A force of 8 N is used to drag a chair 2.5 metres across a room. Calculate the work done to move the chair.

Answers

Answer 1

Answer:

Explanation:

GIVEN

Force (F) = 8 N

Distance (d) = 2.5 metres

Work done = ?

WE know we have the formula

work done = F * d

Work done = 8 * 2.5

                   = 20 Joule

Hope it helps :)

Answer 2

Work is measured in Joule.  The work done by the given chair is 20 J.

Work:

It is defined as the amount change in the energy. It is the product of force applied and displacement due to force.

Formula for work,

W = F x d

Where,

F - Force applied = 8 N  

d - Distance = 2.5 meters  

W - Work done = ?

Put the values in the formula,

[tex]\bold {W = 8\times 2.5 }\\\\\bold {W = 20\ J}[/tex]

Therefore, the work done by the given chair is 20 J.

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Related Questions

Which object has the most momentum if they were all traveling 50 mph? a) A bicycle
b) A car c) A bus d) A train

Answers

Train has the most momentum if they were all traveling 50 mph

Explanation:

A train can have the most momentum even if it is moving slowly because it has a large mass if they were all traveling 50 mph.In simple words, momentum can be described as the amount of motion of a moving body, estimated as a product of its mass and velocity.Momentum= mass x velocity

Study the flowchart. The flowchart shows you all the steps to building a good working model of something. What steps lead to making improvements to a model?

Answers

testing the model and analyzing the results of the tests

Explanation:

One example of a physical change is
a. burning paper.
b. baking cookies.
c. the rusting of iron.
d. mixing a milkshake.

Answers

Answer:

Mixing a milkshake

Explanation:

Becuse it’s physics becuse your using muscle and moving it and changing it by force.

An aircraft as loaded weighs 4,954 pounds at a CG of 30.5 inches. The CG range is 32.0 inches to 42.1 inches. Find the minimum weight of the ballast necessary to bring the CG within the CG range. The ballast arm is 162 inches.

Answers

Answer:

57.16

Explanation:

Final answer:

The minimum weight of the ballast required to bring the CG of the aircraft within the given range is approximately 15.2 lbs. This is calculated using the weight of the aircraft, the distances the CG needs to move and the distance from the CG to the location where the ballast is placed.

Explanation:

The subject matter of this problem is physics, specifically dealing with the concept of Centres of Gravity (CG). In order to find the minimum weight of the ballast required to bring the CG within the given CG range, we first need to establish how far the current CG is from the lower limit of the range. That is 32.0 - 30.5 = 1.5 inches. We can assume that the weight required to move the CG 1 inch can be found using the formula: (Weight of aircraft X Distance CG needs to move) / Distance from CG to ballast. So, in this case, the minimum ballast weight will be: (4954 lbs * 1.5 inches) / (162 inches - 30.5 inches). Carrying out the calculations, this gives us a ballast weight of approximately 15.2 lbs.

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How can you connect four of them to produce an equivalent resistance of 0.25 kω?

Answers

Answer:

Question incomplete

This is the complete question

You have a collection of 1.0 kΩ resistors.

How can you connect four of them to produce an equivalent resistance of 0.25 kΩ?

Explanation:

Given that,

We have collection of resistor

R = 1.0 kΩ.

And we want to design an equivalent resistance of

R = 0.25kΩ

We know that,

Series connection of add up resistance and this increases the value of resistance, I.e

Req = R1 + R2 + R3 +.....

While

Parallel resistance add up the reciprocal of resistance and this reduces resistance

So, 1/Req = 1/R1 + 1/R2 + 1/R3 +......

Now, we want to design 0.25kΩ

Connect four of the resistor in parallel will give a equivalent resistance of 0.25 kΩ

1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4

R1 = R2 = R3 = R4 = 1.0 kΩ

1/Req = 1/1 + 1/1 + 1/1 + 1/1

1/Req = 1+1+1+1

1/Req = 4

Taking reciprocal of both sides..

Req = ¼ kΩ

This is the required equivalent resistance

So, connect the four of the resistor in parallel will give the required 0.25 kΩ resistance

A trough is filled with a liquid of density 835 kg/m3. The ends of the trough are equilateral triangles with sides 4 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)

Answers

Final answer:

The hydrostatic force on one end of the trough can be found using the formula F = pghA, where p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. In this case, the ends of the trough are equilateral triangles with sides 4m long.

Explanation:

To find the hydrostatic force on one end of the trough, we can use the formula F = pghA, where F is the force, p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. Since the ends of the trough are equilateral triangles, each side has a length of 4m. The area of an equilateral triangle is given by A = sqrt(3)/4 * s^2, where s is the length of the side. In this case, A = sqrt(3)/4 * (4m)^2 = 4sqrt(3)m^2. Plugging in the values, we have F = (835kg/m^3)*(9.8m/s^2)*h*(4sqrt(3)m^2).

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at constant speed, it would pass 24000 km from the center of Earth. However, it will be attracted to Earth and might hit our planet. What is the minimum speed the asteroid should have so it will just graze the surface of the Earth?

Answers

the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

To calculate the minimum speed the asteroid should have so it will just graze the surface of the Earth, we can use the concept of gravitational potential energy and kinetic energy.

At the point where the asteroid just grazes the surface of the Earth, its gravitational potential energy will be converted entirely into kinetic energy. The minimum speed required will ensure that this kinetic energy is just enough to overcome the gravitational attraction and reach the Earth's surface.

The gravitational potential energy (U) of an object at a distance (r) from the center of the Earth is given by the formula:

[tex]\[ U = -\frac{GMm}{r} \][/tex]

Where:

- ( U ) is the gravitational potential energy,

- ( G ) is the gravitational constant [tex](\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]

- ( M ) is the mass of the Earth[tex](\( 5.972 \times 10^{24} \, \text{kg} \)),[/tex]

- ( m ) is the mass of the asteroid (which cancels out in this context), and

- ( r ) is the distance from the center of the Earth.

The kinetic energy (K) of an object moving at a speed (v) is given by the formula:

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

At the point of grazing, the gravitational potential energy is equal to the kinetic energy:

[tex]\[ -\frac{GMm}{r} = \frac{1}{2} mv^2 \][/tex]

The mass of the asteroid (m) cancels out from both sides of the equation. We can rearrange this equation to solve for the minimum speed (v):

[tex]\[ v = \sqrt{\frac{2GM}{r}} \][/tex]

Substitute the known values:

[tex]- \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),\\- \( M = 5.972 \times 10^{24} \, \text{kg} \), and\\- \( r = 24000 \, \text{km} = 24000 \times 10^3 \, \text{m} \).[/tex]

[tex]\[ v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{24000 \times 10^3}} \][/tex]

Calculate the minimum speed (v):

[tex]\[ v \approx \sqrt{8.89 \times 10^8} \]\[ v \approx 9426 \, \text{m/s} \][/tex]

So, the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

Look at the diagram showing resistance and flow of electrons. A top box labeled X contains 2 circles with plus signs and 2 circles with minus signs. A bottom box labeled Y contains 4 circles with minus signs and 8 circles with plus signs. An arrow Z runs from the bottom box to the top box. Which labels best complete the diagram?

Answers

Answer:

The answer is D.  The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Explanation:

The electrons will flow from an area of high potential energy to an area of low potential energy.

What is potential energy?

Potential energy is the energy that an item retains as a result of its location in relation to other objects, internal tensions, electric charge, or even other elements. Although it has connections towards the Greek philosopher Aristotle's notion of potentiality.

The gravitational potential energy of such an item, the elastic potential energy of a stretched spring, as well as the electric potential energy of such an electric charge inside an electric field are examples of common forms of potential energy. The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Therefore, the electrons will flow from an area of high potential energy to an area of low potential energy.

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use the image to answer questions about the right-hand rule

Answers

I would love to help

Answer:

The current is flowing To The Left. The magnetic field is flowing out of the Bottom of the screen and into the Top of the screen.

A ray diagram is shown.
Which letter represents the location of the image produced by the lens?

W
X
Y
Z

Answers

Answer:

X

Explanation:

X represents the location of the image produced by the lens.

Where is the location of an object to produce an image?

A converging lens produced a virtual image when the object is placed in front of the focal point. For such a position, the image is magnified and upright, thus allowing for easier viewing.

Which of the following represents an image that is located behind a mirror?

Virtual images are always located behind the mirror. Virtual images can be either upright or inverted. Virtual images can be magnified in size, reduced in size, or the same size as the object. Virtual images can be formed by concave, convex, and plane mirrors.

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What two-word term describes an event when melting ice piles up on a lakeshore after being shoved there by the wind?

Answers

Answer:

The two word term is "ICE SHOVES"

Explanation:

An ice shove also known as ice surge, ice heave, ivu, or shoreline ice pileup is simply a surge of ice which occurs from an ocean or large lake and drops on the shore. These ice shoves are usually caused by ocean currents, strong winds, or temperature differences which tend to push the ice to the shore. The ice shoves can create ice piles up to 12 metres high and more.

Ice shoves has been recorded in at least three lakes in the past years. They are Lake Dauphin (Manitoba, Canada), Mille Lacs Lake (Minnesota), Lake Winnebago (Wisconsin), etc.

Answer:

ICE SHOVE

Explanation:

An Ice Shove is a situation or phenomenon that occurs when the wind carries or pushes melting ice from frozen lakes down to the lake shore.

Agents involved in the pushing of melting ice down to the lake shore includes: currents in the ocean, very strong winds as well as variations and fluctuations in the temperature of the lake.

Ice shoves can result in a pile up and accumulation of ice on the lake shore to be at about 30-40 feet high.

Ice shoves can result in negative effects in the lives of living organisms. Such negative effects includes:

a. Damages to the shoreline of the lakes

b. Damages to the habitats of any living organisms by the lake.

c. Damages to structures (buildings) that can be found at the lake shore.

A guitar string produces 3 beats/s when sounded with a 352-hz tuning fork and 8 beats/s when sounded with a 357-hz tuning fork.

Answers

Explanation:

The fluctuating sound heard when two objects vibrate with different frequencies is called beats. It is given that guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beats/s when sounded with a 357 Hz tuning fork.

It is assumed to find the vibrational frequency of the string.

For 3 beats/s, beat frequency can be :

352 - 3 or 352 + 3 = 349 Hz or 355 Hz

For 8 beats/s, beat frequency can be :

357 - 8 or 357 + 8 = 349 Hz or 365 Hz

It means that the vibrational frequency is 349 Hz.

A 14kg Box rests on a frictionless surface. It is attached to a 8kg weight by a thin wire that passes over a frictionless pulley. The pulley is a uniform solid disk of mass 3kg and diameter 1m. After the box is released find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force on the pulley

Answers

Answer:

(a) Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component  =130.19 N

Explanation:

Refer attached figure for details.

[tex]T_1\ \&\ T_2\ are\ tensions\ in\ the\ string\ and\ a\ is\ the\ acceleration\ of\ the\ masses.[/tex]

Applying Newton's 2 law of motion for 14 kg block in horizontal direction

[tex]T_1 = 14\ a[/tex]-----------(i)

Similarly, applying Newton's 2 law of motion for 8 kg block in vertical direction

[tex]8 g - T_2 =8 a[/tex]-----(ii)

Consider the case of pulley,

[tex]\tau_e_x_t= I\alpha--------(iii)\\\\Where,\\\tau_e_x_t =Torque\ acting\ on\ the\ pulley\\I=moment\ of\ inertia\ of\ pulley\\\alpha= angular\ acceleration[/tex]

where,

[tex]I= \frac{MR^2}{2} (for\ pulley\ disk)[/tex]

[tex]I=\frac{3\cdot0.5^2}{2} =0.375\ kgm^2[/tex] (since mass of the pulley = 3 kg & Radius = 0.5 m)

&[tex]\tau_e_x_t= Net\ force \cdot Distance\ from\ application\ point[/tex]

Hence [tex]\tau_e_x_t = (T_2-T_1) \cdot \frac{1}{2} = 0.375\cdot\alpha[/tex]

[tex]T_2-T_1=0.75\cdot\alpha[/tex]--------(iv)

Relation between linear acceleration (a) and angular acceleration (α) is as follows,

[tex]a = R\alpha=0.5\cdot\alpha \ (R\ is \ radius\ of \ pulley)[/tex]

[tex]\alpha=2a[/tex]--------------------(v)

Putting the value of (v) in to (iv)

[tex]T_2 -T_1= 1.5 a[/tex]---------(vi)

adding equation (i),(ii) & (vi) gives

8g =22 a + 1.5 a

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

now putting the value of a in equation (i) & (ii) we get

[tex]T_1=46.706 N[/tex]

[tex]T_2=46.706 +1.5 \cdot 3.336[/tex] = 51.710 N

(a) Hence Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component = [tex]T_2+8\cdot g[/tex] =51.710 + 8 x 9.81 =130.19 N

Answer:

Tension, T1 = 46.2 N and T2 = 52 N, where as acceleration = 3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

Explanation:

Given:

Mass of the box on rest,[tex]m_1[/tex] = 14 kg

Mass of the attached box,[tex]m_2[/tex] = 8 kg

Mass of the pulley, [tex]m_3[/tex] = 3 kg

Diameter of the pulley, [tex]d[/tex] = 1 m

Radius of the pulley, [tex]r[/tex] = 0.5 m

Here we will be using the concept of net force ([tex]F_n_e_t[/tex]),net torque ([tex]\tau_n_e_t[/tex]) and acceleration of the pulley .

A FBD is attached with.

Lets find the tension on the wire using Fnet.

⇒ [tex]T_1=m_1(a)[/tex]                                                                                       ...for m1

⇒ [tex]m_2g-T_2=m_2(a)[/tex] can be written as [tex]T_2=m_2g-m_2(a)[/tex]                ...for m2

Considering clockwise torque as negative and anticlockwise torque as positive.

Moment of inertia (I) of the disk/pulley = [tex]\frac{m_3r^2}{2}[/tex] and [tex]\alpha=\frac{a}{r}[/tex] .

Now using net torque on the pulley we can say that.

⇒ [tex](T_2-T_1)r=I\alpha[/tex]

⇒ [tex](T_2-T_1)r=\frac{m_3r^2}{2}\times \frac{a}{r}[/tex]

⇒ [tex](T_2-T_1)=\frac{m_3a}{2}[/tex]

⇒ Plugging T1 and T2 .

⇒ [tex]m_2g-m_2a-m_1a=\frac{m_3a}{2}[/tex]

⇒ Isolating a from the rest.

⇒ [tex]m_2g=\frac{m_3a}{2}+m_2a+m_1a[/tex]

⇒ [tex]m_2g=a\ [\frac{m_3}{2}+m_2+m_1][/tex]

⇒ [tex]\frac{m_2g}{\frac{m_3}{2} +m_2+m_1} =a[/tex]

⇒ Plugging the numeric value

⇒ [tex]\frac{(8\times 9.8)}{(\frac{3}{2} +8+14)} =a[/tex]

⇒ [tex]3.3 =a[/tex]

⇒ Acceleration = 3.3 [tex]ms^-^2[/tex]

So,

(a).

Tension in the wire

⇒ [tex]T_1=m_1(a)=14\times 3.3 =46.2\ N[/tex]

⇒ [tex]T_2=m_2g-m_2(a)=8(9.8-3.3)=52\ N[/tex]

(b).

The acceleration of the box is 3.3 ms^-2.

(c).

Forces on the pulley.

Horizontal force, [tex]P_H[/tex] = [tex]T_1[/tex] = [tex]46.2\ N[/tex]

Vertical force,[tex]P_V[/tex] = [tex]T_2+m_3g[/tex] = [tex]52+3(9.8)[/tex] = [tex]81.4\ N[/tex]

The values are as follows:

Tension as T1 = 46.2 N and T2 = 52 N ,where as acceleration =3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

A steel railroad track has a length of 40 m when the temperature is −5 ◦C. What is the increase in the length of the rail on a hot day when the temperature is 35 ◦C? The linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1 . Answer in units of m.

Answers

Answer:

0.0176m

Explanation:

Given that,

railroad track has a length of 40 m

temperature is T₁ −5 ◦C

temperature is T₂ 35 ◦C

linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1

Lo = 40 m

T₁ = -5° C

T₂ = 35° C

dT = T₂ - T₁

    = 35 - (-5)

     = 40°C

L = Lo*(1+alpha*dT)

dL = Lo*alpha*dT

dT =  40°C

alpah = 11 x 10⁻⁶

Lo = 40 m

dL = 40 × 11 x 10⁻⁶  × 40

 = 0.0176m

Answer:

ΔL = 0.0176m

Explanation:

We are given;

Length of railroad track; L = 40 m

First Temperature; T1 = −5 ◦C

Second temperature; T2 = 35 ◦C

linear expansion coefficient of steel; α = 11 × 10^(−6) (◦C)^(-1)

The increase in length is given by the equation;

ΔL = α•L•ΔT

Where,

α is linear coefficient

L is length

ΔT is change in temperature.

ΔT = first temperature - Second Temperature

Thus, ΔT = 35 - (-5) = 35 + 5 = 40°C

Thus,plugging in relevant values,

ΔL = α•L•ΔT = 11 × 10^(−6)•40•40

ΔL = 0.0176m

Find fq, the vertical force that pier q exerts on the right end of the bridge. express the vertical force at q in terms of m and g

Answers

Answer:

[tex]F_Q=\frac{Mg}{3}[/tex]

Explanation:

We are given that

Length of beam=L

Mass of object=M

We have to find the FQ , the  vertical force that pier exerts on the right end of the bridge .

Torque about P

[tex]F_Q(3L)-MgL=0[/tex]

[tex]F_Q(3L)=MgL[/tex]

[tex]F_Q=\frac{MgL}{3L}[/tex]

[tex]F_Q=\frac{Mg}{3}[/tex]

Hence, the force that exerted pier Q exerts on the right end  of the bridge is given by

[tex]F_Q=\frac{Mg}{3}[/tex]

. An electric sander consisting of a rotating disk of mass 0.7 kg and radius 10 cm rotates at 15 rev/s. When applied to a rough wooden wall the rotation rate decreases by 20%. (a) What is the final rotational kinetic energy of the rotating disk

Answers

The final rotational kinetic energy of the rotating disk is 9.95 J

Explanation:

Given data,

mass 0.7 kg radius 10 cm rotates at the speed of 15 rev/sec

We have the formula,

Kf= 1/2 I ω f²

I=1/2 Mr²2

ωf= (0.8) ωo

substitute in the formula we get

Kf= 1/2 (1/2 MR²2) (0.8) ² ( ωo)²

=(0.16) M R²2 ωo²2

=(0.16)(0.7)(0.10)²(15[2π)] ²

Kf=9.95 J

The final rotational kinetic energy of the rotating disk is 9.95 J

What is true when an object is moved farther from a plane mirror?

Answers

Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

Identify the number of significant figures for the following values: a. 72.1 b. 4,525.25 c. 1.999

Answers

1. 3 significant figures

2. 6 significant figures

3. 4 significant figures

What characteristics would an Earth-like planet need to have to support life?

Answers

Answer:

water, air, vegetation, and light. also a good temperature thats not too hot or too cold

Explanation:

it would need oxygen, trees, places for animals to build habitats, and water. The temperature also cant be too high.

How much work does a 65 kg person climbing a 2000 m high cliff do?

Answers

The answer for the following question is explained below.

Therefore the work done is 130 kilo Joules.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

  W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

  g = 10 m/ s²

h represents the height of the cliff

From the above formula;

  W = 65 × 10 × 2000

 W = 130,000 J

  W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

Question 10 (2 points)
Formula
TK = Tc + 273
Oxygen boils at 90.0 Kelvin. What is the temperature in degrees Celsius?
Units are already included (please just type in the number)
***Round your answer to the nearest whole number ***
**If your answer is negative be sure to include a dash like this -
Blank 1:
Question 11 (2 points)

Answers

the answer is -183 degree celcius

Answer:

90= C+273

C=273-90= - 183

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 5500 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force.What was the tension in the cable when the craft was being lowered to the seafloor?

Answers

Answer:

[tex]T = 12910.5\,N[/tex] for a craft with a mass of 1500 kg.

Explanation:

Let consider that craft has a mass of 1500 kg. The submersible craft is modelled after the Newton's Laws, whose equation of equilibrium is:

[tex]\Sigma F = T - W +F_{D} = 0[/tex]

The tension experimented by the cable while the craft is lowering to the seafloor is:

[tex]T = W - F_{D}[/tex]

[tex]T = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)-1800\,N[/tex]

[tex]T = 12910.5\,N[/tex]

A person walks in the following pattern: 3.0 km north, then 2.1 km west, and finally 4.2 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?

Answers

Answer:

(a) 2.42 km

(b) 119.74°

Explanation:

(a)

The shortest distance from starting to end point is the hypotenuse

[tex]C=\sqrt{a^{2}+b^{2}}[/tex]

Where a is is base and b is height. Substituting 1.2 km for a and 2.1km for b then

[tex]C=\sqrt{1.2^{2}+2.1^{2}}=2.41867732448956 km\approx 2.42 km[/tex]

(b)

The angle is given by the [tex]\theta[/tex] as indicated in the sketch

[tex]\theta=\frac {2.1}{1.2}=60.2551187030578\approx 60.26^{\circ}[/tex]

Towards East it will be 180-60.26=119.74°

2. An N-type sample of silicon has uniform density (Nd = 1019/cm–3 ) of arsenic, and a P-type silicon sample has a uniform density (Na = 1015 /cm–3 ) of boron. For each sample, determine the following: (a) The temperature at which the intrinsic concentration ni exceeds the impurity density by factor of 10.

Answers

Answer: The temperature at which the intrinsic concentration exceeds the impurity density by factor of 10 is 636 K.

Explanation:

The given data is as follows.

         [tex]N_{d} = 10^{19} per cm^{-3}[/tex]

         [tex]N_{a} = 10^{15} per cm^{-3}[/tex]

As we are given that [tex]n_{i}[/tex]exceeds impurity density by a factor of 10.

Therefore,   [tex]n_{i} = 10N_{d}[/tex]

   [tex]10^{20} = 3.87 \times 10^{6} \times T^{\frac{3}{2}}e^({\frac{-7014}{T}})[/tex]

[tex]T^{\frac{3}{2}}e^({\frac{-7014}{T}}) = \frac{10^{20}}{3.87 \times 10^{6}}[/tex]

          T = 1985 K

Also,  [tex]n_{i} = 10N_{d}[/tex]

       [tex]10^{6} = 3.87 \times 10^{16} \times T^{\frac{3}{2}}e^({\frac{-7014}{T}})[/tex]

                T = 636 K

Thus, we can conclude that the temperature at which the intrinsic concentration exceeds the impurity density by factor of 10 is 636 K.

The actual numerical solution for the N-type sample yields a temperature of approximately 277 K, and for the P-type sample, the temperature is approximately 160 K.

To find the temperature at which the intrinsic concentration [tex]\( n_i \)[/tex]exceeds the impurity density by a factor of 10, we use the relation:

[tex]\[ n_i^2 = N_d N_a \times 10^2 \][/tex]

where [tex]\( n_i \)[/tex] is the intrinsic carrier concentration,[tex]\( N_d \)[/tex] is the donor density (for N-type), and [tex]\( N_a \)[/tex] is the acceptor density (for P-type). The intrinsic carrier concentration [tex]\( n_i \)[/tex] is temperature-dependent and can be approximated by:

[tex]\[ n_i(T) = n_i(T_{ref}) \left( \frac{T}{T_{ref}} \right)^{3/2} \exp \left( -\frac{E_g}{2k} \left( \frac{1}{T} - \frac{1}{T_{ref}} \right) \right) \][/tex]

where [tex]\( T \)[/tex] is the absolute temperature, [tex]\( T_{ref} \)[/tex]is a reference temperature (usually 300 K), [tex]\( E_g \)[/tex] is the energy bandgap of silicon at the reference temperature (approximately 1.12 eV for silicon), and [tex]\( k \)[/tex] is Boltzmann's constant [tex](\( 8.617 \times 10^{-5} \) eV/K)[/tex].

For the N-type silicon sample, we have [tex]\( N_d = 10^{19} \) cm\( ^{-3} \)[/tex], and we want to find[tex]\( T \)[/tex] such that[tex]\( n_i = 10 \times N_d \)[/tex]. We can rewrite the equation as:

[tex]\[ n_i^2(T) = (10 \times N_d)^2 \][/tex]

[tex]\[ n_i(T_{ref})^2 \left( \frac{T}{T_{ref}} \right)^3 \exp \left( -\frac{E_g}{k} \left( \frac{1}{T} - \frac{1}{T_{ref}} \right) \right) = 100 \times N_d^2 \][/tex]

We know that[tex]\( n_i(T_{ref}) \)[/tex] is approximately [tex]\( 1.5 \times 10^{10} \) cm\( ^{-3} \) at \( T_{ref} = 300 \)[/tex] K. Plugging in the values, we get:

[tex]\[ \left( \frac{T}{300 \text{ K}} \right)^3 \exp \left( -\frac{1.12 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K}} \left( \frac{1}{T} - \frac{1}{300 \text{ K}} \right) \right) = \frac{100 \times (10^{19} \text{ cm}^{-3})^2}{(1.5 \times 10^{10} \text{ cm}^{-3})^2} \][/tex]

Solving this equation numerically for [tex]\( T \)[/tex] gives us the temperature for the N-type silicon sample.

For the P-type silicon sample, we follow the same steps but with [tex]\( N_a = 10^{15} \) cm\( ^{-3} \)[/tex], and we find the temperature at which [tex]\( n_i = 10 \times N_a \)[/tex]. The equation to solve is:

[tex]\[ \left( \frac{T}{300 \text{ K}} \right)^3 \exp \left( -\frac{1.12 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K}} \left( \frac{1}{T} - \frac{1}{300 \text{ K}} \right) \right) = \frac{100 \times (10^{15} \text{ cm}^{-3})^2}{(1.5 \times 10^{10} \text{ cm}^{-3})^2} \][/tex]

Solving this equation numerically for [tex]\( T \)[/tex] gives us the temperature for the P-type silicon sample.

The actual numerical solution for the N-type sample yields a temperature of approximately 277 K, and for the P-type sample, the temperature is approximately 160 K. These are the temperatures at which the intrinsic carrier concentration exceeds the impurity density by a factor of 10 for each sample.

This diagram shows how loud certain frequencies must be in order for people of different ages to hear them. The needed intensity is called the "hearing level." A graph with age in years on the x axis from 20 to 70 and hearing level in decibels from 30 to 0. There are 5 graphs that are arcs that all start from point (25, 0) and end at points (70, 10), (70, 14), (70, 15), (70, 20), (70, 25) and (67, 35) and are labeled 500 Hertz, 1000 Hertz, 2000 Hertz, 3000 Hertz, 4000 Hertz and 6000 Hertz respectively. According to the graph, what is the maximum age at which a 4000 Hz sound wave with an intensity of 20 dB would be heard? 55 60 65 70

Answers

Answer:

65 years

Explanation:

The arc that ends at 4000 Hz intersects the 20 db hearing intensity at 65 years.

Answer:

65

Explanation:

A 4.00kg counterweight is attached to a light cord, which is would around a spool. The spool is a uniform solid cylinder of radius 8.00cm and mass 2.00kg. (a) What is the net torque on the system about the point O (the origin)? (b) When the counterweight has a speed v, the pulley has an angular speed ω=v/R. Determine the total angular momentum of the system about O. (c) Using the fact that τ=dL/dt and your result from (b), calculate the acceleration of the counterweight. slader

Answers

Answer:

Explanation:

Given that,

Mass of counterweight m= 4kg

Radius of spool cylinder

R = 8cm = 0.08m

Mass of spool

M = 2kg

The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:

Then we have,

τ(net) = R~ × T~

τ(net) = R~•i × mg•j

τ(net) = Rmg• k

τ(net) = 0.08 ×4 × 9.81

τ(net) = 3.139 Nm •k

The magnitude of the net torque is 3.139Nm

b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:

L~ = R~ × m~v + I~ω

L = mRv + MR v

L = (m + M)Rv

L = (4 + 2) × 0.08

L = 0.48 Kg.m

C. τ =dL/dt

mgR = (M + m)R dv/ dt

mgR = (M + m)R • a

a =mg/(m + M)

a =(4 × 9.81)/(4+2)

a = 6.54 m/s

Answer:

a) τnet = 3.1392 N-m

b) L = (0.48 Kg-m)*v

c) a = 6.54 m/s²

Explanation:

Given

m = 4 Kg

R = 8 cm = 0.08 m

M = 2 Kg

a) τnet = ?

b) L = ?

c)  a = ?

Solution

a) We use the formula

τnet = R*m*g*Sin 90°

τnet = 0.08 m*4 Kg*9.81 m/s²*(1)

τnet = 3.1392 N-m

b) We apply the equation

L= R*m*v + R*M*v = R*(m + M)*v

then

L = (0.08 m)*(4 Kg + 2 Kg)*v = (0.48 Kg-m)*v

c) We use the relation

τ = dL/dt = d((0.48 Kg-m)*v)/dt = (0.48 Kg-m)*dv/dt

τ = (0.48 Kg-m)*a

then

τ/(0.48 Kg-m) = a

⇒   a = 3.1392 N-m/((0.48 Kg-m)

a = 6.54 m/s²

what is the speed of a sound wave that takes 0.5 s to travel 750 m?
A) 3,000 m/s
B) 375 m/s
C) 1,500 m/s

Answers

Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

Feel free to message me if you need more help! :)

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same spring but a different mass (mass m2, spring constant k) but the period is twice as long. Compare m2 to m1 (calculate the ratio m2/m1)

Answers

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of [tex]m_2[/tex] would be four times that of [tex]m_1[/tex].

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is [tex]T[/tex], then the angular velocity of the SHM would be

[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].

Assume that the mass starts with a zero displacement and a positive velocity. If [tex]A[/tex] represent the amplitude of the SHM, then the displacement of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{x}(t) = A\sin(\omega\cdot t)[/tex].

The velocity of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)[/tex].

The acceleration of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)[/tex].

Let [tex]m[/tex] represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time [tex]t[/tex] would be:

[tex]\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)[/tex],

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant [tex]k[/tex] will be equal to:

[tex]\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}[/tex].

Since [tex]\displaystyle \omega = \frac{2\pi}{T}[/tex], it can be concluded that:

[tex]\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}[/tex].

For the first mass [tex]m_1[/tex], if the time period is [tex]T_1[/tex], then the spring constant would be:

[tex]\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2[/tex].

Similarly, for the second mass [tex]m_2[/tex], if the time period is [tex]T_2[/tex], then the spring constant would be:

[tex]\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Since the two springs are the same, the two spring constants should be equal to each other. That is:

[tex]\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Simplify to obtain:

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

Why is it important to have only one set of chemical symbols in the world ?

Answers

Answer:

As you learned, scientists standardized the short-hand way we represent elements, by their chemical symbol. One of the main reasons this was developed was because using letters was the easiest way to represent the elements. Another reason that we use chemical symbols is to allow us to write chemical formulas easily

Or

One set of chemical symbols means that every science student and scientist -- no matter what languages they speak -- can identify these chemicals.

I hope this helps

As you taught, scientists standardized the use of an element's chemical symbol as a shorthand for representing it. This was created in part because representing the elements with letters was the simplest method. Utilizing chemical symbols also makes it easier to formulate chemical formulas.

What is a Periodic Table?

All identified chemical elements are arranged in rows (referred to as periods) and columns (referred to as groups) in the periodic table of chemical elements, also known as the periodic table, in ascending order of atomic number.

The periodic table is used by scientists to quickly refer to details about an element, such as its atomic mass and chemical symbol. Scientists can identify trends in element properties like electronegativity, ionization energy, and atomic radius thanks to the periodic table's arrangement.

To get more information about Periodic Table :

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What is the underlying principle of seismograph construction?

Answers

Answer:

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

Explanation:

A seismograph is an instrument used to record earthquake waves

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

The first earthquake waves arrive at a seismograph station, a short time after the earthquake occurs.

Answer:

Weights, vibrating rod, pendulum : sensitive to vibrations.

Explanation:

Seismograph is an instrument used to measure earthquakes by recording seismic waves. It provides us all details about earthquake - centre, time, depth, energy.

The device is sensitive to vibrations. It consists of a vibrating rod connected to a pendulum, that vibrates due to earthquake shaking. The weight is also complementary attached with rotating drum & pen, to record ground motion. The seismograph output is then recorded & processed on paper.

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