Answer:
a) T = 2838.6 N, b) W = 1003.2 J, c) W = 6.22 10⁴ J, d) W = 2.79 10³ J
e) v_f = 2.65 m / s
Explanation:
a) To find the tension of the cable let's use Newton's second law
T - W = m a
T = W + ma
T = m (g + a)
let's calculate
T = 285 (-9.8 - 0.160)
T = 2838.6 N
b) net work is stress work minus weight work
W = F d
W = (T-W) d
W = (m a) d
W = (285 0.160) 22
W = 1003.2 J
c) the work done by the cable
W = T d cos 0
W = 2838.6 22.0
W = 6.22 10⁴ J
d) The work done by the weight
the displacement is upwards and the weight points downwards, so the angle is 180º
W = F. d
W = F d cos 180
W = -285 22.0
W = 2.79 10³ J
e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy
W = ΔK
as part of rest K₀ = 0
W = ½ m v_f²
v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]
v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]
v_f = 2.65 m / s
If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?
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)
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
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.
Learn more about Air here:
https://brainly.com/question/636295
#SPJ2
4. Water stands 12.0 m deep in a storage tank whose top is open to the atmosphere at
1.00 atm. The density of water is given as 1000 kg/m² and some pressure conversion
are 1 Pa = 1 N/m² while 1 atm = 101 325 Pa.
a) What is the absolute pressure at the bottom of the tank?
b) What is the gauge pressure at the bottom of the tank?
[4]
[4]
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]
A scooter is accelerated from rest at the rate of 8m/s
. How long will it take to cover
a distance of 32m?
Explanation:
time=Distance/speed
t=32/8
t=4 seconds
Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . ms
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
A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sled attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
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]
HELPPP PLSS!!!!!!
What is the chemical formula for iodine trichloride?
A. 12C|
B. ICI3
C. 3ICI
D. |1C13
Answer:
B
Explanation:
Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away and the other is 10 blocks away. Which of the following is true? [Note that pitch = frequency.]a) The siren from the closer ambulance sounds higher pitched to you.b) The siren from the farther ambulance sounds higher pitched to you.c) The pitch of the two sirens sounds the same to you.d) The siren from the farther ambulance sounds higher pitched, until the closer ambulance passes you.
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:
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
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.
A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)
Answer:
Distance = speed * time
55*5
275 meters.
The train would have covered a distance of 275 m
What is distance ?
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
learn more about distance
https://brainly.com/question/15172156?referrer=searchResults
#SPJ2
PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
You sit with friends around a campfire, roasting marshmallows. Which
transfer of thermal energy involved in this system is an example of radiation
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.
What is thermal Energy ?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.
To know more about Energy :
https://brainly.com/question/1932868
#SPJ2.
What is the "best" explanation for why the universe is the way it is?
A) god created the universe
B) there is a multiverse and this one happens to be perfect for life.
C) this is the only universe and it happens to be perfect for life.
D) It is all in illusion and none of it exists.
E) none of the above, they are all just guesses.
I know the answer I just wanna see what you guys think.
i will give brainly if you get it right.
Planets closer to a star will have what type of average temperature
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
Explain how the Laws of planetary motion and Newton’s laws allow the hotel to keep moving in space.
Answer:
Explanation:
i am sorry i needed points
a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her energy?
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
Learn more about "Friction" here:
brainly.com/question/2122951
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
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.
What is meant by kinetic friction ?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.
To learn more about kinetic friction, click:
https://brainly.com/question/29100737
#SPJ2
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
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
How many atoms of carbon, C, are in 0.020 g of carbon?
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
Identify the reactants in the combustion of methane: CH4 + O2 CO2 + O°H
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
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]
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume.
Required:
a. Find the highest temperature attained by the gas.
b. Find the lowest temperature attained by the gas.
c. Find the highest pressure attained by the gas.
d. Find the lowest pressure attained by the gas.
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
What is needed to Run A Brushless DC motor
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!!
#CARRYONLEARNING>>> JAZMINE
A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: , where , , , and is not given. Which of the following represents a simplified expression for ?tanθ− agTo avoid making mistakes, the expression should not be simplified until the numerical values are substituted.gsinθ−agcosθThe single equation has two unknowns and cannot be solved with the information given.
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]
Select the correct answer.
Which figure shows a correct pattern of field lines?
A. Figure A
B. Figure B
C. Figure C
D. Figure D
The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
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]
Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
a. Sine wave;
b. Square wave,
c. Triangle wave.
Calculate the period of a waveform with the frequency of:
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
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]
In which states of matter will a substance have a fixed volume?
O A. Liquid and solid
O B. Solid and gas
O C. Plasma and gas
O D. Liquid and gas
Answer:
A. liquid and solid
Explanation:
A local FM radio station broadcasts at a frequency of 100.8 MHz. Calculate the energy of the frequency at which it is broadcasting. Energy
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].
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
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
In the mirror diagram shown, which is the normal?
А
В
С
D
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
