Answer:
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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 ]
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.
a horizontal turntable with radius 0.8 m is rotation about a vertical axis at ts center. a small block sits on the turntable at a distance of 0.4 m from the acis and rotates in a circular path at a speed of 0,800 m/s. what minim coeeficent of static friction between the blok and the turntable is required for the block not to slip
Answer: 0.04
Explanation:
Given
Radius of turntable [tex]r=0.8\ m[/tex]
Block is present at a distance of [tex]r_o=0.4\ m[/tex]
Turntable rotates at a speed of [tex]v=0.8\ m/s[/tex]
angular speed of turntable [tex]\omega =\dfrac{v}{r}[/tex]
[tex]\Rightarrow \omega =\dfrac{0.8}{0.8}\\\Rightarrow \omega =1\ rad/s[/tex]
block will experience a force i.e. centripetal force equal to [tex]m\omega ^2r[/tex]. This must balance the friction force [tex]\mu mg[/tex]
[tex]\Rightarrow m\omega^2 r_o=\mu mg\\\Rightarrow 1^2\times 0.4=\mu \times 9.8\\\Rightarrow \mu =0.04[/tex]
Thus, the coefficient of static friction force is [tex]0.04[/tex]
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 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.
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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
A puck moves 2.35 m/s in a -22.0 direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50 degree direction. What was the direction of the acceleration?
Answer:
48.9 is the answer I think !
Answer:
28.4
Explanation:
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
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
explanation on energy from air pressure light from water pressure
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:
Prepare a stock available at home following the procedure you learned in this lesson. Have a picture of the finished product and a narration on how you do it
Answer:
NO, 1. is stocks
2. is also stocks
bouquit GARNI IS NUMBER 3
NUMBER 4 IS ACID PRODUCTS
number 5 is brown stock
Explanation:
stocks are bones
which statement summarized the difference between mass and weight?
Answer:
The second statement.
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 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.
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
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
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 :)
The geometric shape of the molecule H2,O. A. bent B. Trigonal planar C. linear D. t shape
Answer:
A. bent
Explanation:
Water molecule (H2O) is said to contain four valence electron pairs (2 bond pairs and 2 lone pairs). The presence of lone pair of electrons in the water molecule influences its molecular geometry or shape.
Since water has two lone pairs of electrons, which repel each other according to the VSEPR theory, water molecule is said to have a BENT molecular geometry.
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.
A supernova is a large explosion that takes place at the end of a star's life cycle. Although supernovas only burn for a short amount of time, they can tell scientists a lot about the universe. Scientists use electromagnetic waves to detect supernovas. Why are electromagnetic waves used instead of sound waves?
A. Electromagnetic waves are created by physical vibrations.
B. Electromagnetic waves travel slower than sound waves.
C. Electromagnetic waves can travel in vacuums.
D. Electromagnetic waves cannot be absorbed.
Electromagnetic waves can travel in vacuums.
To find the correct statement among all the options, we need to know more about the electromagnetic wave.
What is electromagnetic wave?Electromagnetic wave is a transeverse wave in which both the electric field and magnetic field travel simultaneously.The electric field and magnetic field are perpendicular to each other and perpendicular to the direction of prapagation of the EM wave.How is the electromagnetic wave created?Electromagnetic wave is created from the oscillation of any charge particles like electron, proton, ions etc. .
What is the speed of electromagnetic waves?The speed of electromagnetic wave in vacuums is maximum that is 3×10⁸ m/s. (Speed of EM wave is much more than that of sound (340m/s)).In any other medium, its speed changes and it depends on the refractive index of that medium.How is the electromagnetic wave affected by medium?In any medium, the electromagnetic wave is absorbed to some extent except ideal vacuum. Also in vacuum in space, it's absorbed; because the vacuum in space is not completely vacuum since it contains few molecules per cubic kilometers.However, the electromagnetic wave can travel in any medium even in vacuum while the sound wave can travel only in particle medium.Thus we conclude that scientists use electromagnetic waves to detect supernovas because electromagnetic waves can travel in vacuums.
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While taking a psychology exam, Lori got up and shut the classroom door due to outside noise that affected her concentration. Which theory explains Lori’s behavior?
Group of answer choices
Drive-reduction theory
Expectancy theory
Instinct theory
Arousal theory
Answer:
drive reduction theory
Explanation:
I would say that because of all the cars beeping and making A Lot of cachos on the street so that will definitely affect her taking her exam
Theory explains Lori’s behavior is b) Drive-reduction theory
What is Drive-reduction theory?
It is based on the idea that the primary motivation behind all human behavior is to reduce drives . A drive is a state of discomfort which is triggered by a person's physiology or biological need .
so , she removed discomfort that can happen by the noise outside the classroom in order to maintain student's concentration
correct answer is b) Drive-reduction theory
learn more about Drive-reduction theory
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chứng minh V=kq/r từ mối liên hệ giữ E và V
Answer:V=Aq=KQr
Explanation:
Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?
Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q
Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.
- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2
Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:
dA=dF.dx=KQ.qx2dx
Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:
A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr
Theo đúng định nghĩa: V=Aq=KQr
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
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
When rebuilding her car's engine, a physics major must exert 405 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force in newtons between the piston and cylinder
Answer:
[tex]N=675N[/tex]
Explanation:
From the question we are told that:
Force [tex]F=405N[/tex]
Generally the equation for Normal force in this case is is mathematically given by
[tex]F=\mu_s N[/tex]
Where
Static Friction=[tex]\mu_s[/tex]
[tex]\mu_s=0.6[/tex]
Therefore
[tex]N=\frac{F}{\mu_s}[/tex]
[tex]N=\frac{405}{0.6}[/tex]
[tex]N=675N[/tex]
Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v of the car, assuming that the tires are not slipping against the ground.
Answer:
the linear speed of the car is 28.83 m/s
Explanation:
Given;
radius of the car, r = 0.33 m
angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s
The linear speed of the car is calculated as;
V = ωr
V = 87.35 rad/s x 0.33 m
V = 28.83 m/s
Therefore, the linear speed of the car is 28.83 m/s
bowling ball weighing 710 N attached to the ceiling by a rope of length 3.87 mThe ball is pulled to one side and released it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.51 m/s
What is the acceleration of the bowling ball, in magnitude and direction, at this instant?
What is the tension in the rope at this instant?
Answer:
a = 5.256 m / s² with the vertical direction upwards and T = 1090.8 N
Explanation:
to find the acceleration we must use Newton's second law, in the lowest part of the motion
T - W = m a
where the acceleration is centripetal
a = v² / r
let's calculate
a = 4.51² / 3.87
a = 5.256 m / s²
As the acceleration is centripetal it is directed towards the center of the circle, which in the lower part coincides with the vertical direction upwards.
Let's find the tension of the rope with the first equation
T = W + m a
W = m g
let's calculate
T = 710 + 710 5.256 / 9.8
T = 1090.8 N
Instantaneous speed is...
a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.
PLEASE HURRY
Answer:
The speed attained at a particular instant in time.
Explanation:
Instantaneous speed is the speed attained at a particular instant in time.
It is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
It is equal to the rate of change of speed.
It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.
Hence, the correct option is (b).
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.
