The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3

Answers

Answer 1

Answer:

1) M = 11.3V2) 195.49 grams

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.


Related Questions

g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.

Answers

Answer:gamma ray

Explanation:

A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination.​ Question 96 options:

Answers

Answer:

"Endoscope" is the correct answer.

Explanation:

A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.

You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *

Answers

Answer:

 cost = $ 243.00

Explanation:

This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m

                 #_tables = 3 m (1 / 0.20 m)

                #_tables = 15 tables

Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?

              Cost = 15 (16.20 / 1)

              cost = $ 243.00

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment

Answers

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification

Answers

Answer:

The magnification is  [tex]m = 12[/tex]

Explanation:

From the question  we are told that

   The object distance is [tex]u = 36.2 \ cm[/tex]

     The focal length is  [tex]v = 39.5 \ cm[/tex]

From the lens equation we have that

         [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]

=>     [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]

substituting values

       [tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]

       [tex]\frac{1}{v} = -0.0023[/tex]

=>   [tex]v = \frac{1}{0.0023}[/tex]

=>   [tex]v =-433.3 \ cm[/tex]

The magnification is mathematically represented as

         [tex]m =- \frac{v}{u}[/tex]

substituting values

        [tex]m =- \frac{-433.3}{36.2}[/tex]

         [tex]m = 12[/tex]

         

a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g

Answers

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.

a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above

Answers

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.

Answers

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

       

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Learn more about the average density here:

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Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?

Answers

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]

(b) The angular velocity of disk 2 is :

[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]

(c) The moment of inertia for the two disk system is given by :

[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]

Hence, this is the required solution.

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.

Answers

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?

Answers

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?

Answers

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]

So, the force has a magnitude of 4212 N

What is the difference between matter and energy

Answers

Answer:

Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.

Explanation:

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

Answers

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA

Answers

Answer and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?

Answers

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

pls give brainliest

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      [tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]

Substitute,  [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]

                   [tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]

Hence, The new length of aluminum rod is 17.435 cm.

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Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg

Answers

Answer:

They both have the same acceleration

A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______

a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.

Answers

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz

Answers

Answer:

380 kHz

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

where v is the speed of a wave

λ is the wavelength of the wave

f is the frequency of the wave

f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ 380 kHz

Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?

Answers

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?

Answers

Answer:

The wavelength is  [tex]\lambda = 589 nm[/tex]

Explanation:

From the question we are told that

    The  distance of the mirror shift  is  [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]

      The number of fringe shift is  n =  792

       

Generally the wavelength producing this fringes is mathematically represented as

               [tex]\lambda = \frac{ 2 * k }{ n }[/tex]

substituting values

              [tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]

             [tex]\lambda = 5.885 *10^{-7} \ m[/tex]

            [tex]\lambda = 589 nm[/tex]

Calculate the density of the following material.

1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³

Answers

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

0.179 kg/m³

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Density = 0.179 kg/m³

Hope this helps you

1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field

Answers

Answer:

The electric field is [tex]E = 5.25 V/m[/tex]

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]

Generally the peak magnitude of the electric field is mathematically represented as

         [tex]E = c * B[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

       [tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]

       [tex]E = 5.25 V/m[/tex]

The peak magnitude of the electric field will be "5.25 V/m".

Magnetic field

According to the question,

Magnetic field's peak magnitude, B = 17.5 nT or,

                                                           = 17.5 × 10⁻⁹ T

Speed of light, c = 3.0 × 10⁸ m/s

We know the relation,

→ E = c × B

By substituting the values, we get

      = 3.0 × 10⁸ × 17.5 × 10⁻⁹

      = 5.25 V/m

Thus the above approach is appropriate.

Find out more information about magnetic field here:

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Please help!
Much appreciated!​

Answers

Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force

Answers

Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm

Answers

Answer:

a

   [tex]F = 67867.2 \ N[/tex]

b

  [tex]\Delta L = 2.6 \ mm[/tex]

Explanation:

From the question we are told that

      The Young modulus is  [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]

      The stress is  [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]

      The  diameter is  [tex]d = 2.40 \ cm = 0.024 \ m[/tex]

The radius is mathematically represented as

       [tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]

The cross-sectional area is  mathematically evaluated as

        [tex]A = \pi r^2[/tex]

         [tex]A = 3.142 * (0.012)^2[/tex]

        [tex]A = 0.000452\ m^2[/tex]

Generally the stress is mathematically represented as

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

=>     [tex]F = \sigma * A[/tex]

=>    [tex]F = 1.50 *10^{8} * 0.000452[/tex]

=>    [tex]F = 67867.2 \ N[/tex]

Considering part b

      The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]

Generally Young modulus is mathematically represented as

           [tex]E = \frac{ \sigma}{ strain }[/tex]

Here strain is mathematically represented as

         [tex]strain = \frac{ \Delta L }{L}[/tex]

So    

       [tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]

        [tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]

=>     [tex]\Delta L = \frac{\sigma * L }{E}[/tex]

substituting values

       [tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]

       [tex]\Delta L = 0.0026[/tex]

Converting to mm

      [tex]\Delta L = 0.0026 *1000[/tex]

      [tex]\Delta L = 2.6 \ mm[/tex]

Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R

Answers

Answer:

The value of resistance of each resistor, R is 2.25 Ω

Explanation:

Given;

voltage across the three resistor, V = 1.5 V

power dissipated by the resistors, P = 3.00 W

the resistance of each resistor, = R

The effective resistance of the three resistors is given by;

R(effective) = R/3

Apply ohms law to determine the current delivered by the source;

V = IR

I = V/R

I = 3V/R

Also, power is calculated as;

P = IV

P = (3V/R) x V

P = 3V²/R

R = 3V² / P

R = (3 x 1.5²) / 3

R = 2.25 Ω

Therefore, the value of resistance of each resistor, R is 2.25 Ω

Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have

Answers

Answer:

The planet that is farther from the star must have a time period greater.

Explanation:

We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:

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

Where:

r: is the radius of the circular orbit of the planet and the star

T: is the period

G: is the gravitational constant

M: is the mass of the planet

From equation (1) we have:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex]   (2)          

Where k is a constant

From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.

I hope it helps you!

"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"

Answers

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

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