IV. The table below shows how the speed of a car changes with time. 1 What is the acceleration of the car? please help! :(​

Answer:

C) 2.00 m/s^2

Explanation:

F = m*a

150N = 75kg(a)

a = 150N/75kg

a = 2.0m/s²

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

Answer:

[tex]{ \bf{momentum = mass \times velocity}} \\ \\ { \tt{35 = m \times 10}} \\ { \tt{mass = 3.5 \: kg}}[/tex]

Answer:

vận tốc bằng quãng đường chia thời gian

Explanation:

v=s/t

Answer:

It is streamlined

Explanation:

The speed of a car as well as its air resistance and fuel consumption has a lot to do with its shape.

A car that is streamlined glides through air with less resistance, moves with a higher speed and consumes less fuel.

The essence of streamlining is to reduce the aerodynamic drag on the car thereby increasing its speed, decreasing its air resistance and fuel consumption.

Answer:

it is streamlined

Explanation:

Answer:

The accerlation is a derived from the other quantities like change in velocity/time take etc.

Answer:

W = 1.8 J

Explanation:

The amount of work required to move the given charges can be found by using the following formula:

[tex]W = \frac{kq_1q_2}{\Delta r} \\\\[/tex]

where,

W = Work done = ?

k = Colomb's constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 6 μC = 6 x 10⁻⁶ C

q₂ = magnitude of second charge = 4 μC = 4 x 10⁻⁶ C

Δr = change in distance = 18 cm - 6 cm = 12 cm = 0.12 m

Therefore,

[tex]W = \frac{(9\ x\ 10^9\ Nm^2/C^2)(6\ x\ 10^{-6}\ C)(4\ x\ 10^{-6}\ C)}{0.12\ m}[/tex]

W = 1.8 J

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

1. set up the equation

2. multiply the variables

3. analyze your results

La respuesta correcta para esta pregunta abierta es la siguiente.

A pesar de que no anexas opciones o incisos para responder, podemos comentar lo siguiente.

La importancia de las capacidades físico-motriz que se desarrollan en el futbol de salón son determinantes para desarrollar o maximizar las actividades propias de este deporte con objeto de rendir al máximo y aspirar al mejor de los resultados.

Estas capacidades físico-motrices son las que le permiten a un jugador realizar su máximo esfuerzo, mejorar su desempeño físico y conseguir resultados positivos.

Estamos hablando de la fuerza, la velocidad y la resistencia.

La velocidad es la aceleración que el jugador de futbol necesita para aumentar su velocidad de un punto A, a un punto B, en el menor tiempo posible.

La resistencia es la capacidad del jugador de futbol para mantener ese nivel de aceleración y desempeño, sin bajar su rendimiento. Su capacidad física debe ser resistente para ser constante en su rendimiento físico.

La fuerza es la potencia con la que desempeña los movimiento físicos dentro de la cancha.

Answer↷

c) 5N

➦I hope it seemed helpful to you!

Answer:

D. 2.7 N

Explanation:

Applying

F = kq'q/r²................ Equation 1

Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge

From the question,

Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²

F = 26.94×10⁻¹ N

F = 2.694 N

F ≈ 2.7 N

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Answer:

Work input = 1620

Explanation:

Given the following data;

Work output = 810

Efficiency = 50%

To find the work input;

Mathematically, efficiency is given by the formula;

[tex] Efficiency = \frac {Out-put \; work}{In-put \; work} * 100 [/tex]

Substituting into the formula, we have;

[tex] 50 = \frac {810}{WI} * 100 [/tex]

Cross-multiplying, we have;

[tex] 50I = 810 * 100 [/tex]

[tex] 50I = 81000 [/tex]

[tex] WI = \frac {81000}{50} [/tex]

Work input = 1620

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

The work done by friction is 500 J.

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.There are several types of friction:

Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact. Dry friction is subdivided into static friction ("stiction") between non-moving surfaces, and kinetic friction between moving surfaces. With the exception of atomic or molecular friction, dry friction generally arises from the interaction of surface features, known as asperities

Fluid friction describes the friction between layers of a viscous fluid that are moving relative to each other.

Lubricated friction is a case of fluid friction where a lubricant fluid separates two solid surfaces.Skin friction is a component of drag, the force resisting the motion of a fluid across the surface of a body.

Internal friction is the force resisting motion between the elements making up a solid material while it undergoes deformation.

When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat).

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Answer:

500 kW

Explanation:

Given that,

Energy of the train, E = 2 MJ = 2×10⁶ J

Time, t = 4 seconds

We need to find the power of the train. We know that,

Power = energy/time

So,

[tex]P=\dfrac{2\times 10^6}{4}\\\\P=500000\ W\\\\or\\\\P=500\ kW[/tex]

So, the power of the train is equal to 500 kW.

Melting: the substance changes back from the solid to the liquid

Evaporation: the process by which water changed from a liquid to a gas.

Boiling: the process by which a liquid turns into a vapor when it is heated to it's boiling point.

Condensation: the substance changed from a gas to a liquid

Freezing: the substance changed from a liquid to solid.

Answer:

Melting is a process that causes a substance to change from a solid to a liquid.

Evaporation is the process of turning from liquid into vapour.

Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the

Condensation is the process of water vapor turning back into liquid water

Freezing is a phase transition where a liquid turns into a solid when its temperature is lowered below its freezing point

Answer:

valenc e shell

Explanation:

Answer: 100 V

Explanation:

Given

Resistance [tex]R=1\ k\Omega[/tex]

Potential difference [tex]V=100\ V[/tex]

If the resistor is placed across the potential difference, then it receives a voltage of 100 V as it parallel to the source.

Voltage drop across resistor is 100 V.

Solution :

Let us consider the Gaussian surface that is in the form of a cylinder having a radius of r and a length of A which is [tex]$\text{coaxial with the charged cylinder}$[/tex].

The charged enclosed by the cylinder is given by,

[tex]$q=\rho V$[/tex]       (here, V = [tex]$\pi r^2l$[/tex]  is the volume of the cylinder)

  [tex]$=\pi r^2lp$[/tex]    

If [tex]$\rho$[/tex] is positive, then the electric field lines moves in the radial outward direction and is normal to Gaussian surface which is distributed uniformly.

Therefore, total flux through Gaussian cylinder is :

[tex]$\phi=EA_{cyl}$[/tex]

   [tex]$=E(2\pi rl)$[/tex]

Now using Gauss' law, we get

[tex]$2\pi \epsilon_0rlE = \pi r^2lp$[/tex]

or [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Therefore, the electric field is [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Hence, option (d) is correct.

Answer:

Cold blooded animals are those animals that cannot regulate their body temperature and their temperature keeps changing according to their environment. ... Warm blooded animals are the animals that have a constant body temperature and can easily adapt to extreme temperatures as they can regulate their body temperature.

Answer:

B. low melting points and low densities

Explanation:

Alkali metals are any of the monovalent elements found in Group IA of the periodic table. They readily lose their one valence electron to form ionic compounds with nonmetals. Examples of alkali metal are Potassium (K), Lithium (L), and Sodium (Na).

Compared to most other metals, the chemical properties that alkali metals have are low melting points (28.5°C) and low densities that is typically less than 1 grams per cubic centimeters.

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

Answer:

electrons

In substances such as iron, cobalt, and nickel, most of the electrons spin in the same direction. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field, hence it acts as a magnet.

the materials that can be magnetized this way are called ferromagnetic materials, but it is not permanent and after some times it loses all the magnetic properties.

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Magnets are formed from iron, nickel, or cobalt when the electrons in their atoms' outer energy levels line up in the same direction that is present in Option A, as the alignment creates a magnetic field that can attract or repel other magnets or magnetic materials.

At the atomic level, electrons orbit the nucleus in shells or energy levels, and electrons have a property called spin, which creates a tiny magnetic field around them. Normally, electrons in an atom have random spin directions, so their magnetic fields cancel each other out and the atom has no overall magnetic field. In some materials, such as iron, nickel, and cobalt, the electrons in the outer energy level can line up in the same direction, creating a net magnetic field. This is known as ferromagnetism.

Hence, magnets are formed from iron, nickel, or cobalt when the electrons in their atoms' outer energy levels line up in the same direction that is present in Option A,

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question is incomplete, complete question is below

Magnets are formed from iron, nickel, or cobalt when the ______________________ line up in the same direction.

A)electrons in their atoms' outer energy levels

B)proton explodes

Answer:

water

mercury in thermometer

volume gas thermometer

jar lids

power lines