Complete the following statement:

Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different:_____

a. volume.
b. specific heat capacities.
c. expansion coefficients.
d. mass.
e. length.

Answer:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ  = 60/(25×9.8)

μ = 60/245

μ = 0.245

Answer:

Here, "v" is the velocity of electron and "V" is the potential.

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.

4kΩ

Given;

internal resistance, r = 1kΩ

current, I = 10mA = 0.01A

Voltage of full scale, V = 50V

Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;

From Ohm's law

V = IR

R = [tex]\frac{V}{I}[/tex]                 [substitute the values of V and I]

R = [tex]\frac{50}{0.01}[/tex]

R = 5000Ω = 5kΩ

The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance([tex]R_{S}[/tex]) and the internal resistance (r) in the circuit. i.e

R = [tex]R_{S}[/tex] + r

[tex]R_{S}[/tex] = R - r                 [Substitute the values of R and r]

[tex]R_{S}[/tex] = 5kΩ - 1kΩ

[tex]R_{S}[/tex] = 4kΩ

Therefore the series resistance is 4kΩ

Answer:

s= 8.28×10⁻¹⁶m

Explanation:

given

V= 2.25×10³V

from conservation of energy

mv²/2=qΔV

v=√(2qΔV/m)

v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)

=√7.9×10¹⁴m/s

=2.8×10⁷m/s

the deflection of electron beam is

S= gt²/2

recall t= d/v

s=g([tex]\frac{d}{v}[/tex])²/2

s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²

s= 8.28×10⁻¹⁶m

Answer:A

Explanation:

Explanation:

Given that,

Gravitational force = 600 N

Frictional force = 25 N

Pulled by the Force = 250 N

We know that,

The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.

The balance equation is along y-axis

The box will not move in y-axis therefore, the net force in the y-axis will be zero.

Hence, The net force in the y-direction will be zero.

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

Answer:

E.true only when no charge is enclosed within the Gaussian surface.

Explanation:

Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

Answer:

Effective resistance of two equal resistors in series is twice that of a single resistor and in essence will reduce the amount of current flowing in the circuit.

Explanation:

When two resistors are connected in series, their effective resistance is the sum of their individual resistances. For example, given two resistors of resistance values R₁ and R₂, their effective resistance, Rₓ is given by;

Rₓ = R₁ + R₂            --------------(1)

If these resistors have equal resistance values, say R, then equation 1 becomes;

Rₓ = R + R

Rₓ = 2R

This means that their effective resistance is twice of their individual resistances. In other words, when two equal resistors are in series, their effective resistance is twice the resistance of each single one of those resistors.

Now, according to Ohm's law, voltage(V) is the product of current (I) and resistance (R). i.e

V = IR

I = [tex]\frac{V}{R}[/tex]

We can deduce that current increases as resistance decreases and vice-versa.

So, if the two equal resistors described above are connected in series, the amount of current flowing will be reduced compared to having just a single resistor.

Answer:

  I = 0.2934 I₀

Explanation:

The expression that governs the transmission of polarization is

         I = I₀ cos² θ

Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of

        I₁ = I₀ / 2

this is the light that enters the second polarizer

        I = I₁ cos² θ  

we substitute

        I = I₀ / 2 cos² 40

        I = I₀ 0.2934

        I = 0.2934 I₀

Answer:

the electric field is pointing horizontal direction and in south direction

Explanation:

In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Answer: 479. 425 N

Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.

It is given by N= m(g+a)

When it is accelerating downward, the scale reading is less than the true weight.

It so given by N = m(g-a)

The answer to the above questions is in the attached photo

Answer:

the scale will read 476.414 N

Explanation:

Weight = 635 N

mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)

mass m = 635 ÷ 9.81 = 64.729 kg

initial acceleration of the elevator a = 2.45 m/s^2

the force produced by the acceleration of the elevator downwards = ma

your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss

apparent weight = weight - ma

apparent weight = 635 - (64.729 x 2.45)

apparent weight =  635 - 158.586  = 476.414 N

Answer:

at an advanced stage of social and cultural development. "a civilized society"

Explanation:

polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)

Answer:

Explanation:

A )

electric field E = V / d where V is potential difference between plates separated by distance d .

putting the given values

E = 370 / .040  V / m

= 9250 V / m

B )

Force on charged particle of charge q in electric field E

F = q E

F = 2.4 x 10⁻⁹ x 9250

= 22200 x 10⁻⁹

= 222 x 10⁻⁷ N .

C ) since field is uniform , force will be constant

work done by electric field putting up this force

= force x displacement

= 222 x 10⁻⁷  x 40 x 10⁻³

= 888 x 10⁻⁹ J

D )

change in potential energy

= q ( V₁ - V₂ )

= 2.40 X 10⁻⁹ x 370

= 888 x 10⁻⁹ J .

(a) The magnitude of electric field in the region between the plates is 9,250 V/m.

(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].

(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].

The given parameters;

(a) The magnitude of electric field in the region between the plates is calculated as;

[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]

(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;

F = Eq

F = 9,250 x 2.4 x 10⁻⁹

F = 2.22 x 10⁻⁵ N

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;

[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]

(d) the change of the potential energy is calculated as;

[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]

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Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

v = 1.7 m/s

Answer:

1.99*10-4sec

Explanation:

Signal propagation speed=0.82∗2.46∗108m/s

d=2000 m

Tp=20000/0.82∗2.46∗108 sec

ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8

= 1.99* 10^-4seconds

Answer:

T = 2.06h

Explanation:

In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:

[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex]         (1)

T: time for a complete orbit = ?

r: radius of the orbit

G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2

Mm: mass of the moon = 7.34*10^22 kg

The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:

[tex]r=R_m+160km\\\\[/tex]

Rm: radius of the moon = 1737.1 km

[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]

Then, you replace all values of the parameters in the equation (1):

[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]

In hours you obtain:

[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]

The time that the Apollo takes to complete an orbit around the moon is 2.06h

Complete Question

The complete question is  shown on the first uploaded image  

Answer:

The electric field at that point is  [tex]E = 7500 \ N/C[/tex]

Explanation:

From the question we are told that  

       The  radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]

        The  radius of the outer circle is  [tex]r_o = 1.20 \ m[/tex]

       The  charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]

      The magnitude of the point charge at the center is  [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 [tex]E = \frac{k * q_c }{x^2}[/tex]

substituting values  

                  [tex]E = \frac{k * q_c }{x^2}[/tex]

where  k is  the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

     substituting values

                  [tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]

                 [tex]E = 7500 \ N/C[/tex]

Answer:

(a) P = 103064 Pa = 103.064 KPa

(b) P = 102476 Pa = 102.476 KPa

Explanation:

(a)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)

Pg = 1764 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1764 Pa

P = 103064 Pa = 103.064 KPa

(b)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)

Pg = 1176 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1176 Pa

P = 102476 Pa = 102.476 KPa

The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.

Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.

In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.

The absolute pressure in the bulb can be calculated using the following formula:

P = P₀ + ρgh

where:

P is the absolute pressure in the bulb,

P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),

ρ is the density of the sauce (1200 kg/m³),

g is the acceleration due to gravity (9.8 m/s²), and

h is the height of the sauce in the tube.

(a) When h = 0.15 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)

P ≈ 1.013 x 10⁵ Pa + 1764 Pa

P ≈ 1.031 x 10⁵ Pa

(b) When h = 0.10 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)

P ≈ 1.013 x 10⁵ Pa + 1176 Pa

P ≈ 1.025 x 10⁵ Pa

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Answer:

A = A₀ ,   w = w₀/√2

Explanation:

This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point

                   [tex]F_{e}[/tex] = m_total a

the elastic force is

                   F_{e} = - k x

acceleration is

                   a = d²x / dt²

we substitute

                   - k x = m_total   d²x / dt²

                     d²x / dt² + (k / m_total) x = 0

we substitute

                     d²x / dt² + (k /2m) x = 0

the solution to this differential equation is

                    x = A cos (wt + Ф)

where

                  w = √ (k / 2m)

to find the constant Ф we use the velocity

                    v = dx / dt = - Aw sin (wt + Ф)

At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0

                   0 = - A w sin Ф

for this expression to be zero the sine must be zero therefore Ф = 0

when replacing

                  x = A cos (wt)

                  w = 1 /√2  √ (k / m)

if we want to relate to the initial movement (before placing the block)

                 w₀ = √ (k / m)

                 w = w₀ /√ 2

The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement

                  A = A₀

the subscript is used to refer to the oscillations before placing the second block

we substitute to have the final equation

                 x = A₀ cos (w₀ t /√2)

                 A = A₀

                 w = w₀/√2

Answer:

The magnitude of the magnetic field will act in a direction towards me.

Explanation:

When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.

Answer:

A. b) Directed towards center

B. [tex]v = 7.854\ m/s[/tex]

C. [tex]a_c = 12.337\ m/s^2[/tex]

D. [tex]w = 1.57\ rad/s[/tex]

Explanation:

The "force" that they feel pressing their backs against the wall is because the reaction to the  centripetal acceleration .

A.

This acceleration has its direction towards the center of the circle. (option b)

B.

Their linear speed can be calculated with the equation:

[tex]v = (\theta/t)*r[/tex]

Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:

[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]

C.

The centripetal acceleration is given by the equation:

[tex]a_c = v^2/r[/tex]

[tex]a_c = 7.854^2/5[/tex]

[tex]a_c = 12.337\ m/s^2[/tex]

D.

Their angular speed is given by the equation:

[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]

Answer:

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.

Friction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.

Pressure exists as force over an area: P=F/A

Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

let's divide P₁ by P₂

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that, we can say:

P₁=(4/0.9025)P₂=4.4P₂ or

P₂=(0.9025/4)P₁=0.23P₁

Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,

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Answer:

[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]

Explanation:

Hello,

In this case, since pressure is defined as the force applied over a surface:

[tex]P=\frac{F}{A}[/tex]

We can associate the force with the weight of the needle computed by using the acceleration of the gravity:

[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]

And the area of the the tip (circle) in meters:

[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]

Thus, the pressure exerted on the record turns out:

[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]

Which is truly a large value due to the tiny area on which the pressure is exerted.

Best regards.

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  [tex]v = 11.76 \ m/ s[/tex]

Explanation:

From the question we are told that

   The  total distance traveled is  [tex]d = 1.2 \ m[/tex]

    The mass of the block is  [tex]m_b = 0.3 \ kg[/tex]

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   [tex]PE = KE[/tex]

Where  PE  is the potential energy which is mathematically represented as

      [tex]PE = m * g * h[/tex]

substituting values

     [tex]PE = 3 * 9.8 * 0.60[/tex]

      [tex]PE = 17.64 \ J[/tex]

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          [tex]KE = \frac{1}{2} * m v^2[/tex]

So

      [tex]\frac{1}{2} * m* v ^2 = PE[/tex]

substituting values  

  =>    [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]

=>       [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]

=>    [tex]v = 11.76 \ m/ s[/tex]

Answer:

a) 3344 N

b) 3344 N

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = 3344 N

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is 3344 N

Answer:

μs = 0.30

μk = 0.24

Explanation:

In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.

You use the following formula to find the coefficient of static friction:

[tex]F_1=\mu_s Mg[/tex]       (1)

F1 = 75N

μs: coefficient of static friction = ?

M: mass of the block = 25.0kg

g: gravitational acceleration = 9.8m/s^2

You solve for μs in the equation (1):

[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]

For the coefficient of kinetic friction you have:

[tex]F_2=\mu_k Mg[/tex]       (2)

F2 = 60N

μk: coefficient of kinetic friction = ?

You solve for μk in the equation (2):

[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]

Then, you have:

coefficient of static friction = 0.30

coefficient of kinetic friction = 0.24