What is the relationship between color and wavelength for light?

Answer:

c gravity acts on all objects in the universe

Explanation:

Answer:

work done on the object will be 10 J

Explanation:

As we know by work energy theorem that work done by all the forces is equal to change in kinetic energy

It is given as

[tex]K_f - K_i = W[/tex]

here we know that

[tex]K_i = 40 J[/tex]

now its 10 J of potential energy is converted into kinetic energy

Which means the kinetic energy finally becomes

[tex]K_f = 40 + 10 = 50 J[/tex]

now by above equation we know that

[tex]W = K_f - K_i[/tex]

[tex]W = 50 - 40 = 10 J[/tex]

so work done on the object will be 10 J

Answer:

increase the temperature.

Explanation:

Answer : Density of the substance is [tex]8.9\ g/cm^3[/tex]              

Explanation :

The volume of the substance, [tex]V=10\ cm^3[/tex]

Mass of the substance, [tex]M=89\ g[/tex]

The density of a substance is defined as the mass per unit volume.

[tex]\rho=\dfrac{m}{V}[/tex]  

[tex]\rho=\dfrac{89\ g}{10\ cm^3}[/tex]

[tex]\rho=8.9\ g/cm^3[/tex]            

So, the correct option is (b).

Hence, this is the required solution.              

Final answer:

The four steps in the machine cycle of a four-stroke gasoline engine are Intake, Compression, Combustion (Power), and Exhaust, often termed the Otto cycle.

Explanation:

Steps in the Machine Cycle:

The four steps in the machine cycle for a common four-stroke gasoline engine are Intake, Compression, Combustion (Power), and Exhaust. During the Intake step, a mixture of gasoline and air is drawn into the engine's combustion chamber. Next is the Compression step, where the piston compresses the mixture, increasing its temperature. The Combustion (Power) step follows, which ignites the compressed mixture, causing a rapid expansion of gases that push the piston. Lastly, the Exhaust step releases the combustion gases from the chamber. This process is often referred to as the Otto cycle and is central to the operation of many internal combustion engines.

The three different kinds of graphs that are often used to plot information are Bar graphs, Pie charts, and Line graphs.

A graph is a way to represent a lot of data in such a visual format that it is easy for the user to understand the complete information in one go. Usually, the line of the graph is a function that follows the graph.

The three different kinds of graphs that are often used to plot information are,

Hence, the three different kinds of graphs that are often used to plot information are Bar graphs, Pie charts, and Line graphs.

Learn more about Graph here:

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To solve this Physics problem, the equations of projectile motion were utilized. The flight time was calculated to be 2.27 seconds and the range was found to be 78.4 meters.

The subject of your question falls under projectile motion in Physics. If a cannonball is fired with an initial speed of 40 m/s at a launch angle of 30° from a cliff that's 25m tall, it has both horizontal and vertical components of motion.

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The minimum coefficient of static friction required to keep the mug from sliding is 0.31

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

This problem is about Newton's Law of Motion.

Given:

θ = 17°

Unknown:

μ = ?

Solution:

[tex]\Sigma Fy = ma[/tex]

[tex]N - w \cos \theta = m (0)[/tex]

[tex]N - w \cos \theta = 0[/tex]

[tex]\boxed {N = w \cos \theta}[/tex] → Equation 1

[tex]\Sigma Fx = ma[/tex]

[tex]w \sin \theta - f = m (0)[/tex]

[tex]w \sin \theta - f = 0[/tex]

[tex]f = w \sin \theta[/tex]

[tex]\mu N = w \sin \theta[/tex]

[tex]\mu w \cos \theta = w \sin \theta[/tex] ← Equation 1

[tex]\mu = (w \sin \theta) \div ( w \cos \theta )[/tex]

[tex]\mu = \tan \theta[/tex]

[tex]\mu = \tan 17^o[/tex]

[tex]\mu \approx 0.31[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Mug , Friction , Coefficient , Static

The magnitude of frictional force exerted on the mug is [tex]\fbox{\begin\\({2.865}\right))m\end{minispace}}[/tex] and the coefficient of static friction is [tex]\fbox{0.31}[/tex].

Further explanation:

The opposition that every object feels when it moves is known as the frictional force. This is an opposing force. It always acts in the opposite direction of the motion of the object.

Given:

The angle of inclination of plane is [tex]{17^ \circ }[/tex].  

The acceleration due to gravity is [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex].

Concept used:

The frictional force is defined as the opposition of motion of any object or body.  It is also defined as the product of coefficient of friction to the normal force exerted on the body.

The expression for the net force exerted on the mug is given as.

   [tex]\fbox{\begin\\{F_{{\text{net}}}} = mg\sin \theta\end{minispace}}[/tex]                                          …… (1)

Here,[tex]m[/tex] is the mass of the mug, [tex]g[/tex] is the acceleration due to gravity and[tex]\theta[/tex]  is the inclination angle.

The force that keeps the body at rest is known as the static friction force. It also acts in the opposite direction of motion of body.

The expression for the static frictional force is given as.

[tex]{F_s} = {\mu _s}N[/tex]                                                                  ..…. (2)

Here,[tex]{\mu _s}[/tex] is the coefficient of static frictional force and [tex]N[/tex] is the normal force exerted on the body.

When mug is placed on the inclined plane the static frictional force is equal to the frictional force. The vertical component of weight is balanced by the normal reaction of the mug.

The expression for the normal force is given as.

[tex]N-mg\cos\theta=0[/tex]

Rearrange the above expression.

[tex]N=mg\cos\theta[/tex]

Substitute[tex]mg\cos\theta[/tex]  for [tex]N[/tex] in equation (2).

[tex]{F_s} = {\mu _s}\left( {mg\cos \theta } \right)[/tex]

Here static frictional force is same as the frictional force exerted on the mug.

Substitute [tex]mg\sin\theta[/tex] for[tex]{F_s}[/tex] in above expression and rearrange it.

[tex]{\mu _s} = \dfrac{{\left( {mg\sin \theta } \right)}}{{\left( {mg\cos \theta } \right)}}[/tex]  

Rearrange the above expression.

[tex]\fbox{\begin\\{{\mu _s} = \tan \theta}\end{minispace}}[/tex]               …… (3)

Substitute[tex]{17^ \circ }[/tex]for [tex]\theta[/tex] and [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex]in equation (1).

[tex]\begin{aligned}{F_{{\text{net}}}} &= m\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\sin\left({{{17}^\circ}}\right)\\&=\left( {2.865}\right)m \\\end{aligned}[/tex]

Substitute [tex]{17^ \circ }[/tex]for [tex]\theta[/tex] in equation (3).

[tex]\begin{gathered}{\mu_s}=\tan\left( {{{17}^\circ }}\right)\\= 0.306 \\\end{gathered}[/tex]

Thus, the coefficient of static friction is [tex]0.31[/tex] and the magnitude of frictional force is [tex](2.865)m[/tex].

Learn more:

1.  Conservation of momentum https://brainly.com/question/9484203

2.  Motion of a ball under gravity https://brainly.com/question/10934170

3. Motion of a block under friction https://brainly.com/question/7031524

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Friction, acceleration, carpeted floor, mug, inclined plane, force, relative motion, motion, normal reaction, net force, mass, surface, oppose, 9.8 m/s2, 17 degree, 0.67 m/s^2, 0.3006, 0.31, (2.865)m, (2.87)m.

Answer:

The answer is C

Explanation:

I am 100% sure of this take it or leave it, buddy!

To draw a free body diagram for the skier, identify the forces acting on the skier including the gravitational force, normal force, and frictional force. The normal force can be found by taking the component of the gravitational force perpendicular to the incline. The acceleration of the skier down the hill can be determined using Newton's second law.

To draw a free body diagram for the skier, we need to identify the forces acting on the skier. These include the gravitational force (mg) acting straight down, the normal force (N) perpendicular to the incline, and the frictional force (f) parallel to the incline. The acceleration of the skier down the hill can be determined using Newton's second law, which states that the net force (Fnet) equals the mass (m) times the acceleration (a): Fnet = ma.

The normal force can be found by taking the component of the gravitational force perpendicular to the incline, which is given by N = mg*cos(theta).

The acceleration of the skier down the hill can be determined using the equation Fnet = ma, where the net force is the difference between the gravitational force parallel to the incline and the frictional force: Fnet = mg*sin(theta) - f.

The normal force acting on the skier is approximately 550.32 N.The skier's acceleration down the hill is approximately 3.03 m/s².

Here's how to solve the problem:

a. Free body diagram:

The free body diagram for the skier should show the following forces acting on them:

Weight (mg): This is the force of gravity acting on the skier, directed downwards. Its magnitude is given by `mg`, where `m` is the skier's mass (65 kg) and `g` is the acceleration due to gravity (approximately 9.81 m/s²). In this case, the weight would be `65 kg * 9.81 m/s² ≈ 637.65 N`.

Normal force (N): This is the force exerted by the slope on the skier, perpendicular to the slope. Its magnitude depends on the weight and the angle of the slope.

Friction force (f): This is the force opposing the skier's motion, parallel to the slope. Its magnitude is proportional to the normal force and the coefficient of friction (`μ`).

b. Normal force:

The normal force can be calculated using the component of the weight perpendicular to the slope. This can be done using trigonometry:

N = mg * cos(θ)

where `θ` is the angle of the slope (30° in this case).

Therefore, the normal force would be:

N = 637.65 N * cos(30°) ≈ 550.32 N

c. Acceleration:

The acceleration of the skier can be calculated using Newton's second law of motion:

ΣF = ma

where `ΣF` is the net force acting on the skier, `m` is the skier's mass, and `a` is the acceleration.

The net force is the difference between the downhill component of the weight and the friction force:

ΣF = mg * sin(θ) - μ * N

Substituting the known values:

ΣF = 637.65 N * sin(30°) - 0.1 * 550.32 N ≈ 196.81 N

Finally, solving for the acceleration:

a = ΣF / m = 196.81 N / 65 kg ≈ 3.03 m/s²

Therefore, the skier's acceleration down the hill is approximately 3.03 m/s².

Final answer:

Before school, examples of Newton's laws of motion include remaining seated at breakfast due to inertia, lifting a backpack which demonstrates the second law with force and acceleration, and walking to the bus stop where action and reaction forces illustrate the third law.

Explanation:

Examples of Newton's Laws Before School

Newton's laws of motion are fundamental in understanding how forces affect motion in our daily lives. To apply these laws, we start by selecting an object and listing all the forces acting on it. This approach helps us solve problems involving various forces, such as friction, the normal force, and gravitational force.

First Law (Inertia): As you sit at breakfast, your body is at rest. Unless an unbalanced force acts upon you, you will remain stationary. The force of gravity pulls you down in your chair, and the normal force from the chair supports you from falling through.

Second Law (F=ma): When you pick up your backpack, the force you exert is equal to the mass of the backpack multiplied by the acceleration you give it. If the backpack is heavy (has more mass), it requires more force to accelerate.

Third Law (Action-Reaction): As you push off the ground to walk to the bus stop, your feet apply a force to the ground, and the ground pushes back with an equal and opposite force. This reaction force propels you forward.

These everyday examples show the application of Newton's three laws of motion, from the breakfast table to the walk to the bus stop, illustrating friction, the normal force, and gravitational force at play.

To determine the depth to which the second bullet will penetrate the block, we can use the work-energy theorem. By comparing the initial kinetic energy of the bullet-block system to the work done by the average stopping force in each situation, we can find the depth.

The initial situation is when a 7.00 g bullet is fired into a 0.80 kg block of wood held in a vise. The bullet penetrates the block to a depth of 8.20 cm. In the second situation, the block of wood is placed on a frictionless horizontal surface, and a second 7.00 g bullet is fired into the block. The question asks to determine the depth to which the bullet will penetrate in this case.

To solve this problem, we can use the work-energy theorem. Since the wood block is on a frictionless horizontal surface, there is no external force doing work on the block-bullet system. Therefore, the initial kinetic energy of the bullet is equal to the work done by the average stopping force in stopping the bullet. We can equate these two and solve for the unknown depth.

In the initial situation, the bullet penetrated to a depth of 8.20 cm. We can use this known value along with the mass of the bullet and the mass of the block to find the initial kinetic energy of the bullet-bloc system. Then, we can set this initial kinetic energy equal to the work done by the average stopping force in the second situation to find the depth to which the bullet will penetrate.

Final answer:

To find the expression for the velocity of the second car, consider the motion of both cars separately. The velocity of the second car can be described as v2 = 24 - 6t.

Explanation:

To find the expression for the velocity of the second car, we need to consider the motion of both cars separately.

Let's assume that the first car passes the second car at time t = 0. The position of the first car can be described as x1 = 30t, where t is the time elapsed since the first car passed the second car.

The position of the second car can be described as x2 = (24)t - (1/2)(6)t^2, where the negative term represents the deceleration of the second car due to slowing down at 6.0 m/s^2.

Now, we can find the expression for the velocity of the second car by taking the derivative of its position equation with respect to time: v2 = 24 - 6t.

The velocity of the second car, which starts at 24 m/s and decelerates at 6.0 m/s², can be described by the function v(t) = 24 - 6t. This equation shows how the car's velocity decreases over time. For any given time t, you can calculate the velocity using this expression.

To find the expression of the velocity of the second car, we need to consider its deceleration. The second car's initial velocity is 24 m/s, and it decelerates at a rate of 6.0 m/s². The general equation for the velocity of an object under constant acceleration (or deceleration) is:

v(t) = v₀ + at

Here, v(t) is the velocity at time t, v₀ is the initial velocity, and a is the acceleration (note that since it is deceleration, the acceleration will be negative).

Given:

We can substitute these values into the equation:

v(t) = 24 m/s - (6.0 m/s²) * t

So, the expression for the velocity of the second car as a function of time is:

v(t) = 24 - 6t

Final answer:

Neil has a total of 1 1/30 gallons of paint when combining three partially full cans containing 1/3, 1/5, and 1/2 gallon of paint, after converting them to a common denominator and summing them up.

Explanation:

The student is asking about the total quantity of paint Neil has when combining three partially full cans with varying amounts. To find the total, we need to add the fractional quantities together: 1/3 gallon, 1/5 gallon, and 1/2 gallon of paint.

To add these fractions, they must have a common denominator. The least common denominator for 3, 5, and 2 is 30. Converting each fraction to have a denominator of 30, we get:

Adding these together, we have:

10/30 + 6/30 + 15/30 = 31/30

This result simplifies to 1 1/30 gallons. Therefore, Neil has a little more than one gallon of paint in total.

Answer: Third option; "travels slower in solids because particles are close together"

Explanation: While in the air the particles are very freely and the sound does not need to do a lot of work, in a dense material this is not the case. The particles are more close together, and this causes that in a fixed distance, the sound wave needs to move more particles in a solid than in the air. This makes the sound wave to travel slower when it is in a solid.

Final answer:

Thermal energy is required to initiate the distillation process, which separates salt from water by heating the mixture until the water boils and evaporates.

Explanation:

The type of energy required to begin the distilling process for separating salt from water is thermal energy. To start distilling the mixture, you would typically use a heating source, like a Bunsen burner, to provide enough energy to bring the water to its boiling point. During the distillation process, the heat causes the water to evaporate, leaving the salt behind. The vapor then condenses back to liquid form when it comes into contact with a cool surface, and it is collected separately as distilled water while the salt remains in the distillation flask.

Final answer:

The type of energy required to begin the distilling process is heat energy. Distillation involves boiling a liquid and then condensing the resulting vapor to collect the liquid elsewhere. In the case of salt dissolved in water, heating the flask causes the water to boil and separate from the salt.

Explanation:

The type of energy that is required to begin the distilling process is heat energy.

Distillation is a process that involves boiling a liquid and then condensing the resulting vapor to collect the liquid elsewhere. In the case of salt dissolved in water, heating the flask causes the water to boil and the vapors to travel to a cool surface where they condense, leaving behind the solid salt. This process requires the input of heat energy to raise the temperature of the water to its boiling point and initiate the distillation.

For example, if a Bunsen burner is used to heat the flask, initially heating the flask at a rapid rate is required to reach the boiling point of the water. Once boiling or foaming occurs, the burner can be turned down or waved to moderate the distillation rate.