Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is [tex]N _i = 110 \ turns[/tex]
The radius of inner coil is [tex]r_i = 0.014 \ m[/tex]
The current flowing through the inner coil is [tex]I_i = 9.0 \ A[/tex]
The number of turn of outer coil is [tex]N_o = 160 \ turns[/tex]
The radius of outer coil is [tex]r_o = 0.022\ m[/tex]
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
[tex]B_i = \frac{N_i \mu I}{2 r_i}[/tex]
The magnetic field due to inner coils is mathematically represented as
[tex]B_o = \frac{N_o \mu I_o}{2 r_o}[/tex]
Now for magnetic field at center to be zero
[tex]B_o = B_i[/tex]
So
[tex]\frac{N_i \mu I_i}{2 r_i} = \frac{N_o \mu I_o}{2 r_o}[/tex]
=> [tex]\frac{110 * 9}{2 * 0.014} = \frac{160 *I_o}{2 0.022}[/tex]
[tex]I_o = 9.72 \ A[/tex]
consider a 38 degree F object placed in 68 F room write a differential equation for T thetemperature of the object at time t using k>0 for the constant of proportionality and write your equation in terms of T k and t
Answer:
dY/dt = k(Y - Ys).
Explanation:
Okay, this question based on the law known as the Newton's law of cooling. According to Newton's law of cooling, when an object is placed into a specific or particular environment, the object's temperature will definitely be lower than the temperature of the surrounding or the environment.
So, we are given that the temperature of the object = 38 degree F = Y(o) of the object and the temperature of the environment = 68 degree F = Ys
Hence, dY/dt = k(Y - Ys).
dY/ (Y - Ys) =kdt.
Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (1 ft. thick) hollow spherical shell below your feet. Also suppose there was a small hole in the shell, just big enough four you to fit through. Compare the force of gravity on you outside the shell to the force of gravity if you stepped inside the shell.
Answer:
The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.
At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.
Explanation:
For the collapsed earth:
mass = 5.972 × 10^24 kg
radius = 1 ft
according to Newton's gravitation law, the force of gravity due to two body with mass is given as
Fg = GMm/[tex]R^{2}[/tex]
Where Fg is the gravitational force between the two bodies.
G is the gravitational constant
M is the mass of the earth
m is my own mass
R is the distance between me and the center of the earths in each case
For the case where I stand on the shell:
radius R will be 1 ft
Fg = GMm/[tex]1^{2}[/tex]
Fg = GMm
For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.
A girl walks South at 2.7 m/s. What is the y component of her velocity?
Answer: horizontal components
Explanation:use the Cardinals point
Science Seminar Question: Why did Vehicle 2 fall off the cliff in Claire's test of the collision scene but Vehicle 2 did not fall off the cliff in the film Iceworld Revenge?
Complete Question:
Check the file attached to get the complete question
Answer:
In the film Ice word Revenge, vehicle 2 did not fall of the cliff because, [tex]Weight_{vehicle 1} < Weight_{vehicle 2}[/tex] but in Claire's test, vehicle 2 off the cliff because [tex]Weight_{vehicle 1} \geq Weight_{vehicle 2}[/tex]
Explanation:
In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff. In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.
Answer:
no the answer was both friction and mass
Explanation:
The reason that Vehicle 2 fell off the cliff in Claire's test of the collision scene, but Vehicle 2 did not fall off the cliff in the film Iceworld Revenge is because of the friction and mass. A different friction, called foam, is used in Claire's test. The foam has a really low friction that caused vehicle 2 to fall from the cliff. According to evidence card B which states that, “In Iceworld Revenge, Vehicle 2 moves slowly toward the cliff after the collision, halting only before it goes over the side. In Clair's test, Vehicle 2 went over the cliff at full speed”.This evidence confirms my arguments that in the film, vehicle 2 did not drive as quickly as in the film scene of Claire.
The heating element in an electric kettle is rated as 2.0 kW. If the water in the kettle is at 100.0 °C, what volume of water will be converted into steam in one minute? The specific latent heat of vaporization of the water is 2,257,000 J/kg and the
3 density of water is 1,000 kg/m .
Answer:
The volume is [tex]V =5.32 *10^{-5} \ m^3[/tex]
Explanation:
From the question we are told that
The power of the heating element is [tex]P = 2.0 kW = 2.0 *10^3 \ W[/tex]
The temperature of the water in the kettle is [tex]T _w = 100^oC[/tex]
The time to convert water to steam is t = 1 minute = 60 sec
The specific latent heat of vaporization is [tex]H_v = \ 2,257,000 J/kg[/tex]
The density of water is [tex]\rho_w = 1000\ kg/m^3[/tex]
The power of the heating element is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
Where E Energy generated by the heating element in term of heat
[tex]E = Pt[/tex]
substituting values
[tex]E = 2.0 *10^{3} * 60[/tex]
[tex]E = 120000 J[/tex]
Now
The latent heat of vaporization is mathematically represented as
[tex]H_v = \frac{E}{m}[/tex]
Where m is the mass of water converted to steam
So
[tex]m = \frac{E}{H_v}[/tex]
substituting values
[tex]m = \frac{120000}{2257000}[/tex]
[tex]m = 0.0532\ kg[/tex]
The volume of water converted to steam is mathematically evaluated as
[tex]V = \frac{m }{\rho_w}[/tex]
substituting values
[tex]V = \frac{0.0532}{1000}[/tex]
[tex]V =5.32 *10^{-5} \ m^3[/tex]
Proposed Kinematic Exercise I
Consider a body moving on a straight path with a speed given by v=(3t^2-6t) m/s, where t is in seconds. If the body is at x=4m when t=0s, determine (a) its position at t=4s and (b) its acceleration at t=2s. (c) What is the direction of body movement between t=0s and t=4s? Justify your answer.
Answer:
(a) 20 m
(b) 6 m/s²
(c) Between t=0 and t=2, the body moves to the left.
Between t=2 and t=4, the body moves to the right.
Explanation:
v = 3t² − 6t
x(0) = 4
(a) Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 6t) dt
x = t³ − 3t² + C
Use initial condition to find value of C.
4 = 0³ − 3(0)² + C
4 = C
x = t³ − 3t² + 4
Find position at t = 4.
x = 4³ − 3(4)² + 4
x = 20
(b) Acceleration is the derivative of velocity.
a = dv/dt
a = 6t − 6
Find acceleration at t = 2.
a = 6(2) − 6
a = 6
(c) v = 3t² − 6t
v = 3t (t − 2)
The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.
When 0 < t < 2, v < 0.
When t > 2, v > 0.
what happens at night- describing air circulation
Answer:
The environment is warmed by the light throughout the day, such that the temperature increases. The weather is decreasing and the temperature decreases in the night as the sun falls. There was a misunderstanding. Thanks to the density, the atmosphere becomes densest on the earth. The air becomes colder and colder when you move up.
Explanation:
Answer is above
Hope this helps.
A 397-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR2. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 2500 J. Calculate h in meters.
Answer:
h = 14.4 m
Explanation:
The height can be calculated by energy conservation:
[tex] K_{r} + K_{t} - W = E_{p} [/tex]
Where:
W: is the work
[tex]E_{p}[/tex]: is the potential energy
[tex]K_{r}[/tex]: is the rotational kinetic energy
[tex]K_{t}[/tex]: is the transitional kinetic energy
Initially, the wheel has rotational kinetic energy and translational kinetic energy, and then when stops it has potential energy.
[tex] K_{r} + K_{t} - W = E_{p} [/tex]
[tex] \frac{1}{2}I\omega_{0}^{2} + \frac{1}{2}mv^{2} - W = mgh [/tex]
Where:
I: is the moment of inertia = 0.800 mr²
ω₀: is the angular speed = 25.0 rad/s
m: is the mass = P/g = 397 N/9.81 m*s⁻² = 40.5 kg
v: is the tangential speed = ω₀r²
Now, by solving the above equation for h we have:
[tex] h = \frac{\frac{1}{2}(I\omega_{0}^{2} + mv^{2}) - W}{mg} [/tex]
[tex] h = \frac{\frac{1}{2}(I\omega_{0}^{2} + m(\omega_{0}*r)^{2}) - W}{mg} [/tex]
[tex] h = \frac{\frac{1}{2}(0.800*40.5 kg*(0.600 m)^{2}*(25.0 rad/s)^{2} + 40.5 kg*(25.0 rad/s*0.600 m)^{2}) - 2500 J}{40.5 kg*9.81 m/s^{2}} = 14.4 m [/tex]
Therefore, the height is 14.4 m.
I hope it helps you!
Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires? N/m (b) Is the force attractive or repulsive? attractive repulsive
Explanation:
It is given that,
The separation between two parallel wires, r = 5.6 cm = 0.056 m
Current in both the wires is 2.65 A
(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m[/tex]
(b) As the current is in same direction, the wires will attract each other.
A bird is about 6.5 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 88 mi with the wind in the same time it can fly 72 mi against the wind when the wind speed is 2 mph, what is the rate of the bird in still air?
Answer:
20.0 mph
Explanation:
Let x represent the speed in still air, and t represent the time of flight;
Speed of wind = 2 mph
When flying with the wind, it's Speed is;
x + 2
When flying against wind, it's speed is;
x - 2
distance = speed × time;
When flying with the wind, distance is;
(x+2)t = 88 .....1
When flying against wind, distance is;
(x-2)t = 72 .......2
Solving the simultaneous equation;
Divide equation 1 by 2
(x+2)t/(x-2)t = 88/72
(x+2)/(x-2) = 1.222
x+2 = 1.222(x-2)
x+2 = 1.222x - 2.444
x(1.222-1) = 2+2.444
x(0.222) = 4.444
x = 4.444/0.222
x = 20.0 mph
the rate of the bird in still air is 20.0 mph
A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If g
Complete Question
A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If [tex]g=9.8 \ m/s^2[/tex] , what is the acceleration of the block as it slides down the incline plane
Answer:
The acceleration is [tex]a = 3.142 m/s^2[/tex]
Explanation:
From the question we are told that
The distance from top to bottom of the inclined plane measured along the incline is [tex]d = 3.40 \ m[/tex]
The distance from top to bottom of the inclined plane measured along the vertical axis is
[tex]D = 1.90 \ m[/tex]
According to the SOHCAHTOA rule
[tex]sin \theta = \frac{D}{d}[/tex]
=> [tex]\theta = sin ^{-1} [\frac{D}{d} ][/tex]
substituting values
=> [tex]\theta = sin ^{-1} [\frac{1.09}{3.40} ][/tex]
[tex]\theta = 18.699^o[/tex]T
The acceleration of a block on a frictionless inclined plane is mathematically represented as
[tex]a = gsin \theta[/tex]
substituting values
[tex]a = 9.8 * sin(18.699)[/tex]
[tex]a = 3.142 m/s^2[/tex]
How do you solve this?
Answer:
The circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
Explanation:
We'll begin by calculating the total resistance in the circuit. This is shown below:
R1 = 4Ω
R2 = 3Ω
RT =..?
Total resistance in series can be obtained as follow:
RT = R1 + R2
RT = 4 + 3
RT = 7Ω
Next we shall determine the total current flowing in the circuit:
Voltage (V) = 14V
Resistance (R) = 7Ω
Current (I) =..?
V = IR
14 = I x 7
Divide both side by 7
I = 14/7
I = 2A.
Since the circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
At a track meet, a runner runs the 100–m dash in 16 s. What was the runner's average speed?
Answer:
6.25 m/s
Explanation:
Remember that speed(m/s) = distance(m) / time(s)
what is the study of physilogy
Answer:
physiology is the study of mechanisms and functions in a living system
Explanation:
it's basically a part of biology that works with how a living organism or body part functions
Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 612 km and traveled a distance of 3920 km in 4.09 h. (a) What was the speed (in m/s) of the wave
Answer:
V = 266.23 m/s
Explanation:
The speed of the wave can easily be given by the following formula:
V = S/t
where,
V = Speed of the Wave = ?
S = Distance Covered by Wave = 3920 km
S = Distance Covered by Wave = (3920 km)(1000 m/1 km)
S = Distance Covered by Wave = 3.92 x 10⁶ m
t = Time taken by the wave to cover the distance = 4.09 h
t = Time taken by the wave to cover the distance = (4.09 h)(3600 s/1 h)
t = Time taken by the wave to cover the distance = 14724 s
Therefore,
V = (3.92 x 10⁶ m)/(14724 s)
V = 266.23 m/s
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
ОА
The mechanical waves reach Earth, while the electromagnetic waves do not.
ОВ.
The electromagnetic waves reach Earth, while the mechanical waves do not.
Ос. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer:
B: The electromagnetic waves reach earth, while the mechanical waves do not.
correct Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
A spring of negligible mass and force constant k = 410 N/m is hung vertically, and a 0.200 kg pan is suspended from its lower end. A butcher drops a 2.5 kg steak onto the pan from a height of 0.50 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM
1. What is the speed of the pan and steak immediately after the collision?
2. What is the amplitude of the subsequent motion?
3. What is the period of that motion?
Answer:
1) 2.9m/s
2)0.235cm
3)0.509s
Explanation:
1. moments are conserved, when the steak hits the pan:
velocity 'v' of the steak:
v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s
velocity pan + steak is
v =[tex]m_1v_1/(m_1+m_2)[/tex]= 2.5*3.13/(2.7) = 2.9 m/s
2.initial kinetic energy of pan + steak = spring energy
1/2 mv² = 1/2 kA²
where, A = amplitude
2.7 x 2.9² = 410 x A²
A²=22.707/410
A = 0.235 cm
3. T = 2π√(m/k) = 2π√(2.7/410) = 0.509 s
Therefore, the period of that motion is 0.509s
1) The speed of the pan and steak immediately after the collision is 2.9m/s
2) The amplitude of the subsequent motion is 0.235cm.
3) The period of that motion is 0.509s.
Calculation of the speed, amplitude, and the period:1.
Since
v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s
Also,
v == 2.5*3.13/(2.7)
= 2.9 m/s
2.
Now
initial kinetic energy of pan + steak = spring energy
So,
1/2 mv² = 1/2 kA²
Here,
A = amplitude
2.7 x 2.9² = 410 x A²
A²=22.707/410
A = 0.235 cm
3.
Now
T = 2π√(m/k)
= 2π√(2.7/410)
= 0.509 s
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A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?
A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?
If the distance from the source is doubled. The amplitude of the magnetic field is smaller 4 times.
Which of the following statements is true for an elastic collision? A. Both momentum and kinetic energy are conserved. B. Momentum is conserved, but kinetic energy is not conserved. C. Kinetic energy is conserved, but momentum is not conserved. D. The amount of momentum lost by one object is the same as the amount gained by the other object. E. The amount of kinetic energy lost by one object is the same as the amount gained by the other object.
Answer:
A, D and E
Explanation:
The idea behind elastic is take an elastic band for instance if you stretch it, it will assume it's original shape and form when it returns to its original position hence nothing is lost.
Similarly for elastic collision momentum and kinetic energy are conserved hence; the momentum lost by one is gained by the other and the kinetic energy lost by one is gained by the other.
We have that for the Question ,it can be said that the the appropriate option that identifies elastic collision are
Both momentum and kinetic energy are conservedThe amount of momentum lost by one object is the same as the amount gained by the other objectThe amount of kinetic energy lost by one object is the same as the amount gained by the other object.Therefore Option A,D,C
From the question we are told
Which of the following statements is true for an elastic collision?
A. Both momentum and kinetic energy are conserved.
B. Momentum is conserved, but kinetic energy is not conserved.
C. Kinetic energy is conserved, but momentum is not conserved.
D. The amount of momentum lost by one object is the same as the amount gained by the other object.
E. The amount of kinetic energy lost by one object is the same as the amount gained by the other object.
Generally
Elastic Collision
Elastic collision sees an encounter between two bodies have a net kinetic energy between two bodies remains the same.
Therefore
The the appropriate option that identifies elastic collision are
Both momentum and kinetic energy are conservedThe amount of momentum lost by one object is the same as the amount gained by the other objectThe amount of kinetic energy lost by one object is the same as the amount gained by the other object.Therefore Option A,D,C
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A current cannot produce a magnetic field. *
True or false
Answer:
False
Explanation:
False, as a magnetic field is generated whenever current travels through a conductor.
An electromagnet consists of a coil of wire wrapped around a bar of iron. The coil and iron bar get magnetized when electric current flows through the wire. An electromagnet also has north and south magnetic poles. The magnetic field is strongest at either pole of the magnet.
Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 13 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear
Answer:
The time taken by the sound is [tex]3.82\times 10^{-4}\ s[/tex]
Explanation:
We have,
A low-frequency sound source is placed to the right of a person, whose ears are approximately 13 cm apart, and the speed of sound generated is 340 m/s.
It is required to find the time taken by the sound when the sound arrives at the right ear and the sound arrives at the left ear. Let t is the time taken. It can be calculated as :
[tex]v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{0.13}{340}\\\\t=3.82\times 10^{-4}\ s[/tex]
So, the time taken by the sound is [tex]3.82\times 10^{-4}\ s[/tex]
A toy car's movements is measured using photogates.
Answer:
a) the velocity increases then decreases.
A sample of a compound weighs 2.54 grams. how many ounces (oz.) is this equivalent to?
a. 71.9 oz
b. 0.0139 oz
c. 0.0896 oz
d. 30.8 oz
The correct answer is C. 0.0896 oz
Explanation:
Both grams and ounces are units use to measure the mass of objects. Additionally, 1 gram represents a smaller amount of mass than the one represented by 1 ounce. Indeed, 1 ounce is equivalent to 28.349 grams and therefore, 1 gram is equivalent to 0.035 ounces ( 1 ounces x 1 gram / 28.349 grams = 0.03527 ounces).
According to this, you can convert grams to ounces by multiplying the amount of grams given by 0.035, which is the equivalent of 1 gram to ounces. This means, to determine how many ounces there are in 2.54 you multiply it by 0.035 which is equivalent to 0.0896 ounces (2.54 x 0.03527 = 0.0896 oz).
Find the truth table for the circuit shown. Explain the working principle for all the inputs, briefly. Explain why D1 is used in the network.
Answer:
see below for the truth table
Explanation:
Truth Table
As we will see from the description of operation, any input low causes the output to be high. This is the logic of a NAND gate. The truth table is attached.
Working Principle
Pulling any of A, B, or C low will saturate transistor Q1, depriving Q2 of any base current, cutting it off. Then Q5 is also deprived of base current and is cut off. Meanwhile, the current through R2 supplies base current to Q4, allowing it to pull the output high.
If all of A, B, and C are high (or open), base current is supplied to Q2 through the base-collector junction of Q1. Then Q2 saturates, supplying base current to Q3. Diode D1 ensures that the voltage across Q2 will be insufficient to supply any base current to Q4, so it stays cut off.
Smooth and continuous data is a characteristic of a(n)
analog signal
binary signal
digital signal
pulse signal
Answer:
D
Explanation:
Continuous data is a characteristic of analog signals though been smooth is a subject of question. The smooth signal is after it's been processed. However a wave can still stay continuous defined by what we call duty circle.
Smooth and continuous data is one of the major features of an analog signal.
Analog signal isn't as accurate as the digital signal and it contains minimum
and maximum values which could be positive or negative.
Analog signal comprises of smooth and continuous data. This is explained
by one time-varying quantity representing another time-based variable.In
this type of signal , a variable is an analog of the other.
Analog signals examples include the following : Human voice, Thermometers, analog clocks etc
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How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
B. 4J
C. 39.2 J
D. 20 J
Answer:39.2 (B)
Explanation:
Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volume of 0.44 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.1) How much energy is transferred by heating during this process?
2) How much work is done by the gas during this process?
3) What is the pressure of the gas once the final temperature has been reached?
4) What is the average speed of a gas molecule after the final temperature has been reached?
5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?
6) How much work was done on or by the gas during the constant-pressure process?
Answer:
1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ
2) W₁ = 0 J
3) P = 41.66 x 10³ Pa = 41.66 KPa
4) v = 1618.72 m/s
5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ
6) W₂ = - 7.33 KJ
Explanation:
1)
The heat transfer for a constant volume process is given by the formula:
ΔQ₁ = ΔU = n Cv ΔT
where,
ΔQ₁ = Heat transfer during constant volume process
ΔU = Change in internal energy of gas
n = No. of moles = 4.4 mol
Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k
ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k
Therefore,
ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)
ΔQ₁ = 10.97 x 10³ J = 10.97 KJ
2)
Since, work done by gas is given as:
W₁ = PΔV
where,
ΔV = 0, due to constant volume
Therefore,
W₁ = 0 J
4)
The average kinetic energy of a gas molecule is given as:
K.E = (3/2)KT
but, K.E is also given by:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
mv² = 3KT
v = √(3KT/m)
where,
v = average speed of gas molecue = ?
K = Boltzman Constant = 1.38 x 10⁻²³ J/k
T = Absolute Temperature = 500 K
m = mass of a molecule = 7.9 x 10⁻²⁷ kg
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]
v = 1618.72 m/s
3)
From kinetic molecular theory, we know that or an ideal gas:
P = (1/3)ρv²
where,
P = pressure of gas = ?
m = Mass of Gas = (Atomic Mass)(No. of Atoms)
m = (Atomic Mass)(Avogadro's Number)(No. of Moles)
m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)
m = 0.021 kg
ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³
Therefore,
P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²
P = 41.66 x 10³ Pa = 41.66 KPa
5)
The heat transfer for a constant pressure process is given by the formula:
ΔQ₂ = n Cp ΔT
where,
ΔQ₂ = Heat transfer during constant pressure process
n = No. of moles = 4.4 mol
Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k
ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k
Therefore,
ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)
ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ
Negative sign shows heat flows from system to surrounding.
6)
From Charles' Law, we know that:
V₁/T₁ = V₂/T₂
V₂ = (V₁)(T₂)/(T₁)
where,
V₁ = 0.44 m³
V₂ = ?
T₁ = 500 K
T₂ = 300 k
Therefore,
V₂ = (0.44 m³)(300 k)/(500 k)
V₂ = 0.264 m³
Therefore,
ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³
Hence, the work done , will be:
W₂ = PΔV = (41.66 KPa)(- 0.176 m³)
W₂ = - 7.33 KJ
Negative sign shows that the work is done by the gas
A basketball player is 4.22 m from
the goal when she shoots the ball
at 9.20 m/s at a 69.0° angle.
What is the height of the ball
(above the release point)
when it reaches the goal?
(Unit = m)
Enter
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
3.10 m
Explanation:
Given;
Initial speed of ball u = 9.20 m/s
Angle θ = 69°
Horizontal distance from goal d = 4.22m
Resolving the initial velocity into horizontal and vertical components;
The horizontal component of the initial velocity;
uh = ucosθ
Substituting the given values;
uh = 9.2cos69°
uh = 3.30 m/s
The time taken for it to cover the horizontal distance of 4.22 m (to reach the goal);
Time = distance/speed = d/uh
time = 4.22/3.30 =. 1.279s
The time taken to reach the goal is 1.279 seconds.
To determine the height of the ball, we will resolve the vertical component of the initial velocity;
Vertical component of the ball velocity is;
Uv = usinθ
Uv = 9.20sin69°
Uv = 8.59 m/s
Applying the equation of motion;
Height h = ut - 0.5gt^2
Velocity u = Uv = 8.59 m/s
Time t = 1.279s
Acceleration due to gravity g = 9.8 m/s^2
Substituting the values;
h = 8.69(1.279) - 0.5(9.8×1.279^2)
Height h = 3.0988891 = 3.10 m
The height of ball above the release point is 3.10m
A uniform stationary ladder of length L = 2.7 m and mass M = 14 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ = 0.39. The ladder makes an angle θ = 55∘ with respect to the floor. A painter of mass 8M stands on the ladder a distance d from its base.(a) Find the magnitude of the normal force N, in newtons, exerted by the floor on the ladder.(b) Find an expression for the magnitude of the normal force NW exerted by the wall on the ladder.(c) What is the largest distance up the ladder (d) max, in meters, that the painter can stand without the ladder slipping?
Answer:
A) 1236 N
B) Nw = µ_s•N
C) d_max = 1.525 m
Explanation:
From the question, "smooth vertical wall" means that there is no friction there and thus the only vertical forces are the weights of the ladder and painter and the normal force at the floor.
a) Mass of ladder = 14 kg
Mass of painter = 8M = 8 * 14 = 112 kg
Thus, magnitude of normal force is;
N = total mass x acceleration due to gravity = (14 + 112)9.8
N = 1236 N
(b) Sum of the moments about the base of the ladder:
ΣH = 0
Nw - µ_s•N = 0
Nw = µ_s•N
c) Since they are the only two horizontal forces in play, we know that
Nw = Ff where Ff is the friction force at the floor.
Ff = µ_s*N = 0.39 × 1236
Ff = 482.04 N
So, to find maximum distance painter can stand without slipping, we'll use the formula ;
Nw(Lsinθ) = (Mgcosθ)(L/2) + (8Mgcosθ * d_max)
Plugging in the relevant values, we have;
482.04(2.7*sin55) = ((14 × 9.8cos55)*(2.7/2)) + (8*14*9.8cos55 * d_max)
1066.133 = 106.238 + 629.5575*d_max
629.5575*d_max = 1066.133 - 106.238
629.5575*d_max = 959.895
d_max = 959.895/629.5575
d_max = 1.525 m
A) The magnitude of the normal force is 1236 N
B) expression for the magnitude of the normal force Nw = µ_s•N
C) The largest distance up the ladder (d) max, d_max = 1.525 m
What is friction?The friction is defined as when any object slides on a surface then the force in opposite direction of the external applied force is generated which restrict the motion of the object is called as the friction force.
From the question, "smooth vertical wall" means that there is no friction there and thus the only vertical forces are the weights of the ladder and painter and the normal force at the floor.
a) Mass of ladder = 14 kg
Mass of painter = 8M = 8 * 14 = 112 kg
Thus, the magnitude of normal force is;
N = total mass x acceleration due to gravity = (14 + 112)9.8
N = 1236 N
(b) Sum of the moments about the base of the ladder:
ΣH = 0
Nw - µ_s•N = 0
Nw = µ_s•N
c) Since they are the only two horizontal forces in play, we know that
Nw = Ff where Ff is the friction force at the floor.
Ff = µ_s*N = 0.39 × 1236
Ff = 482.04 N
So, to find maximum distance painter can stand without slipping, we'll use the formula ;
[tex]N_{w}(Lsin\theta) = (Mgcos\theta)(\dfrac{L}{2}) + (8Mgcos\theta \times d_{max)[/tex]
Plugging in the relevant values, we have;
[tex]482.04(2.7\times sin55) = ((14 \times 9.8cos55)\times (\dfrac{2.7}{2})) + (8\times 14\times 9.8cos55 \times d_{max)[/tex]
[tex]1066.133 = 106.238 + 629.5575\times d_{max[/tex]
[tex]629.5575\times d_{max} = 1066.133 - 106.238[/tex]
[tex]d_{max} = \dfrac{959.895}{629.5575}[/tex]
d_max = 1.525 m
Thus, the magnitude of the normal force is 1236 N and an expression for the magnitude of the normal force Nw = µ_s•N . The largest distance up the ladder (d) max, d_max = 1.525 m
To know more about friction, follow
https://brainly.com/question/24386803
. How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?
1) On the Moon, the jump height is about 3 times higher than on Earth.
2) On the Moon, the jump height is about 6 times higher than on Earth.
3) On the Moon, the jump height is about 36 times higher than on Earth.
4) On the Moon, the jump height is more than on Earth by about a times.
Answer:
On the Moon, the jump height is about 6 times higher than on Earth.
Explanation:
The maximum height reached by an object is given by the formula as follows :
[tex]H=\dfrac{v^2}{2g}[/tex]
v is the speed of object
g is acceleration due to gravity on Earth's surface
If g' is acceleration on surface of Moon, g' = g/6
Let H' is the height reached by object on Moon, then,
[tex]H'=\dfrac{v^2}{2g'}\\\\H'=\dfrac{v^2}{2\times \dfrac{g}{6}}\\\\H'=6\times \dfrac{v^2}{2g}\\\\H'=6H[/tex]
On the Moon, the jump height is about 6 times higher than on Earth. Hence, the correct option is (2).