Do charges exit a circuit with less energy than they had when they entered the circuit

Answer:

  see below

Explanation:

The position graph will be a horizontal line (of constant position) located at the negative value representing the position to the left of the origin.

The velocity graph will be a horizontal line at 0, since the object is at rest.

The acceleration graph will be a horizontal line at 0, since the object's velocity is not changing.

Answer:

The S.I unit of G must be m³/kg.s² to keep the equation consistent.

Explanation:

We have the equation:

F = Gm₁m₂/r²

where,

F = Gravitational Force between Two Bodies

G = Gravitational Constant

m₁ = Mass of 1st Body

m₂ = Mass of 2nd body

r = Distance between the Bodies

The S.I Units of the quantities are:

F = Newton = kg.m/s²

m₁ = kg

m₂ = kg

r = meter = m

G = ?

Therefore, to find the units of Gravitational Constant (G), we substitute the known units in the formula:

kg.m/s² = G(kg)(kg)/m²

G = m³/kg.s²

Answer:

then, that IKEA would be the authority on flat-pack ... Hardware for the Hopen Bed, via Swedish Furniture Parts.

Explanation:

Answer:

The bulb with higher temperature(4000 K) will be brighter

Explanation:

From the question we are told that

     The color temperature  for first bulb is  [tex]T_1 = 2000K[/tex]

        The color temperature  for second bulb is  [tex]T_2 = 4000K[/tex]

Generally the emission power of black body radiation is mathematically represented as

     [tex]E = \sigma T^4[/tex]

Where  [tex]\sigma[/tex] is the Stefan-Boltzmann constant  with a value [tex]\sigma = 5.67 * 10^{-8} W m^{-2} K^{-4.}[/tex]

Now for [tex]T_1 = 2000K[/tex]

      [tex]E_1 = 5.67*10^{-8} * (2000)^4[/tex]

       [tex]E_1 = 907.2 \ W/m^2[/tex]

At  [tex]T_2 = 4000K[/tex]

       [tex]E_2 = 5.67*10^{-8} * 4000[/tex]

        [tex]E_2 = 14515.2 \ KW/m^2[/tex]

Looking at the result we got we see that  the emission power  for the higher temperature bulb is higher, this means that its power to emit in the visible spectrum range would be higher  

So the bulb with higher temperature will be brighter

Answer:

 x_total = 4.29m

Explanation:

To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.

Let's start by looking for the angles that the laser refracts

        n₁ sin θ₁ = n₂ sin θ₂

where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333

        θ₂ = sin⁻¹ (n₁  sin θ₁/n₂)

        θ₂ = sin⁻¹ (1 sin 27 / 1,333)

        θ₂ = sin⁻¹ 0.34057

        θ₂ = 19.9º

now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water

               tan θ₁ = y₁ / x₁

               x₁ = y₁ / tan θ₁

               x₁ = 1.49 / tan 27

               x₁ = 2,924 m

Now let's look for the waterfall in the water as far as Robin

             tan θ₂₂ = y₂ / x₂

             x₂ = y₂ / tan θ₂

             x₂ = 3.77 / tan 19.9

             x₂ = 1,364

the distance from the edge of the pool to Robin is

              x_total = x₁ + x₂

              x_total = 2,924 + 1,364

              x_total = 4.29m

Answer:

The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave.

Explanation:

Hope this helps : )

Answer:

About 0.7767 seconds

Explanation:

[tex]\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s[/tex]

Hope this helps!

Answer:

a)A-B =(1,0,-3) +(2,-5,-1) =(3, -5, -4);    

b)B-C =(-2-3, 5-1, 1-1) = (-5,4,0);  

c)-A +(B –C) = (-1,0,3) +(-5,4,0) =(-6,4,3);  

d) =(3,0,-9) –(6,2,2) = (-3, -2, -11);  

e) = (-2,0,6) +(-6,15,3) +(-3,-1,-1) =

d) = (2,0,-6) –(-5*3, 4*3, 0) =  

Explanation:

1)C=k*A = (k*a1, k*a2, k*a3);

2)C=-A = -1*A;

3)C= A+B = (a1+b1, a2+b2, c2+c2);

4)C=A-B =A +(-B);  

5)A+B=B+A;  

6)A+B+C =(A+B)+C =A+(B+C

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

[tex]x=x_o+v_1t[/tex]    (1)

xo = 10 km

v1 = 55km/h

car 2:

[tex]x'=v_2t[/tex]    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]

[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]

The position in which both cars coincides is:

[tex]x=(55km/h)(0.33h)=18.33km[/tex]

Answer: the answer is corect of what everone else said

Explanation:

Answer:

2.in columns

Explanation:

Answer:

Explanation:

In a series connected electrical circuit as also seen in all electrical Circuit the switch allows the flow of Electricity through out the circuit therefore causing the lighting of bulbs and also other electrical appliances to receive electricity.

When the switch is Open the flow of electricity in the circuit is disrupted while when the Switch is closed it allows the flow of electricity seamlessly through out the circuit. there are different types of switches but the most important is the MAIN SWITCH  because it is used to shutoff electricity completely from the entire circuit during emergencies or repairs.

Answer:

Current cannot flow through a circuit when the switch is OPEN

Current flows through a circuit and all of the bulbs are lit only when the switch is CLOSED

Explanation:

Answer:

A.) Work function = 2.3 eV

B.) Max. K.E observed = 1.1 eV

Explanation:

A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV. 

work function (f) is the minimum energy required to remove an electron from the surface of the material.

hf = Ø + K.E (maximum)

Where

h = Plank constant 6.63 x 10-34 J s

Ø = work function

hc/λ = Ø + K.E (max)

(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19

4.58×10^-19 = Ø + 8.8×10^-20

Ø = 4.58×10^-19 - 8.8×10^-20

Ø = 3.7 × 10^-19 J

Converting Joule to eV

Ø = 3.7 × 10^-19/1.6×10^-19

Ø = 2.3 eV

B.) When light of wavelength 362.4 m is used

The maximum K.E observed = incident light K.E - (the work function).

Incident K.E = hf = hc/λ

Incident K.E =

(6.63×10^-34 × 3×10^8)/362.4

Incident K.E = 5.5 × 10^-28J

Let's convert joule to eV

Incident K.E = 5.5×10^-28/1.6×10^-19

Incident K.E = 3.4 × 10^-9

Max. K.E observed = 3.4 - 2.3

Max. K.E observed = 1.1 eV

Answer:

the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg

the initial acceleration is a = 753.47 m/s²

the  an instrument package could they carry a required mass of 4.64 kg

Explanation:

a)  What should be the radius of these balloons so they just hover above the surface of Mars?

Given that :

The density of the Martian atmosphere, ρ = 0.0154 kg/m³

The volume of the sphere, V = (4/3)πr³

The area of the sphere, A = 4πr²

The mass of the balloon is m = (4.60 g/m²)A

m = (4.60×10⁻³ kg/m²)(4πr²)

The formula for the buoyant force is expressed as :

F = ρVg

m×g = ρ×V×g

m = ρ×V

Now;

(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³

r = 3(4.60×10⁻³ kg/m²)/ ρ

r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³

r = 0.896 m

Thus; the radius of the balloon r = 0.896 m

The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)

m = (4.60×10⁻³ kg/m²)(4π×0.896²)

m = 4.64 × 10⁻² kg

b) what would be its initial acceleration assuming it was the same size as on Mars?

The density of the air on earth, ρ = 1.20 kg/m³  

The volume of the balloon is V = (4/3)(π)(0.896 m)³

V = 3.01156 m³

Considering the net force acting on the balloon ; we have

ΣF = ρVg - mg = ma

However; making the initial acceleration  a of the balloon the subject ; we have:

a = (ρVg - mg)/m

a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)

a = 753.47 m/s²

c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?

The volume of the total system is V' = (4/3)π(5r)³

V' =  (4/3)π(5)³(0.896 m)³

V' =376.446 m³

The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)

m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²

m = 1.159587 kg

We can then say that  the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:

F = (m + m')g

ρV'g = (m + m')g

ρV' = (m + m')

Thus; the required mass m' is =  ρV' - m

m'  =  ρV' - m

m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)

m' = 4.64 kg

Thus; the  an instrument package could they carry a required mass of 4.64 kg

Answer:

The direction of the magnetic field should be in the x and y coordinates alone, and their should be no change towards the z coordinate. Also the magnitude of the magnetic field in the z direction would be zero if there is no magnetic field in that coordinate.

The coordinate should be:

B = bi + bj

Answer:

A. 1,000

Explanation:

Kilo is the prefix for thousands, thus, the multiplier that kilo represents is 1,000

A. 1,000

Answer:

B. 1,000

Explanation:

plato

Answer:

Electrons can be made to move from one object to another. However, protons do not move because they are tightly bound in the nuclei of atoms.Static charge occurs when electrons build up on an object. Static charge:

can only build up on objects which are insulators, eg plastic or wood

cannot build up on objects that act as conductors, eg metals

Conductors allow the electrons to flow away, forming an electric current.

When a static charge on an object is discharged, an electric current flows through the air. This can cause sparks. Lightning is an example of a large amount of static charge being discharged.

Explanation:

Answer:

when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.

Explanation:

When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.

Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.

Answer:

The value for [tex]c^2 - 4mk[/tex] is : [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]

The position of the mass (m) after t seconds is:

[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]

Explanation:

The spring constant is :

[tex]F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m[/tex]

The value for [tex]c^2 - 4mk[/tex] is :

= [tex]7^2 - 4(6)(3)[/tex]

= [tex]49 - 27[/tex]

= [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]

The differential equation for this system is :

[tex]m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0[/tex]

[tex]6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0[/tex]

and the auxiliary equation for  this differential equation is :

[tex]6m ^2+ 6m + 3 = 0[/tex]

using the quadratic formula :

[tex]\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]

[tex]\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}[/tex]

=  [tex]\frac{-6 \pm \sqrt{-36} }{12}[/tex]

= [tex]-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i[/tex]

The general solution is :

[tex]x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})[/tex]

Initial conditions : [tex]x(0) = 1 \ m , x' (0) = 0\\[/tex]

[tex]x(0) = ( c_1 cos 0 + c_2 sin 0)[/tex]

[tex]1 = c_1[/tex]

[tex]x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)[/tex]

[tex]x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)[/tex]

[tex]x'(0) = c_2 \frac{\sqrt{36} }{12}-c_1 \frac{1}{2}[/tex]

replacing 1 for [tex]c_1[/tex]

[tex]0 = c_2 \frac{\sqrt{36} }{12} -\frac{1}{2}[/tex]

[tex]c_2 \frac{\sqrt{36} }{12} = \frac{1}{2}[/tex]

[tex]c_2 = \frac{6}{\sqrt{36} }[/tex]

The position of the mass (m) after t seconds is:

[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]

Answer:

As with any wave the speed of sound depends on the medium in which it is propagating.

Explanation:

Answer:

Force, F = 124 N

Explanation:

We have,

Mass of bicycle is 12 kg and mass of rider is 50 kg

Total mass of the system is 12 kg + 50 kg = 62 kg

Acceleration of the system is 2 m/s²

It is required to find the force required to accelerate the system. The force acting on an object is given by :

[tex]F=ma\\\\F=62\ kg\times 2\ m/s^2\\\\F=124\ N[/tex]

So, the force of 124 N is acting on the system.  

The correct answer is C. A new substance forms when the batter is cooked.

Explanation:

When a chemical change occurs the properties, and composition of substances change. This means atoms in the substance re-arrange to form a new substance. This only occurs when there is a chemical change, but not when physical changes occur, indeed a physical change only affects the state of the matter, shape, size, etc.

In the case of the pancake, this is an example of a chemical change because though the process of cooking the pancake changes its composition. Due to this, the properties of the cooked pancake, and the butter are not the same as a new substance forms. Also, in this and most chemical changes, reversibility is not possible, that is why you cannot reverse the process and make the cooked pancake batter once again.

Answer:

C. on edge

Explanation:

Answer:

-In all three parts, the dependent variable was density.  

-The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was mass.

-In Part II involving the inflated and deflated footballs, the independent variable was volume.

-In Part III involving the baseball, the independent variable was material.

Explanation:

correct on edge! hope it helps :)

Answers:

in all three parts, the dependent variable was. density

The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was. mass

In Part II involving the inflated and deflated footballs, the independent variable was. volume

In Part III involving the baseball, the independent variable was .material

Explanation:

Answer: They will repel each other.

Explanation:

Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.

Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

Answer:

Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.

Explanation:

Did it on Egde 2020

Answer:

gravity

Approximately 4.5 billion years ago, gravity pulled a cloud of dust and gas together to form our solar system.

Explanation:

Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

Answer: D

X and Y

Explanation:

X The least amount of magnesium Took in magnesium

Y Slightly less magnesium than the cell Took in magnesium

Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.

Answer:

Cells X and Y

Explanation:

Active transport occurs when a substance moves across a membrane against its concentration gradient.

Cells W and Z were placed in a liquid containing more magnesium than the cells. Magnesium therefore, diffuses passively down it's concentration gradient into the cells.

However, cells X and Y were placed in a solution containing less magnesium than the cells, yet these cells took in magnesium against this concentration gradient. This, shows that active transport had taken place.

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

Answer:

The final angular velocity is  [tex]w_f = 2.1994 rad/sec[/tex]

The angular acceleration is  [tex]\alpha = 1.099 \ rad/sec^2[/tex]

Explanation:

From the question we are told that

    The radius of each  plate is  [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]

    The mass of each  plate is  [tex]m_p = 1 \ kg[/tex]

    The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]

  From the law of conservation of momentum

         [tex]L_i = L_f[/tex]

Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as

          [tex]L_i = I_1 w + 0[/tex]

here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as

              [tex]I_1 = \frac{m_pr^2}{2}[/tex]

      and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state

         [tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as

              [tex]L_f = (I_1 + I_2 ) w_f[/tex]

Where  

           [tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate )   and is equal to  [tex]I_1[/tex]

and  [tex]w_f[/tex] is the final angular speed

So we have  

             [tex]I_1 w = (I_1 + I_2)w_f[/tex]

             [tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]

             [tex]w = 2 * w_f[/tex]

substituting values

             [tex]4.3988 = 2 * w_f[/tex]

            [tex]w_f = \frac{4.3988 }{2}[/tex]

            [tex]w_f = 2.1994 rad/sec[/tex]

The the rotational impulse-momentum theorem can be mathematially represented as

        [tex]\tau * \Delta t = 0.09891[/tex]

Where [tex]\tau[/tex] is the torque  and  [tex]\Delta t[/tex] is the change in time  

So at  [tex]\Delta t = 2 \ sec[/tex]

        [tex]\tau = \frac{0.09891}{2}[/tex]

        [tex]\tau = 0.0995 \ Nm[/tex]

now the angular acceleation is mathematically represented as

         [tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]

substittuting values

           [tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]

           [tex]\alpha = 1.099 \ rad/sec^2[/tex]