A 285-kg load is lifted 22.0 m vertically with an acceleration a=0.160g by a single cable. Determine

(a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the

load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started

from rest.​

Answer:

if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)

Answer:

Explanation:

i am sorry i needed points

Answer:

belpw

Explanation:

The distance prior to the sliding friction dispersing her energy would be:

- The distance will remain unaffected by the sliding friction i.e. 354m

As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]            (∵ Work in -ve denotes it is done opposite to friction)

Given that,

m(mass) [tex]= 50 kg[/tex]

v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]

The coefficient of Kinetic Friction [tex]= 0.01[/tex]

g(gravitational force) [tex]= 9.8 m/s^2[/tex]

Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]

[tex]= 8.33 m/s[/tex]

Now by employing the provided values,

[tex]F =[/tex] μ[tex]mg[/tex]

[tex]= (0.01) (50) (9.8)[/tex]

[tex]= 4.9[/tex]

∵ [tex]F = 4.9 N[/tex]

By using the above expression, we will find the distance;

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]

⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]

⇒ [tex]1734.7225 = 4.9S[/tex]

⇒ [tex]S = 1734.7225/4.9[/tex]

∵ [tex]S = 354 m[/tex]

Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]  [tex]= -[/tex] μmgS

⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]

Thus, the distance will remain unaffected by the sliding friction i.e. 354m

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Answer:

[tex]\mu=0.185[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=17kg[/tex]

Force [tex]F=33N[/tex]

Velocity [tex]v=1.6m/s[/tex]

Distance [tex]d= 9.8m[/tex]

Generally the equation for Work done is mathematically given by

 [tex]W=\triangle K.E+\triangle P.E[/tex]

Where

 [tex]\triangle K.E=(F-F_f)*2[/tex]

 [tex]F_f=F+\frac{\triangle K.E}{d}[/tex]

 [tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]

 [tex]F_f=30.8N[/tex]

Since

 [tex]f = \mu*m*g[/tex]

 [tex]\mu= 30.8/(m*g)[/tex]

 [tex]\mu= 30.8/(17*9.81)[/tex]

 [tex]\mu=0.185[/tex]

Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution

answer:

P = 141.21 N

Explanation:

Given data:

Mass of crate = 50 kg

coefficient  of static friction ( μ ) = 0.25

Calculate minimum horizontal force ( P ) that holds the crate from sliding

∑fx = 0

     = P + Fcos θ - N*sinθ = 0

     = P + 0.25N cos 30° - Nsin30°  = 0

∴ P = 0.2835 N = 0

P - 0.2853 N = 0 ------- ( 1 )

∑fy = 0

     - 50g + Ncosθ + Fsinθ

     - 50*9.81 + Ncos30° + 0.25Nsin30°

∴ N = 494.942 N ----- ( 2 )

input 2 into 1

P - 0.2853 ( 494.942 ) = 0

P = 141.21 N

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.

Here,

Mass of the box, m = 150 lb = 68.1 kg

Coefficient of kinetic friction, μ = 0.45

Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,

F(k) = μmg

F(k) = 0.45 x 68.1 x 9.8

F(k) = 300.32 N

Now, the box sits on a ramp inclined at 60°

Coefficient of kinetic friction, μ = 0.45

The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.

So,

Frictional force, F(k)' = μmgcosθ

F(k)' = 0.45 x M x 9.8 x cos 60

F(k)' = 2.2M

Weight of the box acting horizontally,

W = Mgsinθ

W = M x 9.8 x sin60

W = 8.5M

Therefore, net force,

Fn = W - F(k)'

Fn = 8.5M - 2.2M

Fn = 6.3M

The total force acting on the box is

F = F(k) - Fn

ma = 300.32 - 6.3M

Since, the box is moving with constant speed, the acceleration, a = 0

Therefore,

300.32 - 6.3M = 0

6.3M = 300.32

M = 300.32/6.3

M = 47.7 kg = 105.16 pound

Hence,

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

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Answer:

The surface tension is 190.2 N/m.

Explanation:

Initial radius, r = 4 cm

final radius, r' = 6 cm

Work doen, W = 15 J

Let the surface tension is T.

The work  done is given by

W = Surface Tension x change in surface area

[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

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Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.

CORRECT ME IF IM WRONG!!

>>> JAZMINE

Answer:

[tex]E=6.68\times 10^{-26}\ J[/tex]

Explanation:

Given that,

The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz

We need to find the energy of the wave. We know that,

Energy, E = hf

Put all the values,

[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]

So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].

Answer:

C

Explanation:

The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B

Answer:

[tex]P=217600Pa[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1000kg/m^3[/tex]

Depth of Water [tex]d=12.0m[/tex]

Generally the equation for Pressure is mathematically given by

 [tex]P=\rho gh[/tex]

 [tex]P=1000*9.8*12[/tex]

 [tex]P=117600N/m^2[/tex]

Therefore

Absolute Pressure=P+P'

Where

P=Pressure under water

P'=Atmospheric Pressure

Therefore

 [tex]P_A=P+P'[/tex]

 [tex]P_A=117,600+10^5[/tex]

 [tex]P=217600Pa[/tex]

Answer:

9.6352× 10²⁰ C atoms

Explanation:

From the given information,

The molar mass of Carbon = 12 g/mol

number of moles = 0.020g/ 12 g/mol

number of moles = 0.0016 mol

If 1 mole of C = 6.022 × 10²³ C atoms

∴

0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C

= 9.6352× 10²⁰ C atoms

Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms

Air is classified as a mixture. Option D is the correct answer.

Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.

In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.

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Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

Answer:

a) [tex]T=0.01s[/tex]

b) [tex]T=0.001s[/tex]

c) [tex]T=0.00001s[/tex]

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

[tex]T=\frac{1}{f}[/tex]

Therefore

a)

For

[tex]T=100 Hz[/tex]

[tex]T=\frac{1}{100}[/tex]

[tex]T=0.01s[/tex]

b)

For

[tex]F=1kHz[/tex]

[tex]T=\frac{1}{1000}[/tex]

[tex]T=0.001s[/tex]

c)

For

[tex]F=100kHz[/tex]

[tex]T=\frac{1}{100*100}[/tex]

[tex]T=0.00001s[/tex]

Solution :

Given expression :

[tex]$\mu_k$[/tex]mgcosθ = mgsinθ − ma

Here, g = 9.8 [tex]m/s^2[/tex] , a = 3.60 [tex]m/s^2[/tex] , θ = 27°

Therefore,

[tex]$\mu_k mg \cos \theta = mg \sin \theta - ma$[/tex]

[tex]$\mu_k mg \cos \theta = m(g \sin \theta - a)$[/tex]

[tex]$\mu_k g \cos \theta = (g \sin \theta - a)$[/tex]

[tex]$\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$[/tex]

Mow calculating the coefficient of kinetic friction as follows :

[tex]$\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$[/tex]

[tex]$\mu_k=\frac{9.8 \times \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$[/tex]

[tex]$\mu_k=0.097$[/tex]

Answer:

A. liquid and solid

Explanation:

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

Answer:

The answer is c

Thermal energy moves within the air from the flames to the marshmallow.

Explanation:

Hope it helps

You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.

In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.

Hence option C is correct.

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Answer:

Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

The pitch does not depend on the distance of the object from the observer.

As per the given data

pitch = frequency

Frequency = [tex]f_{0}[/tex]  [tex]\frac{V +- V_{0}}{V +- V_{s}}[/tex]

[tex]f^{'}[/tex] = [tex]f_{0}[/tex]  [tex]\frac{V }{V - V_{s}}[/tex]

Hence, the pitch of the two sirens remains the same for the observer.

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

Answer:

a. 1 Newton = 100000 Dyne

b. a represents acceleration.

Explanation:

Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;

1 Newton = 10⁵ Dyne

1 Newton = 100000 Dyne

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Force = mass * acceleration

[tex] F = ma[/tex]

Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

[tex]d=79.9m[/tex]

Explanation:

From the question we are told that:

coefficient of static friction [tex]\mu=0.38[/tex]

Velocity [tex]v=87.9=>24.41667m/s[/tex]

Generally the equation for Conservation of energy is mathematically given by

 [tex]\mu*mgd = 0.5 m v^2[/tex]

 [tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]

 [tex]d=79.9m[/tex]

Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

Answer:

B

Explanation: