Answer:
The dipole always tries to align in the direction of electric field so that the net torque is zero.
Explanation:
When a dipole placed in an electric field at any arbitrary angle, it experiences a torque.
Let the dipole moment is p, electric field is E and the angle is A.
The torque is given by
Torque = p E sin A
where, A is the angle between the electric field and dipole moment.
So, the dipole always tries to align in the direction of electric field so that the net torque is zero.
A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?
Explanation:
kinetic energy was converted to potential energy in the spring.
the answer is in the above image
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
Answer:
3.0−0.12=2.88 or 2.88÷0.04=72
0.04×3.0=0.12 and 0.04+3.0=3.04
Answer: 2.4 J
Explanation: Khan Academy
A box is being pulled to the right over a rough surface. t > fk , so the box is speeding up. Suddenly the rope breaks. What happens? The box:_________.
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
b. stops immediately.
c. continues speeding up for a short while, then slows and stops.
d. continues with the speed it had when the rope broke.
Answer:
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
Explanation:
Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.
Since there is a net force, there is an acceleration and thus an increasing velocity.
When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.
Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.
So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.
So, the answer is a.
in physics If we interchange rows and columns of Matrix A, what is the new matrix known as 'Matrix' A ?
Answer:
The correct answer is (C), as explained below. The transpose of a matrix is created by interchanging corresponding rows and columns.
Switching Rows
You can switch the rows of a matrix to get a new matrix.
Explanation:
If A is an m × n matrix and AT is its transpose, then the result of matrix multiplication with these two matrices gives two square matrices: A AT is m × m and AT A is n × n. ... Indeed, the matrix product A AT has entries that are the inner product of a row of A with a column of AT.
HELP ME PLS
Chlorine (chemical symbol Cl) is located in Group 17, Period 3. Which is
chlorine most likely to be?
A. A metalloid with properties of both metals and nonmetals
B. A gaseous, highly reactive nonmetal
C. A soft, shiny, highly reactive nonmetal
D. A soft, shiny, highly reactive metal
Answer:
Option B
Explanation:
A gaseous, highly reactive non-metal
Answer:B
Explanation:I just took the test
which statement summarized the difference between mass and weight?
Answer:
The second statement.
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite ends of the same rope and pull the other toward him. The magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration. What is the ratio of Hank's mass to Harry's mass?
Answer:
the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1
Explanation:
Given the data in the question;
Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.
We know that from Newton's Second Law;
Force = mass × Acceleration
F = ma
Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.
Now,
Mass[tex]_{Hank[/tex] × Acceleration[tex]_{Hank[/tex] = Mass[tex]_{Henry[/tex] × Acceleration[tex]_{Henry[/tex]
so
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = Acceleration[tex]_{Henry[/tex] / Acceleration[tex]_{Hank[/tex]
given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = 1 / 1.26
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = 0.7937 or [ 0.7937 : 1 ]
Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]
The moon Phobos orbits Mars
(mass = 6.42 x 1023 kg) at a distance
of 9.38 x 106 m. What is its period of
orbit?
Answer:
The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.
(4.56 x 10^-13)-(1.17 x 10^-13)
If a duster is left on the top of the chalkboard, it is said to possess kinetic energy.
Answer:
In the above diagram, we move by horizontally with a velocity to the right. But due to gravity the hand falls a distance as we move .
Therefore the time it takes to fall by a distance is given by =.
If = as →∞, then from the laws of motion
=12()22
Therefore for the whole journey from left to right our muscle has to give a beat up at an amount equal to the gravity force. So, the work supplied by the muscle for one trip would be
∫=∫012()22
A man is going to rub the chalk off a blackboard, he is going to choose a way to rub off the chalk in two ways
Starting from the upper left corner of the board and moving horizontally to the right and moving slightly down and then again moving to the right and then when it reaches the left corner and so on
The second way is given in the following: He starts at the top and rubs down and then up again and so on.
He is going to choose the way that is less tiring in his arms. i.e., less work done by his muscles!
Somehow from vague intuitive notion I am inclined to agree that it would be less tiring to rub off the chalk if he moves the duster beginning at the top and rubbing horizontally and gradually decreasing the height when one reach the corners till everything is off.
I tried to calculate the muscle work by considering the following ideas(please correct me if I'm wrong):
The nerves inside our muscle has to fire its signal continuously throughout the entire interval of the process of rubbing and its has to oppose the gravitational force keeping the hand up the air.
When the hand is at the highest point on the blackboard, the muscle has to work against gravity =, as the muscle continues to work against gravity throughout the interval from left to right (the horizontal path of the first case) I couldn't find the total work withstood by our muscle.
Define reversible change
Answer:
Reversible changes are changes that can be undone or reversed. Melting, freezing, boiling, evaporating, condensing, dissolving and also, changing the shape of a substance are examples of reversible changes. If playback doesn't begin shortly, try restarting your device.
Explanation:
Answer:
A reversible change is a change that can be undone or reversed.
Explanation:
explanation on energy from air pressure light from water pressure
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
What is the resistance of a bulb of 4ow
connected in a line of 220v?
2
Answer:
1210 ohm
Explanation:
Given :
P=40 W
V=220 V
Now,
[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]
Therefore, resistance of bulb will be 1210 ohm
The figure shows a graph of the angular velocity of a rotating wheel as a function of time. Although not shown in the graph, the angular velocity continues to increase at the same rate until t = 8.6 s. What is the angular displacement of the wheel from 0 to 8.6 s?
Answer:
θ = 33.54 rad
Explanation:
This is a circular motion exercise
θ= θ₀ + w₀ t + ½ α t²
suppose that for t = 0 the body is at its initial point θ₀ = 0 and from the graph we see that the initial angular velocity w₀ = -9.0 rad / s
we look for the angular acceleration,
α = [tex]\frac{\Delta w}{ \Delta t}[/tex]
from the graph taken two points
α = [tex]\frac{0 - (-9.0)}{3.0 - 0}[/tex]
α = 3 rad / s²
we substitute in the first equation
θ = 0 -9 t + ½ 3 t²
the displacement is requested for t = 8.6 s
θ = = -9 8.6 + 3/2 8.6²
θ = 33.54 rad
The angular displacement of the wheel from 0 to 8.6s is 33 radians.
Given to us
t = 8.6s
Acceleration of the wheelWe know acceleration is the ratio of velocity and time, therefore, the slope of the velocity-time graph will give us acceleration, therefore,
At point t=3, ω = 0
At point t = 5, ω = 6
Acceleration = slope of the Velocity-time graph = 3 rad/sec²
Angular displacementUsing the equation,
[tex]\theta = \omega_0t+\dfrac{1}{2}\alpha t^2[/tex]
SUbstitute values,
[tex]\theta = (-9.0\times 8.6)+\dfrac{1}{2}(3\times 8.6^2)\\\theta = 33\rm\ radians[/tex]
Hence, the angular displacement of the wheel from 0 to 8.6s is 33 radians.
Learn more about Angular displacement:
https://brainly.com/question/14613604
Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular accelerations of the point.
a. Both are zero.
b. Only the angular acceleration is zero.
c. Only the linear acceleration is zero.
d. Neither is zero
Explain your choice
Answer:
c. Only the linear acceleration is zero.
Explanation:
The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.
The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.
Therefore, the correct option is:
c. Only the linear acceleration is zero.
what is the escape velocity of earth
Answer:
The Escape Velocity Of Earth is
11.19 km/s
Explanation:
Hope it Helps!
Why do you think we see the sun moves across the sky?
answer it for brainlliest
Answer:
From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.
what are parasites? Give some example
Answer:
parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs
Assignment: 06.05 Infections and Health
What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?
Explanation:
the answer is in the image above
The Y-component of a vector A, which is of magnitude 16√2 and at a 45° angle to the horizontal would be 16
What is a vector quantity?The quantities that contain the magnitude of the quantities along with the direction are known as the vector quantities.
Examples of vector quantities are displacement, velocity acceleration, force, etc.
As given in the problem we have to find out the Y-component of a vector A, which is of magnitude 16√12 and at a 45° angle to the horizontal,
Y component of the vector A = 16√2 sin45°
=16√2 ×1/√2
=16
Thus, the Y component of vector A would be 16.
To learn more about the vector quantity here, refer to the link;
https://brainly.com/question/15516363
#SPJ2
What is the acceleration of a 0.30 kilogram ball that is hit
with a force of 27 N?
Answer: 90
Explanation: a=Fnet/M
=27/0.30
=90
the acceleration is 90
Answer:
The acceleration of the ball is 83.333ms2 [forward].
Explanation:
i hope it helps :)
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm. a) What is the wave? b) Wrote the wave function, y(x,t) c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
What is the answer bro idont now
A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its
kinetic energy?
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet.
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
Answer:
1. 5.825 × 10¹⁷ J
2. i. $ 29.125 billion ii. I would not invest in the company
3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. a i. The momentum will not be relativistic
ii. This is because objects with large masses do not move at relativistic speeds
b i. 155 m/s
ii. This speed wouldn't be a problem for the ship.
Explanation:
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)
So, K = (γ - 1)mc²
= (1/√(1 - β²) - 1)mc²
= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²
= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²
= 11.65 × 10¹⁶ kgm²/s²
= 1.165 × 10¹⁷ J
Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have
20 % of E = K
0.2E = K
E = K/0.2
= 1.165 × 10¹⁷ J/0.2
= 5.825 × 10¹⁷ J
2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
i. how much would this launch cost?
Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion
ii. Would you invest in this company?
I would not invest in the company
3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy, then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
i. Will the recoil momentum of the ship be relativistic?
The momentum will not be relativistic.
ii. Justify your argument.
This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
i. At what speed will the ship recoil after it launches a probe?
Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J
Also, E'² = (pc)² + (mc²)² where p = momentum of payload
So, making p subject of the formula, we have
(pc)² = E'² - (mc²)²
pc = √[E'² - (mc²)²]
p = √[E'² - (mc²)²]/c
substituting the values of the variables into the equation, we have
p = √[E'² - (mc²)²]/c
p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s
p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s
p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s
p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s
p = 18.59 × 10¹⁶/3 × 10⁸ m/s
p = 6.20 × 10⁸ kgm/s
From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.
So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship
So, v = p/m'
= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg
= 1.55 × 10² m/s
= 155 m/s
ii. Do you think that this is a problem for the ship?
Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.
A 2.80 kg mass is dropped from a
height of 4.50 m. Find its potential
energy (PE) when it is 3.00 m
above the ground.
Answer:
PE = 82.32J
Explanation:
PE = m*g*h
PE = 2.80kg*9.8m/s²*3m
PE = 82.32J
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Force required to lift the wheelbarrow
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
What happens to the water when you throw rock into a pond
Answer:
The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.
Mark pushes his broken car 140 m down the block to his friend's house. He has to exert a 110 N horizontal force to push the car at a constant speed. How much thermal energy is created in the tires and road during this short trip?
Answer:
Explanation:
This is a work problem...energy is created and used in the form of work.
W = FΔx where W is work, F is the force needed to move the object Δx in meters.
W = 110(140) ∴
W = 15000 J
HELP PLSS I CANT FAIL!!!
Elements from Period 3 of the periodic table are highlighted. Which element
is a metalloid?
A. Sodium
B. Argon
C. Sulfur
D. Silicon
