reference length.

The coefficient of drag (CD) is a dimensionless quantity that relates the drag force experienced by an object to the dynamic pressure of the flow and the reference area. It is calculated by dividing the drag force by the product of the dynamic pressure and the reference area. All three coefficients, CL, CM, and CD, are examples of dimensionless quantities commonly used in aerodynamics to analyze and compare the performance of different airfoils and objects in fluid flows.

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The mass M of the star is mathematically given as

M=24.269kg

Question Parameters

Orbital period P= 14.3*24*60*60 s

P= 1235520s

Speed V=160000

Generally, the equation for the Radius is  mathematically given as

R =\(\frac{ V*t}{ 2*pi}\)

Therefore

R =\(\frac{ ( 160000 )*(1235520 s)}{ 2*3.142}\)

R=\(3.14581795*10^{10}\)

Where

M*v^2/R = GM^2/(2R)^2

Therefore

M=\(=4(v^2)R/G\\\\ =4(240^2)R/G\\\\ = \frac{4*(160000)^2)*3.14581795*10^{10}}{6.67*10^{-11}}\)

M=4.82956159*10^{31}kg

Considering, that the mass of our Sun is 1.99*10^{30} kg.

The mass M of the star is

M\(=\frac{4.82956159*10^{31}}{1.99*10^{30}}\)

M=24.269

In conclusion, the mass M of the star is

M=24.269

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Answer: Be more specific

Answer:

uno thos threse u here me

Explanation:

Answer:

29.436 J/g.°C

Explanation:

We are given;

Mass of tungsten; m_t = 215 g

Initial temperature of tungsten; T_ti = 100°C

Final temperature; T_f = 21.5°C

Mass of water; m_w = 450 g

Initial temperature of water; T_tw = 20°C

From tables, specific heat capacity of water; c_w = 4.184 J/g.C

Thus;

Energy lost by tungsten = energy gained by water

Thus;

215 × c_t × (100 - 21.5) = 450 × 4 184 × (21.5 - 20)

Where c_t is specific heat capacity of tungsten

16877.5c_t = 496800

c_t = 496800/16877.5

c_t = 29.436 J/g.C

The velocity of ball B after the collision is 6.5 m/s.

According to the law of conservation of momentum, The overall momentum of two or more bodies acting on one another in an isolated system stays constant unless an external force is introduced.

Before collision ,

ball A of mass of 3 kg has velocity = 2 m/s.

ball B of mass of 1 kg has velocity= -4 m/s.

After the collision,

Ball A has a velocity = -1.5 m/s.

Let   Ball B has a velocity = v m/s.

Applying   law of conservation of momentum,

Total initial momentum = total final momentum

3×2 +1×-4 = 3×-1.5 + 1×v

⇒ v = 3×2 +1×-4 - 3×-1.5

⇒ v = 6.5 m/s.

hence, the velocity of ball B after the collision is 6.5 m/s.

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This exercise is about Reflection and Mirrors. It explores the relationship between convex mirrors, virtual images, and complex images as well as to object distance d₀.

Any d₀ value greater than 10 for a convex lens will produce a true or real image, while any d₀ value greater than 15 for a complex mirror will produce a real image.

Any d₀ number less than 10 for a convex lens will result in a virtual image. Also, any d₀ value less than 15 for a complex mirror will result in a virtual picture.

The range of d₀ values that would result in a reduced picture in both lenses and mirrors is:

d₀ > 2 f, where is the Focal length.

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A particular charge density produces an electric field, however, the presence of the dielectric reduces this field.

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Bumpers, collapsible frames, airbags, and seat belts are safety features in vehicles designed to protect occupants in the event of an accident.

Bumpers help absorb impact and minimize damage to the vehicle, collapsible frames absorb and distribute energy, airbags rapidly inflate to provide a cushioning effect, and seat belts restrain occupants and prevent them from being thrown forward. All of these safety features work together to enhance occupant safety during accidents.

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Answer:

c

Explanation:

its c

Answer:

B

Explanation:

Kinetic energy is speed so the one with greater speed has more kinetic energy

Answer:A

Explanation:

In a gas-powered car, a battery A) produces chemical potential energy to run the engine.

Answer:Its empty car, small child, and then adult.

Explanation:

i took it

Answer:

Its empty car, small child, and then adult.

Answer:

Conductors of electricity?

The Answer is C metal :)

xXxAnimexXx

The spring constant is 1.74 N/m and the maximum amplitude of the glider's motion is 0.06 m.

The formula to find the force of the spring is

F = -kx,

where k is the spring constant and x is the displacement from the equilibrium position.

Since the glider is pushed 12.0 cm against the spring, or 0.12 m, the force of the spring can be found using the formula:

F = -kxF = -(k)(0.12)

F = -0.12k

To find the spring constant k, we need to know the mass of the glider.

The formula for the period of a spring oscillator is:

T = 2π√(m/k),

where T is the period, m is the mass, and k is the spring constant.

Rearranging this formula, we get:

k = (4π²m)/T²

Plugging in the values we have:

T = 10 s (for 10 oscillations) and m = 0.140 kg, k = (4π²)(0.140)/(10²)k ≈ 1.74 N/m

Now that we know k, we can find the maximum amplitude of the glider's motion.

The formula for the maximum amplitude is:

A = x₀, where x₀ is the initial displacement from the equilibrium position. In this case, x₀ = 0.12 m.

However, we can also use the formula

A = (1/2) x_max

Plugging in the values we have:

A = (1/2) x_maxx_max = 2A = 2(0.12) = 0.24 m

The maximum amplitude of the glider's motion is 0.24 m, or 0.06 m per oscillation.

Hence, the spring constant is 1.74 N/m and the maximum amplitude of the glider's motion is 0.06 m.

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Complete question:

A 140 g air-track glider is attached to a spring. The glider is pushed in 12 cm and released. A student with a stopwatch finds that 10.0 oscillations take 10.0 s. What is the spring constant?

Kirchhoff's rules, consisting of Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL), do not introduce any new physical principles beyond Ohm's law.

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. It provides a fundamental relationship between current, voltage, and resistance. Kirchhoff's rules, on the other hand, build upon Ohm's law and extend it to complex circuits.

Kirchhoff's voltage law (KVL) states that the algebraic sum of the potential differences (voltages) in any closed loop of a circuit is zero. This principle is based on the conservation of energy, where the sum of the potential drops and rises around a closed loop must equal zero. KVL allows us to analyze circuits with multiple voltage sources, resistors, and other components.

Kirchhoff's current law (KCL) states that the algebraic sum of currents at any junction (node) in a circuit is zero. This principle is based on the conservation of charge, where the total current flowing into a node must equal the total current flowing out of it. KCL enables the analysis of current division and combination in complex circuits with multiple branches.

Together, KVL and KCL provide a systematic approach to analyzing and solving electrical circuits, allowing us to determine currents, voltages, and power dissipation in complex circuit configurations. While Kirchhoff's rules do not introduce new physical principles beyond Ohm's law, they are essential tools for understanding and designing circuits that go beyond simple series or parallel configurations.

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From the principle of conservation of linear momentum, the mass of Lee is 50 Kg.

Momentum is the product of mass and velocity. The principle of conservation of linear momentum states that the total momentum before collision is equal to the total momentum after collision.

Let the mass of Lee be m

Hence;

(4.0-kg  × 3.0 m/s) + (m × 3.0 m/s) = (4.0-kg  × 15.5 m/s) + (m × 2.0 m/s)

12 + 3m = 62 + 2m

12 - 62 = 2m - 3m

-50 = -m

m = 50 Kg

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Answer:

Stars create new elements in their cores by squeezing elements together in a process called nuclear fusion. First, stars fuse hydrogen atoms into helium. Helium atoms then fuse to create beryllium, and so on, until fusion in the star's core has created every element up to iron.

Explanation:

Hope this helps.

Answer:

hahahahahaha hahahahaha hahah

When a light ray travels parallel to the principal axis of a curved mirror, the ray will not be reflected back in the opposite direction like in the case of a flat mirror.

The reason why this does not occur is  because the surface of a curved mirror is not a flat surface and it has a varying curvature, which causes the reflected rays to behave differently compared to the flat mirror.

In the case of a concave mirror, when a light ray travels parallel to the principal axis and strikes the mirror, it is reflected back towards the focus of the mirror. This is because the concave mirror surface curves inward, causing the reflected light rays to converge towards the focus.

In the case of a convex mirror, when a light ray travels parallel to the principal axis and strikes the mirror, it is reflected back in a direction away from the focus of the mirror. This is because the convex mirror surface curves outward, causing the reflected light rays to diverge away from the focus.

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