Answer:
The answer is D
A physical system is in the state |α> = cos(α)|+> + sin(α)|->. Two observables  = a(|+><+| + |-><-|) + b(|+><-| + |-><+|) e B = a(|+><+| - |-><-|) are measured. Check the uncertainty relationship for these two operators.
Answer:
sorry dear
Explanation:
sorry dear but ur question is hard to understand can u try to edit it so i can tell u the answer?
Answer the following questions
1. Heat in liquid travels from
a) bottom to top
b) top to bottom
c) left to right
d) right to left
2. The direction of flow of heat is
a) always from a cooler body to a hotter body
b) always from a hotter body to cooler body
c) always from a body at a lower temperature to a body at a higher temperature
d) all the above
3. A cold steel spoon is dipped in a cup of hot milk. The steel spoon transfer the heat to its other end by the process of
a) convection
b) conduction
c) radiation
d) none of the above
What is this sport ⚽⚾
Answer:
sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.
hope it is helpful to you
The lumberjack pulls on the sled with 40 N at an angle of 30 degrees, pulling so the sled moves at a constant velocity. 1) What is the x component of the applied force? 2) What is the y component of the applied force? 3) If the loaded sled has a mass of 65 kg, what is the magnitude of the force of gravity? 4) What is the magnitude of the normal force acting on the sled? 5) What is the coefficient of friction between the snow and the sled?
1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N
4) The net force on the sled acting in the vertical direction is made up of
• the sled's weight, 585 N, pointing downward
• the vertical component of the applied force, 20 N, pointing upward
• the normal force, with magnitude n, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ F = n + 20 N - 585 N = 0 ==> n = 565 N
5) The net force in the horizontal direction consists of
• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)
• kinetic friction, with magnitude f, pointing in the opposite and negative direction
By Newton's second law,
∑ F = 34.6 N - f = 0 ==> f ≈ 34.6 N
Now if µ is the coefficient of kinetic friction, then
f = µn ==> µ = f/n = (34.6 N) / (565 N) ≈ 0.0613
The component of the force is the effective part of that force in that direction.
What is the component of a force?The component of the force is the effective part of that force in that direction.
1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N
2) The vertical component of the force = 40 N sin 30 degrees = 20 N
3) The magnitude of the gravitational force = mg cos 30 degrees = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N
4) The normal force = 551.7 N
5) The coefficient of friction = F/R = 40 N /551.7 N = 0.07
Learn more about component of a force:https://brainly.com/question/15529350
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A person places a cup of coffee on the roof of his car while he dashes back into the house for a forgotten item. When he returns to the car, he hops in and takes off with the coffee cup still on the roof.
(a) If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, what is the maximum acceleration the car can have without causing the cup to slide?
(b) What is the smallest amount of time in which the person canaccelerate the car from rest to 13 m/s andstill keep the coffee cup on the roof?
Answer:
(a) The acceleration is 2.35 m/s^2.
(b) The time is 5.53 s.
Explanation:
coefficient of friction = 0.24
(a) The acceleration of the car is
[tex]a =\mu g\\\\a = 0.24 \times9.8\\\\a = 2.35 m/s^2[/tex]
(b) initial velocity, u =0, final velocity, v = 13 m/s
Let the time is t.
Use first equation of motion
v = u + a t
13 = 0 + 2.35 x t
t = 5.53 seconds
1. An AAMU basketball player is 2.03 meters tall. What is his height given in US customary units of feet and
inches?
Answer:
His height is 6.66 feet or 79.92 inches.
Explanation:
Given that,
An AAMU basketball player is 2.03 meters tall.
Let h is the height.
We know that,
1 m = 3.28 feet
So,
2.03 m = 6.66 feet
Also,
1 m = 39.37 inches
2.03 m = 79.92 inches
Hence, this is the required solution.
The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the chair/rocket would be ?
(80m/s/s)(70kg)=5600N
(80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
(80m/s/s)(70kg)=5600N - Fg = 5600N+(70kg)(9.8N/kg)~5600N-700N=4900N
I need the time
please explain need this ASAP
I assume you're talking about a pilot. If the ejection seat has an acceleration of 8g, then it would exert a normal force of 8g (70 kg) ≈ 5600 N.
(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)
To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the net force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)g ≈ 4900 N.
If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the net force would change to -5600 N - (70 kg)g ≈ -6300 N
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
A 1500kg car is travelling at v=30m/s. The cars kinetic energy is? *
A) 45000J
B) 1350000J
C) 22500J
D)675000J
show your work please
Hi there!
[tex]\large\boxed{\text{D. 675000J}}[/tex]
Use the following formula to solve:
KE = 1/2mv², where:
KE = kinetic energy
m = mass (kg)
v = velocity (m/s)
Therefore:
KE = 1/2(1500)(30)²
KE = 1/2(1500)(900)
KE = 675000 J
A 1200-kg car is being driven up a 5.0o hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f = 524 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 290 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is +150 kJ?
I suppose the hill makes an angle of 5.0° with the horizontal.
• F acts parallel to the road and in the direction of the car's motion, so it contributes a positive amount of work, F (290 m).
• Friction does negative work on the car since it opposes the car's motion. As the car moves up the slope, the work done by friction is (-524 N) (290 m) = -151,960 J.
• The car's weight has components that act parallel and perpendicular to the road. The parallel component has a magnitude of W sin(5.0°) and points down the slope, so it contributes negative work of -(1200 kg) g sin(5.0°) ≈ 1,024.95 J. The perpendicular component of W does not do any work.
• The normal force FN also doesn't do any work to move the car up the slope because it points perpendicular to the road, so we can ignore it, too.
The net work done on the car is then
F (290 m) + (-151,960 J) + 1,024.95 J = 150,000 J
==> F (290 m) ≈ 300,935 J
==> F ≈ (300,935 J) / (290 m) ≈ 1,037.71 N
Determine the values of m and n when the following average magnetic field strength of the Earth is written in scientific notation: 0.0000451 T. Enter m and n, separated by commas.
Answer:
B = 4.51×10⁻⁵ T
Explanation:
Given that,
The average magnetic field strength of the Earth is 0.0000451 T.
We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :
N=a×bⁿ
Where
n is any integer and a is a real no
So,
0.0000451 = 4.51×10⁻⁵ T
So, the required answer is equal to 4.51×10⁻⁵ T.
A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 kg and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
Answer:
(a) vf = 1.51 m/s
(b) vf = 36.22 m/s
Explanation:
The rate of change of momentum is equal to the force:
[tex]F = \frac{mv_f-mv_i}{t}[/tex]
[tex]Ft = m(v_f-v_i)[/tex]
where,
F = Force = 1035 N
t = time = 0.175 s
vi = initial speed = 0 m /s
vf = final speed = ?
(a)
m = mass of body = 120 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\[/tex]
vf = 1.51 m/s
(b)
m = mass of head = 5 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\[/tex]
vf = 36.22 m/s
a car increases its speed as it moves across the floor. which form of energy is increasing for the car?
Answer:
kinetic
Explanation:
i just remember it from last year
Answer:
kinetic energy
Explanation:
expression for kinetic energy is
kinetic energy = (1/2) × mass × (velocity)^2
so , as velocity increases K.E increases
A major artery with a 1.3 cm^2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.6 cm^2. By what factor is the average velocity of the blood reduced when it passes into these branches?
Answer:
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
Explanation:
Given;
initial area of the artery, A₁ = 1.3 cm²
Area of each smaller 18 arteries, a₂ = 0.6 cm²
Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²
Apply flow rate equation;
Q = AV
where;
Q is the flow rate of the blood
V is the average velocity of the blood
If the flow rate is constant, then;
A₁V₁ = A₂V₂
[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
The volume of a liquid is 830m'at 30°C and it is 850m'at 90°C. The
coefficient of volume expansion of the liquid is
Answer:
4.02×10⁻⁴ K⁻¹
Explanation:
Applying,
γ = (v₂-v₁)/(v₁Δt)................. Equation 1
Where γ = coefficient of volume expansion, v₂ = Final volume, v₁ = initial volume, Δt = change in temperature.
From the question,
Given: v₂ = 850 m³, v₁ = 830 m³, Δt = (90-30) = 60°C
Substitute these values into equation 1
γ = (850-830)/(830×60)
γ = 20/(830×60)
γ = 4.02×10⁻⁴ K⁻¹
Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm lần so với
trong chân không
Answer:
B.
Explanation:
sana makatulong sayo
What is the torque in ( lbs-ft ) of a man pushing on a wrench with 65 lbs of force 8 unches from the nut / bolt he is trying to turn?
Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
A pair of butterflies reproduces and has one thousand offspring. All one thousand of the offspring have the alleles Aa. What is the most likely combination of alleles (genotype) for each parent?
Answer:
Alleles AA and aa
Explanation:
For all the offsprings of the butterflies to have the same heterozygous alleles Aa, it means the two parents have different homogyzous alleles. That is, one of the parents had the alleles AA while the other had the alleles aa. Thus, a combination of the two pairs of alleles will produce 100% Aa alleles in the offspring as seen in the image attached.
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
What must be true if energy is to be transferred as heat between two bodies in physical contact?
1-The two bodies must have different volumes.
2-The two bodies must be at different temperatures.
3-The two bodies must have different masses.
4-The two bodies must be in thermal equilibrium.
Answer:
answer is d
Explanation:
i hope this helps you
Write the prime factorization of 32. Use exponents when appropriate and order the factors
from least to greatest
A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2. After the collision the bullet becomes embedded in the block. How much work is being dne by bullet?
Answer:
Work done by the bullet is 612.26 J.
Explanation:
mass of bullet, m = 0.5 kg
initial velocity of bullet, u = 50 m/s
coefficient of friction = 0.2
mass of block, M = 3 kg
let the final speed of the bullet block system is v.
use conservation of momentum
Momentum of bullet + momentum of block = momentum of bullet block system
0.5 x 50 + 3 x 0 = (3 + 0.5) v
v = 7.14 m/s
let the stopping distance is
The work done is given by change in kinetic energy of bullet
initial kinetic energy of bullet, K = 0.5 x 0.5 x 50 x 50 = 625 J
Final kinetic energy of bullet, K' = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J
So, the work done by the bullet
W = 625 - 12.74 = 612.26 J
Got it never mind. The only reason I'm typing more is to fill out the required space
Answer:
hey. i dont know what you tryna say but if u replying to someone else, you should use the comments section. in that way you won't lose points.
What innovation did jethro wood add to plows in the 1800s?
Built by 1800, it was the home of inventor Jethro Wood (1774-1834), whose 1819 invention of an iron moldboard plow revolutionized American agriculture.
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperature of the system 50 C. ( specific heat water= 4200 J/Kg C , specific heat copper= 390 J/Kg C
Answer:
Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)
Explanation:
Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].
Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].
Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].
Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].
The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:
[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].
[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].
Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].
In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].
Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].
Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:
[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].
These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 13[/tex].
Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].
CHEGG A neutron star has a mass of 2.08 × 1030 kg (about the mass of our sun) and a radius of 6.73 × 103 m. Suppose an object falls from rest near the surface of such a star. How fast would it be moving after it had fallen a distance of 0.0093 m? (Assume that the gravitational force is constant over the distance of the fall, and that the star is not rotating.
Let g be the acceleration due to gravity on the surface of the star. By Newton's second law, the gravitational force felt by the object has a magnitude of
F = GMm/r ² = mg
where
• G = 6.67 × 10⁻¹¹ Nm²/kg² is the gravitational constant,
• M = 2.08 × 10³⁰ kg is the mass of the star,
• m is the unknown mass of the object, and
• r = 6.73 × 10³ m is the radius of the star
Solving for g gives
g = GM/r ²
g = (6.67 × 10⁻¹¹ Nm²/kg²) (2.08 × 10³⁰ kg) / (6.73 × 10³ m)²
g ≈ 3.06 × 10¹² m/s²
The object is in free fall with uniform acceleration and starting from rest, so its speed after falling 0.0093 m is v such that
v ² = 2g (0.0093 m)
v = √(2g (0.0093 m))
v ≈ 240,000 m/s ≈ 240 km/s
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field created by the
negative test charge.
ОА.
1.4 x 106 newtons/coulomb
ОВ.
1.9 x 106 newtons/coulomb
OC. 5.4 x 10-1 newtons/coulomb
OD
3.6 x 106 newtons/coulomb
Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1. Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.
2. The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinite
Answer:pp
Explanation:
ii
You are driving home from school steadily at 97 km/h for 190 km . It then begins to rain and you slow to 60 km/h instantly. You arrive home after driving 4.0 hours.
how far is your hometown from school?
Please delete my answer. I made a mistake