someone painted the building last year.into passive​

Answer:

T

beacuse:

Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

Answer:

Turned up side down

Explanation:

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2π / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

Answer:

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

Answer: Mercury

Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

Answer:

A

Explanation:

if it was B it would say from one to another to another

Answer:Unlike fixed joints or cartilaginous joints, where the bones are connected by either connective tissue or cartilage, the bones in synovial joints are not directly joined by anything, which allows for a much greater range of motion.

Explanation: I got a 100% on my test

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

[tex] F = ma [/tex]

[tex] a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2} [/tex]

Now, we can calculate the final speed of the crate at the end of 10.0 m:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{1} [/tex]                  

[tex] v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s [/tex]    

For the next 10.5 meters we have frictional force:

[tex] F - F_{\mu} = ma [/tex]

[tex] F - \mu mg = ma [/tex]

So, the acceleration is:

[tex] a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2} [/tex]

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{2} [/tex]  

[tex] v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s [/tex]  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

Answer: A or B

Explanation: I’m guessing that they even each other out depending on the incline, gravity will help keeping it in place

Answer:

Explanation:

From the given information:

Let the first weight be [tex]m_ 1[/tex] = 80 kg

The weight of the buddy be [tex]m_2[/tex] = 120 kg

The weight of  Bubba be [tex]m_3[/tex] = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

[tex]x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2[/tex]

The length of the boat be [tex]x_2[/tex] = 4 m

∴

We can find the center of mass of the system by using the formula:

[tex]X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}[/tex]

Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.

Given the following data:

Mass A = 5 lb to kg = 2.27 kg.

Mass B = 20 lb to kg = 9.02 kg.

Force = 3.6 lb to N = 16.02 Newton.

In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.

For cylinder A:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]

For cylinder B:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]

For the belt, we have

[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]

Also, we would determine the angular acceleration of cylinder B:

[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]

Next, we would calculate the forces acting on the cylinders:

[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]

[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]

Next, we would determine the force of static friction:

[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]

From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.

Read more on angular acceleration here: https://brainly.com/question/6860269

Answer:

a)  v = 3.116 m / s,  b)  μ = 1.65 10⁻²

Explanation:

a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = M v₀

final instant. After the crash

        p_f = (M + m) v

the moment is preserved

       M v₀ = (M + m) v

       v = [tex]\frac{M}{M+m} \ v_o[/tex]

let's calculate

      v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]

      v = 3.116 m / s

b) for this part let's use the relationship between work and kinetic energy

        W = ΔK

as the body has its final kinetic energy is zero

the work of the friction forces is

          W = - fr x

the negative sign is because the friction forces always oppose the movement

let's write Newton's second law

Y axis  

        N - W_sled -W_person = 0

        N = mg + M g

        N = (m + M) g

X axis

        fr = ma

the friction force has the expression

        fr = μ N

        fr = μ g (m + M)

we substitute

        - μg (m + M) x = 0- ½ (m + M) v²

          μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]

let's calculate

        μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]

        μ = 0.0165

        μ = 1.65 10⁻²

Answer:

The mass formula is also given as m = F / a If acceleration itself is the gravity, then M = F / g

please mark me as brainleist

Answer:

    w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Explanation:

To solve this exercise, let's use the concept of conservation of the angular number.

We create a system formed by the two discs, in this case the forces last the shock are internal

initial instant .. just before shock

           L₀ = I₀ w₀ + I₁ w₁

instnte final. Right after crash

           L_f = (I₀ + I1) w

angular momentum is conserved

           I₀ w₀ + I₁ w₁ = (I₀ + I₁) w

           w = I₀ w₀ + I₁ w₁ / Io + I1

The moment of inertia of a disk with an axis passing through its thermometric center

          I₀ = ½ m² r₀²

          I₁ = ½ m₁ r₁²

         we substitute

          I₀ = ½ 2.0 0.70²

          I₀ = 0.49 kg m

          I₁ = ½ 2.0 0.5²

          I₁ = 0.25

₁  

let's reduce the magnitudes the SI system

          w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s

          w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s

we will assume that the counterclockwise turns are positive

           w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)

            w = (- 4.696 + 1.0995) 102) / 0.74

            w = -197.75 + 0.25

            w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Answer:

hey mate......looks like the question is incomplete

Answer:

h = 5.38 10¹⁶ m

Explanation:

Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy

starting point. Just before ionization

          Em₀ = U = qE

final point. Right after ionization

           Em_f = K = ½ m v²

Energy is conserved

           Em₀ = Em_f

           q E = ½ m v²

           v² = 2qE / m

Now we can use the relationship between net work and kinetic energy

           W_net = ΔK

net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J

           W - Ф = K_f - K₀

we assume that the electron converts all its initial initial kinetic energy to be zero

           W -Ф = ½ m v² - 0

            W = ½ m v² +Ф

we substitute

           W = 1/2 m 2qE/m + E

           W =  qE +Ф

           W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹

            W = 4.8 10⁻¹³ + 3 10⁻¹⁹

           W = 4.8 10⁻¹³ J

When the electron is in air, its kinetic energy can be transformed into gravitational potential energy

As the electron is in the air, all work is transformed into scientific energy

           W = K

starting point Em₀ = K = W

end point Em_F = U = m g h

energy conservation Em₀ = Em_f

                       W = m g h

                       h = [tex]\frac{W}{mg}[/tex]

let's calculate

                       h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]

                       h = 5.38 10¹⁶ m

Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.

Energy before ionization;

[tex]\rm Em_0 = U = qE[/tex]

Energy after ionization;

[tex]Em_f = K = \frac{1}{2} mv^2[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]

The relationship between net work and kinetic energy;

[tex]W_{net} = \triangle K[/tex]

The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.

[tex]W - \phi = K_f - K_0[/tex]

[tex]W = K_f+ \phi[/tex]

[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]

EMF at starting point;

[tex]\rm Em_0 = K = W[/tex]

EMF at the endpoint;

[tex]\rm Em_F = U = m g h[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]

Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.

To learn more about the electric field refer to the link;

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Answer:

C) Kinetic energy

Explanation:

Since the skateboarder is ALREADY going down the hill, he is using kinetic energy.

Kinetic energy is when you're moving and in motion

Answer:

I’m not okay . Me and my bf are bickering once again

Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer:

Explanation:

this is like rubbing a balloon on your head to make your hair stand up.  Do that to the can.    The balloon is filled , ofc,  and then just rub the balloon on the can.  This will charge the can with static electricity.   :P

Answer:

Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.

The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.

Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.

A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.

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To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

Learn more about work here:

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