Anwser this Question I will BrainElist You

Answer:

Mechanical advantage is the measure of the force amplification achieved by using a tool , mechanical device or machine . The machine preserve the input power and supply trade off force. against movement to obtain a desired amplification in the output force .

Answer:

The ratio of load overcome by the machine to the effort applied is called mechanical advantage.

Answer:

a. Time = 3.6 seconds

b. Time = 4.4 seconds

Explanation:

Given the following data;

Distance = 180 m

Speed = 50 m/s

a. To find the time;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

[tex]Speed = \frac{distance}{time}[/tex]

Making time the subject of formula, we have;

[tex]Time = \frac{distance}{speed}[/tex]

Substituting into the equation, we have;

[tex]Time = \frac{180}{50}[/tex]

Time = 3.6 seconds

b. Distance = 220 meters

Speed = 50 m/s

To find the time;

[tex]Time = \frac{distance}{speed}[/tex]

Substituting into the equation, we have;

[tex]Time = \frac{220}{50}[/tex]

Time = 4.4 seconds

Answer:

27°C

Explanation:

We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27 °C

Initial temperature (T₁) = 27 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 2 m³

Initial temperature (T₁) = 300 K

Initial pressure (P₁) = 1 atm

Final pressure (P₂) = 2 atm

Final volume (V₂) = 1 m³

Final temperature (T₂) =?

P₁V₁/T₁ = P₂V₂/T₂

1 × 2 / 300 = 2 × 1 / T₂

2/300 = 2/T₂

1/150 = 2/T₂

Cross multiply

T₂ = 150 × 2

T₂ = 300 K

Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 300 K

T(°C) = 300 – 273

T(°C) = 27°C

Thus, the final temperature is 27°C

Answer:

HONORS PHYSICS

Introduction

Matter & Energy

Math Review

Kinematics

Defining Motion

Graphing Motion

Kinematic Equations

Free Fall

Projectile Motion

Relative Velocity

Dynamics

Newton's 1st Law

Free Body Diagrams

Newton's 2nd Law

Static Equilibrium

Newton's 3rd Law

Friction

Ramps and Inclines

Atwood Machines

Momentum

Impulse & Momentum

Conservation Laws

Types of Collisions

Center of Mass

UCM & Gravity

Uniform Circular Motion

Gravity

Kepler's Laws

Rotational Motion

Rotational Kinematics

Torque

Angular Momentum

Rotational KE

Work, Energy & Power

Work

Hooke's Law

Power

Energy

Conservation of Energy

Fluid Mechanics

Density

Pressure

Buoyancy

Pascal's Principle

Fluid Continuity

Bernoulli's Principle

Thermal Physics

Temperature

Thermal Expansion

Heat

Phase Changes

Ideal Gas Law

Thermodynamics

Electrostatics

Electric Charges

Coulomb's Law

Electric Fields

Potential Difference

Capacitors

Current Electricity

Electric Current

Resistance

Ohm's Law

Circuits

Electric Meters

Circuit Analysis

Magnetism

Magnetic Fields

The Compass

Electromagnetism

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Silicon

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Processing

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Waves & Sound

Wave Characteristics

Wave Equation

Sound

Interference

Doppler Effect

Optics

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Modern Physics

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Models of the Atom

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DYNAMICS

Newton's 1st Law

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Newton's 2nd Law

Static Equilibrium

Newton's 3rd Law

Friction

Ramps and Inclines

Atwood Machines

Silly Beagle

Newtons's 3rd Law of Motion

Newton’s 3rd Law of Motion, commonly referred to as the Law of Action and Reaction, describes the phenomena by which all forces come in pairs. If Object 1 exerts a force on Object 2, then Object 2 must exert a force back on Object 1. Moreover, the force of Object 1 on Object 2 is equal in magnitude, or size, but opposite in direction to the force of Object 2 on Object 1. Written mathematically:

Newton's 3rd Law Equation

This has many implications, some of which aren’t immediately obvious. For example, if you punch the wall with your fist with a force of 100N, the wall imparts a force back on your fist of 100N (which is why it hurts!). Or try this. Push on the corner of your desk with your palm for a few seconds. Now look at your palm... see the indentation? That’s because the corner of the desk pushed back on your palm.

running tiger

Although this law surrounds your actions everyday, often times you may not even realize its effects. To run forward, a cat pushes with its legs backward on the ground, and the ground pushes the cat forward. How do you swim? If you want to swim forwards, which way do you push on the water? Backwards, that’s right. As you push backwards on the water, the reactionary force, the water pushing you, propels you forward. How do you jump up in the air? You push down on the ground, and it’s the reactionary force of the ground pushing on you that accelerates you skyward!

As you can see, then, forces always come in pairs. These pairs are known as action-reaction pairs. What are the action-reaction force pairs for a girl kicking a soccer ball? The girl’s foot applies a force on the ball, and the ball applies an equal and opposite force on the girl’s foot.

How does a rocket ship maneuver in space? The rocket propels hot expanding gas particles outward, so the gas particles in return push the rocket forward. Newton’s 3rd Law even applies to gravity. The Earth exerts a gravitational force on you (downward). You, therefore, must apply a gravitational force upward on the Earth!

Answer:

True

Explanation:

I guess you made a mistake on question.

but I understood what you wanted to say.

Hope this helps... :)

Answer:

¿Podrías poner la pregunta en inglés por favor?

Explanation:

Answer:

Effort = 540 Newton

Explanation:

Given the following data;

Load arm = 60 cm

Effort arm = 40 cm

Load = 360 N

Conversion:

100 cm = 1 meters

40 cm = 40/100 = 0.4 meters

60 cm = 60/100 = 0.6 meters

To calculate the effort that Reha should apply to lift Neha, we would use the expression;

Effort * effort arm = load * load arm

Substituting into the expression, we have;

Effort * 0.4 = 360 * 0.6

Effort * 0.4 = 216

Effort = 216/0.4

Effort = 540 Newton

Answer:

Velocity = distance / time

V = 10/1

V = 10km/h

Answer:10km/h or 2.77m/s.

Explanation:

Distance =10km

Time =1h

Velocity =10/1 =10km/h

Or,

Distance =10km =10000m

Time =1h =60min = 3600s

Velocity =10000/3600 =2.77m/s

Answer:

I = 20 A

Explanation:

The question says that, "A load of 6,000 C is conducted through a cross section in 5 minutes. Determining-if a current is not correct, we will find the value of?"

We have,

Charge, q = 6,000 C

Time, t = 5 minutes = 300 s

We need to find the current. We know that, the charge flowing per unit time is equal to current. So,

[tex]I=\dfrac{q}{t}\\\\I=\dfrac{6000}{300}\\\\I=20\ A[/tex]

So, the current flowing through the circuit is 20 A.

Answer:

h = 1/2gt^2

Is the formula for the height of a free falling object.

Put in the numbers:

h = 1/2*9.8 (gravitational constant) * 10^2

h=490 meters.

Please note that the gravitational constant is approximate, you could also use a more specific value or 10

Answer:

Jupiter and neptune

Explanation:

Answer:

table and chair

Explanation:

In a meeting room there are tables, chairs, and other people. Which of them have temperatures

a) smaller, b) bigger and d) equal to that of air?

the temperature of tables and chairs is same as air.

Explanation:

3: the electricity is generated in the permanent magnet

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

Answer:

a. V₂ = 6 V

b. 360 joules

c. The positive terminals of both the voltmeter and ammeter are connected to the positive terminal of the power source

Please see the attached drawing created with MS Visio

Explanation:

a. The voltmeter readings are;

V₁ = 12 V, V = 18 V

Given that the voltage reading, 'V', is the voltage reading across two loads with voltages, V₁ and V₂ connected in series, we have;

V = V₁ + V₂

V₂ = V - V₁

V₂ = 18 V - 12 V = 6 V

b. The reading of the ammeter, A, I = 0.5 A

The heat energy, Q  = I·V·t

Where;

t = The time = 1 minute (60 seconds)

Therefore, for the lamp L₁, where V = 12 V, we have;

Q₁ = 0.5 A × 12 V × 60 s = 360 Joules

The amount of electrical energy changed into heat and light in lamp L₁, Q₁ = 360 joules

c. The positive terminals of the voltmeter and ammeter are connected to the positive terminal of the power source

Please see attached drawing created with MS Visio

Answer:

w=f×s

w = 16000 J, hope this helps

Answer:

Ionosphere

Explanation:

The thermosphere reaches 600 kilometres just above mesosphere and begins immediately above the mesosphere. This layer is where the aurora and satellites appear.

The ionosphere is the comprehensive career of the mesosphere because most of the thermosphere, located 80–400 kilometres just above ground atmosphere.

Auroras — magnificent flowing streaks of light seen in the night sky – appear in this location.

The gravitational force does change believe it or not, but the explaination for this is because the earths orbit is an oval (or a not circle) the closer it nears its self to the sun.

Answer:

Explanation:

Here's what we know because it was given to us:

a = -9.8 m/s/s and

time = 3.32 seconds

Here's what we know because we rock physics:

v₀ = 0 (because the object was held still before it was dropped).

Here's the equation that ties all that info together in a single one-dimensional equation:

v = v₀ + at

Filling in and solving for v:

v = 0 + (-9.8)(3.32) and

v = -33m/s

The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)

Answer:

power per square meter = 4.593 × 10^(6) W/m²

Wavelength of peak intensity = 9.67 × 10^(-7) m

Explanation:

From Stefan-Boltzmann law, total emitted power per square meter is given as;

P/A = eσT⁴

where;

P is power

A is surface area

σ = Stefan-Boltzmann constant = 5.67 × 10^(-8) W/m².k⁴

T = temperature of the body = 3000 K

e = emissivity of the substance (for ideal radiation, it has a value = 1)

Thus, Plugging in the relevant values we have;

P/A = 1 × 5.67 × 10^(-8) × (3000)^(4)

P/A = 4.593 × 10^(6) W/m²

Let's find the wavelength of peak intensity.

From wiens displacement law, we know that;

λ_m × T = b

where;

λ_m = maximum wavelength

T = Temperature

b is Wien's displacement constant = 2.9 × 10^(−3) m/K

thus;

λ_m = b/T = (2.9 × 10^(−3))/3000 = 9.67 × 10^(-7) m

Explanation:

The difference of electrical potential between two points is called the potential difference.

It is also defined as the work done in the transfer of a unit quantity of electricity from one point to the other.

The SI unit of electric potential is volt. It is denoted by V.

Answer:

O C. A person absorbs most of the light that hits him or her.

Explanation:

Answer:

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:

[tex]p = m\cdot v_{o}[/tex] (1)

Where:

[tex]p[/tex] - Net momentum, in kilogram-meters per second.

[tex]m[/tex] - Mass of the bullet, in kilograms.

[tex]v_{o}[/tex] - Initial speed of the bullet, in meters per second.

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

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

[tex]p = 25\,\frac{kg\cdot m}{s}[/tex]

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 ([tex]K[/tex]), in joules:

[tex]K = \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex] (2)

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

[tex]K = 625\,J[/tex]

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:

[tex]v_{f} = \frac{m\cdot v_{o}}{m+M}[/tex] (3)

Where:

[tex]v_{f}[/tex] - Final speed, in meters per second.

[tex]M[/tex] - Mass of the block, in kilograms.

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

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

[tex]v_{f} = 7.143\,\frac{m}{s}[/tex]

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 ([tex]K[/tex]), in joules, is:

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

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

[tex]K = 89.289\,J[/tex]

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:

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

Where:

[tex]W[/tex] - Work done by friction, in joules.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]s[/tex] - Travelled distance, in meters.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

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

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

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

[tex]s = 13.007\,m[/tex]

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

Answer: The given statement is True

Explanation:

A uniform motion is defined as the motion where an object is moving at a constant speed.

A non-uniform motion is defined as the motion where an object keeps changing its position and does not move at a constant speed.

We are given:

A car takes a full round of journey in a roundabout with constant speed

As the speed remains constant in a circular path, it is considered a uniform motion.

Hence, the given statement is True

Answer:

a. 2:3

b. The data illustrates the law of multiple proportions by showing that the the masses of oxygen that reacts with a fixed mass of sulfur are in a ratio of small whole numbers

Explanation:

The weight of oxygen that combines with 32.06 grams of sulfur in sulfur dioxide = 32 grams

The weight of oxygen that combines with 32.06 grams of sulfur in sulfur trioxide = 48 grams

a. The ration of the weights of oxygen that combine with 32.06 g of sulfur = 32:48 = 2:3

b. The law of multiple proportions states that when two elements are able to interact chemically to form more than one compound, then the (different) weights of one of the element that combines with a fixed weight of the other element are in small whole number ratios

The data demonstrates the law of multiple proportions by showing that the ratios of the weights of oxygen that combine with a fixed weight of sulfur to form sulfur dioxide and sulfur trioxide is in the ratio of 2 to 3 which are small whole number ratios

Answer:

acceleration is the vector quantity because it depends on particular direction and has magnitude