Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal

It does not change the chemical composition of water.

Answer:

Explanation:

A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.

If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).

Answer:

mercury and alcohol

ii) used to test temperatures

i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

ii) It does not wet (cling to the sides of) the tube.

Alcohol:

i) Alcohol has greater value of temperature coefficient of expansion than mercury.

ii) it's freezing point is below –100°C.

Answer:

461.73 K

Explanation:

Given that,

The mass of a bullet, m = 5.7 g

Speed of the bullet, v = 490 m/s

Half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree.

Using the conservation of energy,

[tex]\dfrac{1}{2}(\dfrac{1}{2}mv^2)=mc\Delta T\\\\\Delta T=\dfrac{v^2}{4c}[/tex]

Where

x is the specific heat of lead, c = 130 J/kg K

So,

[tex]\Delta T=\dfrac{(490)^2}{4\times 130}\\\\=461.73\ K[/tex]

So, the increase in temperature of the bullet is 461.73 K.

Answer:

false.

Explanation:

We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:

V = λ*f

So, if the velocity is constant and we increase the frequency to:

f' > f

we will have a new wavelength λ'

Such that:

V = f'*λ'

And V = f*λ

Then we have:

f'*λ' = f*λ

Solvinf for λ', we get:

λ' =(f/f')*λ

And because:

f' > f

then:

(f/f') < 1

Then:

λ' =(f/f')*λ < λ

So, if we increase the frequency, we need to decrease the wavelength.

So, for higher frequency waves, we must have proportionally shorter wavelengths.

Then we can conclude that the given statement:

"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"

is false.

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

[tex]F = BIl \ sin(\theta)[/tex]

(a) When the angle, θ = 60 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N[/tex]

(b) When the angle, θ = 90 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N[/tex]

(c) When the angle, θ = 120 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N[/tex]

Answer:

Option B.

Explanation:

Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.

So, when we analyze the problem of "how long takes an object to hit the ground"

We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.

So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.

And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.

So, both objects will hit the ground at the same time.

(notice that here we are ignoring things like air resistance and other complex forces)

So here the correct option is b:  Neither. Their vertical motion is the same, so they will reach the ground at the same time.

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as

[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]

Substituting (2) into (1), we get

[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]

where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as

[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]

From (5), we can solve for N as

[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]

Substituting (7) into (4) we get

[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]

Collecting similar terms together, we get

[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]

or

[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]

Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Answer:

The direction of magnetic field is along + Z axis.

Explanation:

The direction of motion of electron is along y axis.

The magnetic  force is along - X axis.

The force on the charged particle moving in the magnetic field is

[tex]\overrightarrow{F} = q (\overrightarrow{v}\times \overrightarrow{B})\\\\- F \widehat{i} = - q (v \widehat{j}\times \overrightarrow{B})\\[/tex]

So, the direction of the magnetic field is along + Z axis.

Answer:

A. (b)

B. (a)

Explanation:

The electric dipole moment is the product of charge and the length of the dipole.

The torque on the dipole placed in the external electric field is given by

torque = p E sin A

where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.

When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

So, the option (b) is correct.

Teh energy is given by

U = - p  E cos A

When the angle A is zero , the potential energy is negative and it is minimum.

In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:

A) Letter b

B) Letter a

So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:

[tex]torque = p E sin (A)[/tex]

where:

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

Now the energy is given by:

[tex]U = - p E cos (A)[/tex]

We can say that when the angle A is zero , the potential energy is negative and it is minimum.

See more about dipole at brainly.com/question/12757739

Answer:

Electromechanical transducer and Electrical receiver.

Explanation:

Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]

Why is the temperature constant during the melting of water?

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{ REASON}}}}}}}[/tex]

THE HEAT WE R SUPPLYING TO THE WATER TO RAISE THE TEMP OF THE WATER IS USED BY THE MOLECULES TO BREAK INTERMOLECULAR BONDS WHICH HELP IN THE CHANGING OF THE LATTICE (STRUCTURE) OF THE WATER .

ICE HAS A HEXAGONAL RING LIKE STRUCTURE WHICH IS CONVERTED INTO REGULAR CRYSTALLINE STRUCTURE WHICH CAN ONLY BE FORMED WITH THE HELP OF FORMATION OF NEW BONDS AND BREAKDOWN OF OLDER ONES

THE AMOUNT OF ENERGY WHICH IS USED IN CONVERSATION OF THE STATE OF FROM SOLID TO LIQUID IS KNOWN AS LATENT HEAT OF FUSION.

SO TEMP REMAIN CONSTANT DURING CHANGE IN STATE .

[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

Answer:

increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.

Explanation:

Answer:

a) ΔT₁ = -4.68 N,   ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N

Explanation:

In this exercise we will use Newton's second law.

         ∑F = m a

Let's start with the set of three cars

         F_total = M a

         F_total = M 0.12

where the total mass is the sum of the mass of each charge

          M = m₁ + m₂ + m₃

This is the force with which the three cars are pulled.

Now let's write this law for each vehicle

car 1

         F_total - T₁ = m₁ a

         T₁ = F_total - m₁ a

car 2

         T₁ - T₂ = m₂ a

         T₂ = T₁ - m₂ a

car 3

         T₂ = m₃ a

note that tensions are forces of action and reaction

a) They tell us that 39 kg is removed from car 2 and placed on car 1

         m₂’= m₂ - 39

         m₁'= m₁ + 39

         m₃ ’= m₃

they ask how much each tension varies, let's rewrite Newton's equations

The total force does not change since the mass of the set is the same F_total ’= F_total

car 1

           F_total ’- T₁ ’= m₁’ a

           T₁ ’= F_total - m₁’ a

           T₁ ’= (F_total - m₁ a) - 39 a

           T₁ '= T₁ - 39 0.12

           ΔT₁ = -4.68 N

car 2

           T₁’- T₂ ’= m₂’ a

           T₂ ’= T₁’- m₂’ a

           T₂ '= (T₁'- m₂ a) + 39 a

           T₂ '= T₂ + 39 0.12

           ΔT₂ = 4.68 N

b) in this case the masses remain

            m₁ '= m₁

           m₂ ’= m₂ - 39

           m₃ ’= m₃ + 39

we write Newton's equations

car 3

          T₂ '= m₃' a

          T₂ ’= (m₃ + 39) a

          T₂ '= m₃ a + 39 a

          T₂ '= T₂ + 39 0.12

          ΔT₂ = 4.68 N

car 1

            F_total - T₁ ’= m₁’ a

            T₁ ’= F_total - m₁ a

car 2

            T₁' -T₂ '= m₂' a

            T₁ ’= T₂’- m₂’ a

            T₁ '= (T₂'- m₂ a) + 39 a

            T₁ '= T₁ + 39 0.12

            ΔT₁ = 4.68 N

The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

Tension is the pulling force carried by the flexible mediums like ropes, cables and string.

Tension in a body due to the weight of the hanging body is the net force acting on the body.

At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.

The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,

[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]

On solving the above 3 equation, we get the values of tension in each bar as,

[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]

The tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]

The tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]

Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

Learn more about the tension here;

https://brainly.com/question/25743940

Answer:

 μ = 0.15

Explanation:

Let's start by using Hooke's law to find the force applied to the block

          F = k x

          F = 87.0 0.065

          F = 5.655 N

Now we use the translational equilibrium relation since the block has no acceleration

          ∑ F = 0

          F -fr = 0

          F = fr

the expression for the friction force is

          fr = μ N

if we write Newton's second law for the y-axis

          N -W = 0

          N = W = mg

we substitute

          F = μ mg

          μ = F / mg

          μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]

          μ = 0.15

Answer: objective lens

Explanation:

Light enters a refra

Light enters a telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it.

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + ua + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u ,  z = - v

Answer:

a)[tex]V=1.067\: m/s[/tex]

b)[tex]v=434.65\: m/s [/tex]  

Explanation:

a)

Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.

[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]

Where:

Then, let's find the initial speed of the bullet-block system.

[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]

[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]

[tex]V=1.067\: m/s[/tex]

b)

Using the conservation of momentum we can find the velocity of the bullet.

[tex]mv=MV[/tex]

[tex]v=\frac{MV}{m}[/tex]

[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]

[tex]v=434.65\: m/s [/tex]  

I hope it helps you!

Answer:

Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.

Explanation:

Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

Answer:

[tex]I_2=30.9A[/tex]

Explanation:

From the question we are told that:

Wire segment [tex]l_s=2.9m[/tex]

Initial Current [tex]I_1=1400A[/tex]

Force [tex]F=2.00N[/tex]

Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]

Generally the equation for Force is mathematically given by

 [tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]

 [tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]

 [tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]

 [tex]I_2=30.9A[/tex]

Answer:

C

Explanation:

20 cm = 0.2m

since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty

therefore it's 0.200m

Answer:

D) the two spheres remain of equal size.

Explanation:

Answer:

The inductance is 17784.96 ohm and rms current is 4.77 mA.

Explanation:

Voltage, V = 120 V

frequency, f = 60 Hz

Inductance, L = 47.2 H

The rms  voltage is

[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]

The reactance is given by

[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]

The rms current is

[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]

Answer:

The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.

Explanation:

The direction of the electric field due to the dipole on the axial line is same as  the direction of dipole moment.

The magnitude of the electric field due to an electric dipole on its axial line is

[tex]E=\frac{2kp}{r^3}[/tex]

where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.

The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.