The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
On putting estimate values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]
⇒ [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]
⇒ [tex]v=1.94 \ cm[/tex]
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]
⇒ [tex]{h_{1}}=1.16 \ mm[/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.
. 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).
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!
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
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
A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline. (a) What was the magnitude of the ball's acceleration, assume it to be constant
Answer:
a) a = 2.35 m/s^2
Explanation:
(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
x: position of the ball after t seconds = 87 m
t: time = 8.6 s
a: acceleration of the ball = ?
vo: initial velocity of the ball = 0 m/s
You solve the equation (1) for a:
[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]
You replace the values of the parameters in the previous equation:
[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]
The acceleration of the ball is 2.35 m/s^2
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.
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.
A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm2 and a large one of cross-sectional area 300 cm2. What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg
Answer:
735.75 N
Explanation:
We are given;
Area of small lift; A1 = 5 cm²
Area of large lift; A2 = 300 cm²
Mass required to lift large lift;m = 4500 kg
Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N
By Pascal's law,
F1/A1 = F2/A2
We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.
Thus;
F1 = (F2 × A1)/A2
Plugging in the relevant values, we have;
F1 = (44145 × 5)/300
F1 = 735.75 N
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).
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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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.
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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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.
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]
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 charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.
Required:
a. Determine the direction of the force on a charge.
b. Determine the direction of the force on a charge.
1. along the line between the charge and the center of the square outward of the center
2. along the line between the charge and the center of the square toward the center
3. along the side of the square toward the other charge that lies on the side
4. along the side of the square outward of the other charge that lies on the side
Answer:
Explanation:
Given that:
length of side , a = 0.5 m
charge , q = 6.65 mC
length of diagonal , d = 0.5 * sqrt(2)
d = 0.707 m
F is the force due to adjacent particle ,
F1 is the force due to diagonal particle
Now , for the net charge on a particle
Fnet = 2 * F * cos(45) + F1
Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2
Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)
Fnet = 3.05 *10^6 N
the magnitude of net force acting on each particle is 3.05 *10^6 N
part B)
for the direction of particle
d) along the line between the charge and the center of the square outward of the center
an object with momentum must also have
a. acceleration
b. kinetic energy
c. impulse
d. force
An object with momentum must also have impulse.
An object with momentum must also have: c. impulse.
Momentum can be defined as the product of the mass possessed by an object and its velocity. Also, it is a vector quantity and as such has both magnitude and direction.
Mathematically, momentum is giving by the formula;
[tex]Momentum = Mass \times Velocity[/tex]
In Physics, the impulse experienced by an object is always equal to the change in momentum of the object. This is due to the force acting on an object.
[tex]Impulse = Change\;in\;momentum\\\\Force \times time = m \Delta V[/tex]
In conclusion, an object with momentum must also have impulse.
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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.
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
The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate
(i) the instantaneous velocity when t = 1
(ii) the instantaneous acceleration when t =1
Answer:
(i) v = 44 m/s
(ii) a = 72 m/s^2
Explanation:
You have the following equation for the potion of a car:
[tex]x=20t+6t^4[/tex]
(i) The instantaneous velocity is the derivative of x in time:
[tex]\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3[/tex]
for t = 1 is:
[tex]v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s[/tex]
(ii) The instantaneous acceleration is the derivative of the velocity:
[tex]\frac{dv}{dt}=24(3)t^2=72t^2[/tex]
for t = 1
[tex]a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2[/tex]
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:
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
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
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)
Calculate the average speed of a car between 0 And 10 second give the unit
Answer:
I think the distance should be given in the question because the formula for getting the
average speed= total distance/total time taken
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.
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]
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.0 m/s
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m