The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
Answers
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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Related Questions
A person is losing thermal energy through the skin at a rate of 120 W when his skin temperature is 30°C. He puts on a sweater, and his skin temperature rises to 33°C. The effective thermal conductivity between his core and the environment changes from 0.22 W/m·K to 0.18 W/m·K. At what rate (J/s) is he now losing thermal energy?
Answers
The rate at which the person is now losing thermal energy is 99 J/s.
Thickness of the person's skin
The thickness of the person's skin is calculated as follows;
Q = k(ΔT)h
where;
k is thermal conductivityh is thickness of the person's skinΔT difference in temperature across the skin = 30 °C = 303 Kh = Q/k(ΔT)
h = (120) / (0.22 x 303)
h = 1.8 m
Rate at which the person is now losing thermal energyQ = k(ΔT)h
where;
ΔT is new temperature difference across the skin = 33 °C = 306 KQ = (0.18)(306)(1.8)
Q = 99 J/s
Thus, the rate at which the person is now losing thermal energy is 99 J/s.
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An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a length of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)
Answers
The the tension in the horizontal rope is 7,019.4 N.
What is tension?
Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.
Tension can also be described as the action-reaction pair of forces acting at each end of said elements.
The unit of tension is Newton (same unit as force, since tension is also a force).
Tension in the horizontal ropeThe tension in the horizontal rope is calculated as follows;
Apply the principle of torque;
T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ
T = (M + 2m)g sinθ/cosθ
T = (M + 2m)g tanθ
let θ = 66⁰ (this value should be given in the question)
T = (M + 2m)g tanθ
T = (74.9 + 2 x 122)(9.8)(tan 66)
T = 7,019.4 N
Thus, the the tension in the horizontal rope is 7,019.4 N.
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In the Roman soldier model for refraction, what happens to the distance between the rows of each soldier once they enter the stream?
A. Not enough info
B. They are closer together
C. They stay the same distance
D. They are further apart
Answers
They pick up speed and move away from the usual.
Using meter sticks, a group of pupils makes a straight line (shoulder to shoulder) and connects to their closest neighbor. Two media are separated in the room by a strip of masking tape. Students can be seen moving normally in one of the media (on one side of the tape). The kids in the other medium (or on the opposite side of the tape) move very slowly and in baby stages. The class moves in the union in a straight line in the direction of the diagonal piece of masking tape. As they get close to the masking tape, the pupils stay in line. Each pupil immediately changes the speed of her or his walk as she or he reaches the tape.
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An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 200 N applied to its edge causes the wheel to have an angular acceleration of 0.844 rad/s2.
What is the moment of inertia of the wheel (in kg · m2)?
________kg · m2
What is the mass (in kg) of the wheel?
_______kg
The wheel starts from rest and the tangential force remains constant over a time period of 4.00 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
_______rad/s
Answers
(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Moment of inertia of the wheel
Apply principle of conservation of angular momentum;
Fr = Iα
where;
F is applied forcer is radius of the cylinderα is angular accelerationI is moment of inertiaI = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
Mass of the wheelI = ¹/₂MR²
where;
M is mass of the solid cylinderR is radius of the solid cylinderI is moment of inertia of the solid cylinder2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
Angular speed of the wheel after 4 secondsω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
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Driving space is the space around your car_____
vulnerable to collision.
O
To the side
All of the above
To the rear
To the front
in which you are
Answers
Answer:
to the rear as there will be an elastic collision
Driving space is the space around your car to the rear vulnerable to collision. the correct answer is option(c).
Driving space indicates that you have a space or a larger area to drive in that is unrestricted. You see only broad, open spaces on the road ahead and there are no limits around your automobile, so you are free to travel forward or change lanes without encountering any obstacles.
Any driver must have room around their car, preferably in the front, back, and on the sides, in order to react to problems as they arise. Your separation from the car in front of you is the most important distance to leave. Drivers' Speeding Patterns Federal and state organizations that research various elements of speeding have taken notice of speeding as the most frequent cause of deadly collisions. The NHTSA has discovered tendencies among drivers who speed recently.
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a)Calculate the angular velocity of the Earth in its orbit around the Sun.
Express your answer using three significant figures.
b) Calculate the angular velocity of the Earth about its axis.
Express your answer using three significant figures.
Answers
The earth's orbital angular velocity around the Sun is 0.986° per day, and its axial angular velocity is 15.041° per hour.
Calculation:
The speeds, not the velocities, are what we require in this case, therefore that is how I will approach it.
Regular speed is calculated by dividing the distance traveled by the travel time.
So, speed = [tex]\frac{distance -travelled}{time}[/tex]
The angular velocity is essentially the same.
Hence, it's divided by the amount of time it took to turn the angle.
(a) Calculation of Earth's orbit around the Sun :
The Earth orbits the Sun in a circle that is 2 radians in diameter (360 degrees). It also takes a year, as we are aware (approx 365 days)
360°/365.25636 days = 0.986°/day ≈ 1° per day
(b) Next, we have to calculate Earth's angular velocity on its axis:
determining the Earth's angular velocity as it fulfills a complete revolution on its axis (a solar day)-
This one requires more precision because a day does not always have 24 hours. Depending on how we define a day,If the day is defined as the time between the Sun's highest and lowest points in the sky, a year's worth of data equals an average of 24 hours.If we define a day as the duration of time it requires for a planet to get to the same location in the sky the following night, then it is 23 hours 56 minutes 4.09 seconds.According to this, the angular speed of rotation = 360°/ 23h 56min 4.09s = 15.041° / hour.
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A 940-kg sports car collides into the rear end of a 2500-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
What was the speed sports car at impact?
Express your answer to two significant figures and include the appropriate units.
Answers
The speed of the sport car at the time of impact is 6.61 m/s.
What is the frictional force of the two cars?Frictional force of the two cars = coefficient of kinetic frictin × mass × acceleration of gravity
= 0.8 × (2500+940) × 9.8
= 26970N
What is the acceleration of the skidded cars?As per Newton's second law of motion, force = mass × accelerationAcceleration= force / mass= 26,970/3440
= 7.8 m/s²
What is the velocity of the sport car at the time of impact?As per Newton's equation of motion, V² - U² = 2aSHere, V = 0 m/s, a= -7.8 m/s², S= 2.8 mSo, 0²-U²= 2×(-7.8)×2.8=> U = √43.68
= 6.61 m/s
Thus, we can conclude that the speed of the sport car at the time of impact is 6.61 m/s.
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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
Answers
(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
Kinetic energy of the projectile at maximum height
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
m is mass of the projectilev₀ₓ is the initial horizontal component of the velocity at maximum heightNote: At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
Work done in firing the projectileBased on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
v is the resultant velocityv = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
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In the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C. What is the electric
field at point P? Include a + or - sign to
indicate the direction.
P
91
0.424 m-
0.636 m
-
92
(Remember, E points away from + charges,
and toward charges.)
(Unit = N/C)
Answers
E =+823.12N/C is the electric field at point P
Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. in the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C hence E = 823.12 N/C
A region of space surrounding andd P Include a + or - sign toindicate the direction.P910.424 m-0.636 m- electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.
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Please help!! will give brainliest if correct!
B: Scientists assume the universe is a vast single system in which basic laws are consistent.
I ____________ with this statement
Explain your answer, using an example from your investigation about habitable worlds.
Answers
Scientists assume the universe is a vast single system in which basic laws are consistent. I agree with this statement
What is the name of the theory that states that our universe is from a single point?The Big Bang Theory is known as one of the greatest consensus that was said to have occurred among scientists, astronomers and also that of cosmologists.
Note that the theory is one that states the Universe as we all know it was formed as a result of a massive explosion that is not only formed by the majority of matter, but also with the use of the physical laws that works in our ever-expanding cosmos.
Therefore, In the case above, i do agree with the fact that Scientists assume the universe is a vast single system in which basic laws that are consistent.
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The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed of an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens
....What is the magnification of the image?
Answers
The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
Location of the imageThe image distance formed by the lens is calculated as follows;
1/v + 1/u = -1/f
where;
v is the image distanceu is the object distancef is the focal length of the lens1/v = -1/f - 1/u
1/v = -(-1/26) - 1/63
1/v = 1/26 - 1/63
1/v = 0.022588
v = 1/0.022588
v = 44.27 cm
What is magnification of lens?The magnification of a lens is defined as the ratio of the height of an image to the height of an object.
Magnification of the image formedThe magnification of the image is calculated as follows;
Magnification = image distance/object distance
M = 44.27/63
M = 0.7
Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
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A 92-kg fullback is running at 3.6 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west.
a)Calculate the original momentum of the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the momentum is to the east, and negative value if the direction of the momentum is to the west.
b)Calculate the impulse exerted on the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
c)Calculate the impulse exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
d)Calculate the average force exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is to the east, and negative value if the direction of the force is to the west.
Answers
a) The original momentum of the fullback is 331.2 kgm/s.
b) The impulse exerted on the fullback is - 331.2 kgm/s
c) The impulse exerted on the tackler is 331.2 kgm/s
d) The average force exerted on the tackler is 389.64 N
Given:Mass of the fullback, m = 92 kg
Initial velocity of the fullback , u = 3.6 m/s
Time of the motion of the fullback , t = 0.85s
The original momentum of the fullback ;
[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]
The impulse exerted on the full-back;
J = ΔP = [tex]m v_f - m v_i[/tex]
J = [tex]m ( v_f - v_i)[/tex]
J = 92 ( 0- 3.6)
J = - 331.2 kgm/s
The impulse exerted on the tackler;
[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]
The average force exerted on the tackler;
F= Δmv/t
[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]
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If little herbert does something uncon
sciously he will probably
Answers
If little herbert does something unconsciously he will probably, he's definitely very unware of he has done
Doing something unconscious may have some implications which may not be favourable.
What is unconscious?Unconscious simply means a state of someone being absent minded of what is happening to him or her at a particular period of time
So therefore, if little herbert does something unconsciously he will probably, he's definitely very unware of he has done
Complete question:
What happens if little herbert does something unconsciously he will probably
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The Kennedy Space Center was in charge of the Apollo 13 mission.
True
False.
Answers
Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant digits? Make sure to account for the travel from you to the wall and from the wall back to you.
Answers
It will take you 0.23 seconds to hear your echo
What is Echo ?Echo can be simply defined as the reflection of sound wave.
Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.
The speed of sound V = 340 m/sThe distance D to and fro = 2 x 38.6 = 77.2 mThe time taken T = ?Speed V is the distance per time
V = D / T
Substitute all the parameters into the formula
340 = 77.2 / T
Make T the subject of formula
T = 77.2 / 340
T = 0.2270
T = 0.23 s to two significant digits
Therefore, it will take you 0.23 seconds to hear your echo.
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An electron is moving the east with a speed of 5.0 × 106 m/s. There is an electric field of
3.0 kV/m in the east direction. What will be the electrons speed after it has moved 1.00 cm?
Answers
The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
What is Electric field?Electric field is the physical field that surrounds a charge.
How to find final velocity of the electron when it moves some distance in a certain electric field?From Newton's second law, the acceleration the electron will bea=F/m=qE/m
where q= charge of electronE= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
The kinematics equation v²=v0²+2a(Δx)where v=final velocity of the electronv0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2=25×10¹²+10.52×10¹²
=35.52×10¹²
Now velocity of electron=5.95×10⁶m/s.Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
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Respond to the following prompts in your initial post:
1. Is Felisha's argument sound or unsound? Use the concepts discussed in your textbook: a valid
structure, true premises, and a sound or unsound argument.
2. Using the fallacies discussed in chapter 11 of your book, identify one fallacy (by name) in
Felisha's argument and explain why this fallacy is an illogical way to argue her conclusion.
3. Describe a time when you successfully argued a point of view and explain why you believe
you were successful.
Answers
Based on the textbook, Felisha's argument is known to be unsound because not all the premises that were given was true.
2. She stated that the “The Cowboys are popular and using only their huge fan base to tell is not right and if they are are better as she claim, why are they not the best but she stated that they are among the best.
3. A time when when i successfully argued a point of view was during my high school debate on the topic that doctors are better than lawyers and i gave a lot of valid reasons why the world cannot do without doctors.
What is fallacies and its types?The term fallacies are seen as flawed or deceptive and false kind of arguments that is known to be be proven wrong with the aid of reasoning.
Note that there are two key types of fallacies which are the formal fallacy and an informal fallacy.
Therefore, Based on the textbook, Felisha's argument is seen as an unsound argument because not all the premises that were given was true and they were not standard enough.
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**please look at attached photo for proper understanding* Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 295 N/C.
What is the magnitude of the acceleration of the electron (in m/s2)?
_____m/s2
The electron is initially at rest. What is its speed (in m/s) after 8.50 ✕ 10−9 s?
________m/s
Answers
The magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
Acceleration of the electronThe magnitude of the acceleration of the electron is calculated as follows;
Force on the electron, F = ma = Eq
ma = Eq
a = Eq/m
where;
E is electric fieldq is charge of electronm is mass of electrona = (295 x 1.6 x 10⁻¹⁹) / (9.1 x 10⁻³¹)
a = 5.187 x 10¹³ m/s²
Speed of the electronThe speed of the electron at the given time is calculated as follows;
v = at
v = ( 5.187 x 10¹³)(8.5 x 10⁻⁹)
v = 4.41 x 10⁵ m/s
Thus, the magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
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A 100 N force pulls a box but it doesn’t move. How much work is done?
A. 0 J
B. 100 J
C. 50 J
D. Not enough info
Answers
Answer:
A
Explanation:
Work = force * distance
= 100 n * 0 = 0 work done
*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 300 N/C, pointing right.
What is the magnitude (in N) and direction of the electric force on the sphere?
magnitude_______N
Direction?
- toward the right
- toward the left
- the magnitude is zero
(b) What is the work (in J) done on the sphere by the electric force as it moves from A to B?
__________J
(c) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.)
PEB − PEA = ______J
(d) What is the potential difference (in V) between A and B?
VB − VA = ________V
Answers
(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
(d) The potential difference between A and B is 56.7 V.
Electric force on the sphere
The electric force on the sphere is calculated as follows;
F = Eq
where;
E is electric fieldq is the chargeF = 300 x (44 x 10⁻⁹)
F = 1.32 x 10⁻⁵ N
The direction of the force is towards the right.
Work done on the sphereW = Fd
W = 1.32 x 10⁻⁵ N x 0.189 m
W = 2.5 x 10⁻⁶ J
Change of the electric potential energyThe change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.
Potential difference between A and BVB − VA = Ed
VB − VA = 300 N/C x 0.189 m
VB − VA = 56.7 V
Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
The potential difference between A and B is 56.7 V.
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Consider the f(x) = cos(x) + D function shown in the figure in blue color. What is the value of parameter D?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
Answers
The value of parameter D is -3.
What is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
f(x) = Acos(x) + D
where;
A is amplitude of the waveB is phase difference of the waveWhat is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
From the blue colored graph, y = -2 at x = 0
-2 = cos(0) + D
-2 = 1 + D
-2 - 1 = D
-3 = D
Thus, the value of parameter D is -3.
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A constant net force F acts on a body during a time interval t. If u and v are the initial and final velocity of the body respectively, the impulse Ft of this force is given by the equation Ft = mv-mu.
Answers
Answer:
Impulse is defined as change in momentum of an object divided by time interval.
at t= 0s
initial velocity = u , initial momentum = mu
at some time t .
final velocity = V, final momentum= mv.
now, change in momentum= ( final - initial) = ( mv-mu)
time interval = (t-0) = t
impulse force = (mv-mu)/ ( t)
Ft = (mv-mu) proved .
this law is known as Newton's second law.
How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km?
Answers
1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
What is Speed ?Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s
In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.
What are the parameters to consider ?
The parameters are;
The distance S = 3.85 × [tex]10^{5}[/tex] kmThe Speed of Light C = 3 × [tex]10^{8}[/tex] m/sThe time taken t = ?Speed = distance S ÷ Time t
Convert kilometer to meter by multiplying it by 1000
C = S/t
3 × [tex]10^{8}[/tex] = 3.85 × [tex]10^{8}[/tex] / t
Make t the subject of formula
t = 3.85 × [tex]10^{8}[/tex] / 3 × [tex]10^{8}[/tex]
t = 1.2833
t = 1.3 s
Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
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The CM of an empty 1300-kg car is 2.45 m behind the front of the car.
How far from the front of the car will the CM be when two people sit in the front seat 2.70 m from the front of the car, and three people sit in the back seat 3.65 m from the front? Assume that each person has a mass of 60.0 kg .
Express your answer to three significant figures and include the appropriate units.
Answers
The center of mass of the car with all people seated will be 3.20 m far from the front of the car.
Taking the front of the car as reference:
m₁x₁ + m₂x₂ + m₃x₃ = mx
where,
m₁ = mass of empty car = 1300 kg
x₁ = distance of center of mass of empty car from front = 2.45 m
m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg
x₂ = distance of center of mass of 2 people sitting in front from the front = 2.7 m
m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg
x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m
m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg
x = distance of center of mass of the car from the front when all are seated
Therefore,
(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x
x = 4166 kg.m/1300 kg
x = 3.20 m
Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.
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The distance between a carbon atom ( m = 12 u ) and an oxygen atom ( m = 16 u ) in the CO molecule is 1.13 × 10-10 m .
How far from the carbon atom is the center of mass of the molecule?
Express your answer to two significant figures and include the appropriate units.
Answers
6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.
The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.
The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms, [tex]x_{c}[/tex] and [tex]x_{o}[/tex] their distances relative to C atom.
Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )
[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]
[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m
Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.
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A converging lens has a focal length of 20cm. find graphically the image location for an object at each of the following distances from the lens: (a) 50cm (b) 20cm (c) 15cm (d) -40cm: determine the magnification in each case. Check your results by calculating the image position and lateral magnification from Eqs. (10) and (11), respectively.
Answers
Answer:
4
Explanation:
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4 answer
A force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
A). What is the potential energy of the spring when it is stretched 0.700 m?
B). What is its potential energy when it is compressed 8.00 cm?
Answers
Answer:
399j and 456j
Explanation:
(a)spring energy= force * distance
spring energy=570 * 0.700= 399 joules
(b) convert 8.00 cm to m we have 0.08m
E =f * s
E=570 *0.08= 456j
The potential energy of the spring when it is stretched 0.700 m would be 199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.
What is the spring constant?The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.
The mathematical relation for calculating the spring constant is as follows
F = - Kx
As given in the problem, if a force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
k = 570 / 0.7
k = 814.28 N / m
The potential energy stored in spring when it is stretched 0.700 m
E = 1/2 kx²
=0.5×814.28×0.7²
= 199.5 Joules
The potential energy stored in spring when it is stretched 8 cm
E = 1/2 kx²
=0.5×814.28×0.08²
=2.60 Joules
Thus, the potential energy of the spring when it is stretched 0.700 m would be 199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.
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What is the mass of the sun plus all the planets in solar system?
Answers
Why is glass classified the way it is?
A. The bulb lights up when the switch is closed.
B. Free electrons can carry charge.
C. Delocalized electrons create a current.
D. Not enough charge carriers.
Answers
The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.
An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.
High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.
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a dog runs back and forth between its two owners, who are walking toward one another. the dog starts running when the owners are 10 m apart. if the dog runs with a speed of 3 m/s and the owners each walk with a speed of 1.3 m/s how far has the dog traveled when the owners meet
Answers
Answer:
11.54 meters
Explanation:
The owners take 10m / (1.3 + 1.3 m/s) = 3.85 seconds to meet
The dog is running at constant speed of 3 m/s for these 3.85 seconds
3 m/s * 3.85 sec = 11.54 meters