I suppose the hill makes an angle of 5.0° with the horizontal.
• F acts parallel to the road and in the direction of the car's motion, so it contributes a positive amount of work, F (290 m).
• Friction does negative work on the car since it opposes the car's motion. As the car moves up the slope, the work done by friction is (-524 N) (290 m) = -151,960 J.
• The car's weight has components that act parallel and perpendicular to the road. The parallel component has a magnitude of W sin(5.0°) and points down the slope, so it contributes negative work of -(1200 kg) g sin(5.0°) ≈ 1,024.95 J. The perpendicular component of W does not do any work.
• The normal force FN also doesn't do any work to move the car up the slope because it points perpendicular to the road, so we can ignore it, too.
The net work done on the car is then
F (290 m) + (-151,960 J) + 1,024.95 J = 150,000 J
==> F (290 m) ≈ 300,935 J
==> F ≈ (300,935 J) / (290 m) ≈ 1,037.71 N
When an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt). Determine the following quantities. (a) maximum voltage V (b) rms voltage V (c) rms current A (d) peak current A (e) Find the current when t = 0.0045 s.
Explanation:
the answer is in the above image
(a) The maximum voltage V is 190 Volts.
(b) The rms voltage V is 95√2 Volts.
(c) The rms current in Amperes is 7.9 A.
(d) The peak current Amperes is 11.18 A.
(e)The current when t = 0.0045 s is 7.26 A.
What is current?The current is the stream of charges which flow inside the conductors when connected across the end of voltage.
Given is an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt).
(a) From the given voltage equation, maximum voltage V is 190 Volts.
(b) rms voltage =Vmax/√2
Put the values, we get
Vrms = 190/√2 = 95√2 Volts
(c) rms current = Vrms/Resistance
Put the values, we get
Irms = 95√2 /17
Irms = 7.9 Amperes.
(d) peak current =√2 Irms
Substitute the values, we get
Peak current = 7.9 √2 = 11.18 A
(e) The current when t = 0.0045 s is written as
I = (190 V)sin(50πt)/R
Substitute the values, we have
I = (190 )sin(50π x0.0045)/17
I = 7.26 A.
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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
Match the atmospheric energy transfer process that best fits each of the following scenarios:
Warming of the Earth's surface on a sunny day
[ Choose ] convection conduction radiation advection
On a sunny afternoon, you watch cumulus clouds forming
[ Choose ] convection conduction radiation advection
A very shallow layer of air in contact with the ground is warmed
[ Choose ] convection conduction radiation advection
A south wind carries warm air into the central United States
[ Choose ] convection conduction radiation advection
Answer:
a) RADIATION, b) CONVECTION, c) CONDUCTION, d) CONVECTION
Explanation:
In the heating processes there can be three types: conduction, convention and radiation.
The conduction process occurs when the movement of atoms or thermal agitation of molecules creates the transfer of thermal energy.
The convention process occurs when there is a movement of matter creating the transfer of energy
The process of Radiation an electromagnetic wave indexes on a material and is absorbed, creating the process of energy transfer.
Now let's examine each situation>
a)Warming of the Earth's surface on a sunny day
in this case the sunlight heats the earth as it is absorbed, which is why it is a RADIATION process
b) On a sunny afternoon, you watch cumulus clouds forming
in this case the nines rise from the surface, since it is a moving mass, the process is CONVECTION
c) A very shallow layer of air in contact with the ground is warmed
in this case the thermal movement of the layer molecules heat the earth, for which the process of CONDUCTION
d) A south wind carries warm air into the central United States
As we have a movement of a mass of matter, the process is CONVECTION.
Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
Answer:
the law of conservation of energy.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.
The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.
Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.
One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.
Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.
This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.
What happens to the water when you throw rock into a pond
Answer:
The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.
a system absorbs 500J of heat and at the same time 400J of work is done on the system what is change in internal energy
This is the solution of the problem. The problem is calculated from a closed system energy balance equation
State two (2) examples of osmosis occurring in everyday life
Short-term memory is active, while long-term memory is:
A dynamic
B
reflective.
c) passive
D
recessive.
Answer:
the answer is b. reflective
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm.
a) What is the wave?
b) Wrote the wave function, y(x,t)
c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
Explanation:wave=wavelength×frequency,
You are helping to design a new electron microscope to investigate the structure of the HIV virus. A new device to position the electron beam consists of a charged circle of conductor. This circle is divided into two half circles separated by a thin insulator so that half of the circle can be charged positively and the other half can be charged negatively. The electron beam will go through the center of the circle. To complete the design your job is to calculate the electric field in the center of the circle as a function of the amount of positive charge on the half circle, the amount of negative charge on the other half circle, and the radius of the circle.
Answer:
The electric field in the center is [tex]\frac{2k}{\pi R^{2}}\left ( Q_{2}-Q_{1 } \right )[/tex].
Explanation:
The engine in the car from question 1 uses a force of 1200 N to cause the car to accelerate at 3.5 m/s2. What is the car's mass?
Answer:
Mass of car's engine = 342.85 kg (Approx.)
Explanation:
Given values:
Force applied by car = 1,200 N
Acceleration of car' engine = 3.5 m/s²
Find:
Mass of car's engine
Computation:
⇒ Mass = Force / Acceleration
⇒ Mass of car's engine = Force applied by car / Acceleration of car
⇒ Mass of car's engine = 1,200 / 3.5
⇒ Mass of car's engine = 342.85 kg (Approx.)
The first charged object is exerting a force on the second charged object. Is the second charged object necessarily exerting a forcer on the first?
Answer:
Explanation:
Of course because it's Newton's Law that if body A exerts force on body B, then body B will exert equal but opposite force on body A.
HAPPY LEARNING:)
The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction, from that we can understand when force is exerted on first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first.
The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.
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Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very few collisions with other molecules. Express your answer using two significant figures.
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ [tex]v=\frac{502.1}{\sqrt{3} }[/tex]
[tex]=289.9 \ m/s[/tex]
The time will be:
⇒ [tex]t=\frac{d}{v}[/tex]
[tex]=\frac{2\times 6}{289.9}[/tex]
[tex]=\frac{12}{289.9}[/tex]
[tex]=0.041 \ sec[/tex]
hence,
⇒ [tex]N=\frac{1}{t}[/tex]
[tex]=\frac{1}{0.041}[/tex]
[tex]=24.39 \ per \ sec[/tex]
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
Assignment: 06.05 Infections and Health
Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?
Answer:
Rahul's weight
Explanation:
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
Using formulas, Rahul's weight is equal to
W=mg
where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).
A man standing on a frictionless ice throws a 1.00kg mass at 20m/s at an angle elevation of 40.0 degrees. What was the magnitude of the mans momentum immediately after the the throl
Answer:
Explanation:
1.00kg×20m/s×cos40=15.3
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
What is used to represent the magnitude of the force in an FBD?
Answer:
See explanation below
Explanation:
The length of the arrow represents the magnitude/size of the force. The longer it is, the higher the force's magnitude is.
Use a projectile motion kinematic equation for the vertical motion to find the time t that the ball is in the air, from when it leaves the track until it strikes the floor.
Answer:
The time of flight is [tex]T = \frac{2 u sin A}{g}[/tex].
Explanation:
Let the initial velocity is u and the angle of projection is A.
Use first equation of motion for vertical motion
Let the time to reach the maximum height is t.
[tex]v = u - gt\\\\0 = u sin A - gt \\\\t = \frac{ u sin A}{g}[/tex]
Total time of flight is
T =2 t
[tex]T = \frac{2 u sin A}{g}[/tex]
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down? Assume the speed of sound is 343 m/s.
Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
Now imagine that you are a Haitian taptap driver and want a more comfortable ride. You decide to replace the springs with new springs that can handle the typical heavy load on your vehicle. What spring constant do you want your new spring system to have?
The new springs should have a spring constant that is (slighty larger, substantially larger, slightly smaller, substantially smaller) substantially larger slightly larger slightly smaller substantially smaller than the spring constant of the old springs.
Answer:
We use a spring of large spring constant.
Explanation:
The spring constant is defined as the force applied on the spring per unit extension or compression in length.
F = k x
where, F is the force, x is the extension, k is the spring constant.
Its unit is N/m.
To get the comfortable ride, we use the spring of large spring constant, so that the spring gets stiffer and we get comfort.
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
What is the potential energy of a 7kg object 4m off the ground ?
please show your work
Answer:
Gravitational potential energy is mass of the object times the gravitational constant times the height of the object:
U = mgh (I will use 10 for the gravitational constant but you can use 9.8 or 9.81 or something even more accurate)
U = 280
The gravitational potential of the object is 280 joules
What is a reasonable measurement for the distance to Neptune?
30 light years
30 kilometers
30 parsecs
30 Astronomical Units
Answer:
30 kilometers is a reasonable measurement
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
Identifying the factors contributing to and acting as determinant factors of health disparities during the program theory and development process is a means of culturally tailoring the program.
a) true
b) false
Answer:
a) True
Explanation:
A program-specific message provided to an individual or group with the intention of raising awareness of a health condition, motivating behavior change, removing perceived barriers to participating in a health habit, or something else relating to the program's aims and objectives. The most effective intervention messages are usually theory-based and culturally adapted.