Answer:
108 NmExplanation:
Given data
mass m= 2 kg
radius r= 0.5 m
angular velocity ω= 12 rad/s
distance d= 0.75 m
we know that
[tex]Force= mass * acceleration\\\ acceleration= w^2r\\\\ Force= m*w^2r\\\\Force =2*12^2*0.5= 144 N[/tex]
we know that torque = F*d= 144*0.75= 108 Nm
g A point mass of 1.5kg is attached to a spring and set to oscillate through simple harmonic oscillations. If the period of the oscillation is 10s, find the spring constant.
Answer:
k = 0.6 N/m
Explanation:
The time period of a spring mass oscillation system is given by the following formula:
T = 2π√(m/k)
where,
T = Time Period of Oscillation = 10 s
m = Mass attached to the spring = 1.5 kg
k = spring constant = ?
Therefore,
10 s = 2π√(1.5 kg/k)
squaring on both sides we get:
100 s² = 4π²(1.5 kg/k)
k = 6π² kg/100 s²
k = 0.6 N/m
Two moons orbit a planet in nearly circular orbits. Moon A has orbital radius r, and moon B has orbital radius 16r. Moon A takes 10 days to complete one orbit. How long does it take moon B to complete an orbit
Answer:
Kepler's Third Law: The square of the period of any planet about the sun is proportional to cube of its mean distance from the sun.
Mathematically: T^2 = K R^3
So (TA / TB)^2 = (RA / RB)^3
TB^2 = TA^2 * (RB / RA)^3
TB^2 = 10^2 * 16^3
TB = (409600)^1/2 = 640 days
A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
To know more about experiment and hydrochloric acid
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what is quantic fisic
Answer:
it is the physics that explains how everything works. The best description we have of the. nature of the particles that make up matters and the forces with which they interact. It underlines how atoms work, and so why chemistry and biology work as they do
A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is
Answer:
The kinetic energy at a displacement of half the amplitude is 37.5 J
Explanation:
Given;
total energy on the spring, E = 50 J
When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.
E = K + U
Where;
K is the kinetic energy
U is the elastic potential energy
K = E - U
K = E - ¹/₂KA²
When the displacement is half = ¹/₂(A) = A/₂
K = E - ¹/₂K(A/₂)²
K = E - ¹/₂K(A²/₄)
K = E - ¹₄(¹/₂KA²)
Recall, E = ¹/₂KA²
K = ¹/₂KA² - ¹₄(¹/₂KA²) (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)
K = 1(¹/₂KA²) - ¹₄(¹/₂KA²) = ³/₄(¹/₂KA²)
K = ³/₄(¹/₂KA²)
But E = ¹/₂KA² = 50J
K = ³/₄ (50J)
K = 37.5 J
Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J
The kinetic energy when the displacement is half the amplitude
Given the following data:
Total energy = 50 Joules.Displacement, x = [tex]\frac{A}{2}[/tex]To find the kinetic energy when the displacement is half the amplitude:
The total energy of the system of a block and a spring is the sum of the spring's elastic potential energy and kinetic energy of the block and it's proportional to the square of the amplitude.
Mathematically, the total energy of the system of a block and a spring is given by the formula:
[tex]T.E = U + K.E[/tex] .....equation 1.
[tex]T.E = \frac{1}{2} kA^2[/tex]
Where:
T.E is the total energy.U is the elastic potential energy.K.E is the kinetic energy.A is the amplitude.Making K.E the subject of formula, we have:
[tex]K.E = T.E - U[/tex] .....equation 2.
But, [tex]U = \frac{1}{2} kx^2[/tex] ....equation 3.
Where:
k is spring constant.x is change in position (displacement).Substituting the eqn 3 into eqn 2, we have:
[tex]K.E = T.E - \frac{1}{2} kx^2[/tex]
[tex]K.E = T.E - \frac{1}{2} k(\frac{A}{2})^2\\\\K.E = T.E - \frac{1}{2} k(\frac{A^2}{4})\\\\K.E = T.E - \frac{1}{4} (\frac{1}{2} kA^2)\\\\K.E = T.E - \frac{1}{4} (T.E)\\\\K.E = 50 - \frac{1}{4} (50)\\\\K.E = 50 - 12.5[/tex]
K.E = 37.5 Joules.
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A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 42.0 cm from its center.
Answer:
-1.4x10^6N/C
Explanation:
Pls see attached file
The magnitude of the electric field.
Magnitude is the size of the object in properties that is determines the size of the object. It also displays the result of the order of class of the object. The direction of the electric field tells us about the position of the field in four different directions. As per the question, the answer is 1.4x10^6N/C.
The rod of 16cm of total length is given. Has a charge of a total of -25.0uc. The rod's axis is pointed at 42.0 cm from its center and is given in the question. The rod Length will be then 0.16m and the total change will be 25x10 cm and point where the electricity will be calculated is shown by the axis of the rod at the distance of 42 cms.The magnitude and direction will be calculated based on the measure of the formula of E. This answer to the question will be 1.4x10^6N/C.Learn more about the uniformly charged.
brainly.com/question/12088419.
In his experiments on "cathode rays" during which he discovered the electron, J.J. Thomson showed that the same beam deflections resulted with tubes having cathodes made of different materials and containing various gases before evacuation.
A) Are these observations important? Explain.
B) When he applied various potential differences to the deflection plates and turned on the magnetic coils, alone or in combination the fluorescent screen continued to show a single small glowing patch. Argue whether his observation is important.
C) Do calculations to show that the charge-to-mass ratio Thomson obtained was huge compared with that of any macroscopic object or of any ionized atom or molecule.
D) Could Thomson observe any deflection of the beam due to gravitation?
Answer:
A) his observation is of little importance ,
B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force
C) see that in this second case it is 4 times less
D) the force of gravity is of the order of 10⁻⁴⁰
therefore it is 10²⁸ times less than the electric force,
Explanation:
A) This observation is of little importance since the cacodylate ray tube always emits electrons, regardless of the material of which it is made.
B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force
C) the elect's load is 1.6 10⁻¹⁹ C its mass is 9.1 10⁻³¹ kg, let's look for its relation
e / m = 1.6 10⁻¹⁹ / 9.1 10⁻³¹
e / m = 0.1 758 10 10¹² N
look for this in the case of an atom, let's use the lightest atom hydrogen
the homogenize have an electron of charge 1.6 10⁻¹⁹ C
and a mass of 1.6735575 10⁻²⁷ ka
e / M = 1.6 10⁻⁻¹⁹ / 1.67 10⁻²⁷
e / M = 0.96 10⁸ N
We see that in this second case it is 4 times less
D) the force of gravity is of the order of 10⁻⁴⁰
therefore it is 10²⁸ times less than the electric force, therefore it should not contribute to the movement of the light beam
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
Answer:
Explanation:
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac slow twitch intermediate fast twitch
Answer:
Dead lifting uses tho muscle fundamentals
Explanation:
Answer:
Fast twitch
Explanation:
Edmentum
A ball is thrown horizontally from the top of a 41 m vertical cliff and lands 112 m from the base of the cliff. How fast is the ball thrown horizontally from the top of the cliff?
Answer:
4.78 second
Explanation:
given data
vertical cliff = 41 m
height = 112 m
solution
we know here time taken to fall vertically from the cliff = time taken to move horizontally ..........................1
so we use here vertical component of ball
and that is accelerated motion with initial velocity = 0
so we can solve for it as
height = 0.5 × g × t² ........................2
put here value
112 = 0.5 × 9.8 × t²
solve it we get
t² = 22.857
t = 4.78 second
ball thrown horizontally from the top of the cliff in 4.78 second
A ball bouncing against the ground and rebounding is an example of an elastic collision. Describe two different methods of evaluating this interaction, one for which momentum is conserved, and one for which momentum is not conserved. Explain your answer.
Answer:
Momentum is conserved when there are no outside forced present and it has an equal and opposite reaction, also momentum is conserved the ball's momentum is transferred to the ground. This first instance is the case of a Closed system.
The second case where momentum is not conserved is when there is a variation or difference in the moment of the ball because of influence of external forces
A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the earth. The earthʹs radius is 6.38 × 10^{6} m. The acceleration of this satellite is closest to
Answer:
a = 0.43 m/s²
Explanation:
First, we need to find the velocity of the satellite:
Velocity = V = Distance Covered/Time Taken
here,
Distance = 1 revolution = 2π(1.5 x 10⁷ m) = 9.43 x 10⁷ m
Time = (8.65 hours)(3600 s/1 hour) = 31140 s
Therefore,
V = (9.43 x 10⁷ m)/(31140 s)
V = 3028.26 m/s
Now, the acceleration of the satellite will be equal to the centripetal acceleration, with the center of circular motion as the center of earth:
a = V²/R
where,
R = 1.5 x 10⁷ m + 0.638 x 10⁷ m
R = 2.138 x 10⁷ m
Therefore,
a = (3028.26 m/s)²/(2.138 x 10⁷ m)
a = 0.43 m/s²
A 0.140-kg baseball is thrown with a velocity of 27.1 m/sIt is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?
Answer:
The bat and the ball were in contact for 1.8 x 10⁻³ s
Explanation:
Given;
mass of baseball, m = 0.14 kg
initial velocity of the baseball, u = 27.1 m/s
applied force in opposite direction, F = -5000 N
final velocity in opposite direction, v = -37 m/s
Note: The applied force and final velocity are negative because they act in opposite direction to the initial velocity.
impulse received by the body = change in momentum of the body
Ft = Δmv
Ft = mv - mu
Ft = m(v-u)
t = m(v-u) / F
[tex]t = \frac{0.14(-37-27.1)}{-5000} \\\\t = \frac{0.14(-64.1)}{-5000} \\\\t = \frac{-8.974}{-5000} \\\\t = 0.0018 \ s\\\\t = 1.8*10^{-3} \ s[/tex]
Therefore, the bat and the ball were in contact for 1.8 x 10⁻³ s
A tube of water is open on one end to the environment while the other end is closed. The height of the water relative to the base is 100 cm on the open end and 40 cm on the closed end. What is the absolute pressure of the water at the top of the closed end in units of atm
Answer:
1.06 atm
Explanation:
On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water
The pressure due to a height of water = ρgh
where ρ is the density of water = 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
h is the height of the water column
The height of water column on the open end = 100 cm = 1 m
pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa
Atmospheric pressure = 101325 Pa
The total pressure on the open end = 101325 Pa + 9810 Pa = 111135 Pa
The pressure due to the water column on the closed end = ρgh
The height of the water in the closed end = 40 cm = 0.4 m
The pressure due to this column of water = 1000 x 9.81 x 0.4 = 3924 Pa
The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa = 107211 Pa
In atm unit, this pressure = 107211/101325 = 1.06 atm
Two 10-cm-diameter charged rings face each other, 18.0 cmcm apart. Both rings are charged to 30.0 nCnC . What is the electric field strength
Answer:
E=7453.99 V/m
Explanation:
The electric field on the charged is given by
E= Kqx/(r^2 +x^2)^3/2
Where;
K= constant of Coulomb's law
q= magnitude of charge= 30.0×10^-9 C
r= radius of the rings= 5 cm or 0.05m
x= distance between the rings = 18cm = 0.18 m
Substituting values;
E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2
E= 48.6/(2.5×10^-3 + 0.0324)^3/2
E= 48.6/(0.0025 + 0.0324)^3/2
E= 48.6/6.52×10^-3
E=7453.99 V/m
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate
For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
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Which of the following is not a benefit of improved cardiorespiratory fitness
Answer:
C - Arteries grow smaller
Explanation:
The option choices are:
A. Faster post-exercise recovery time
B. Lungs expand more easily
C. Arteries grow smaller
D. Diaphragm grows stronger
Explanation:
There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.
What would be the Roche limit (in units of Earth radii) if the Earth had the same mass, but its radius was increased to 1.5 Earth radii?
First calculate the density of this new, larger, Earth. Now use this new density and the new radius in the calculator above to determine the Roche limit for this new larger 'Earth.
Answer:
Roche limit = 1.89 of earth radius
Explanation:
We know that,
Mass of earth = 5.972 × 10²⁷ g
New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸
Density of earth = 5.5132 g/cm³
New density of earth = Mass of earth / (4/3)πr³
New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³
New density of earth = 1.634 g/cm³
Roche limit = [2(Density of earth)/(New density of earth)]¹/³r
Roche limit = 1.89 of earth radius
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.
What must be the diameter of a cylindrical 120-m long metal wire if its resistance is to be ? The resistivity of this metal is 1.68 × 10-8 Ω • m.
Answer:
The diameter is [tex]d = 6.5 *10^{-4} \ m[/tex]
Explanation:
From the question we are told that
The length of the cylinder is [tex]l = 120 \ m[/tex]
The resistance is [tex]\ 6.0\ \Omega[/tex]
The resistivity of the metal is [tex]\rho = 1.68 *10^{-8} \ \Omega \cdot m[/tex]
Generally the resistance of the cylindrical wire is mathematically represented as
[tex]R = \rho \frac{l}{A }[/tex]
The cross-sectional area of the cylindrical wire is
[tex]A = \frac{\pi d^2}{4}[/tex]
Where d is the diameter, so
[tex]R = \rho \frac{l}{\frac{\pi d^2}{4 } }[/tex]
=> [tex]d = \sqrt{ \rho* \frac{4 * l }{\pi * R } }[/tex]
[tex]d = \sqrt{ 1.68 *10 ^{-8}* \frac{4 * 120 }{3.142 * 6 } }[/tex]
[tex]d = 6.5 *10^{-4} \ m[/tex]
When we describe electric flux, we say that a surface is oriented in a certain direction with respect to an electric field. When we try to calculate how much electric field passes through the surface, we make use of the:_________.
1. Wedge Product
2. Dot Product
3. Cross Product
Answer:
2. Dot Product
Explanation:
The calculation of the electric flux gives an scalar result.
When we tray to calculate how much electric field passes trough a surface, we are calculating a scalar value. Furthermore, the concept of flux requires the calculation of a scalar value.
Also it is necessary to take into account that the magnitude of the flux trough a surface depends of the inclination of the surface respect to the direction of the electric field. This is taken into account sufficiently by a dot product.
Then, the answer is:
2. Dot Product
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other? What would the general shape of the field lines look like? What would the field lines look like in between the two pieces?
Answer:
Explanation:
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Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
When a spinning bike wheel is placed horizontally, hung from a pivot at one end, the axis of rotation of the wheel will swing in a horizontal circle. In which direction does it turn?a) upwardb) downwardc) horizontally, CWd) horizontally, CCW
Answer:
answer is D
Explanation:
horizontally, CCW
A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system
Answer:
E = 0.645J
Explanation:
In order to calculate the total mechanical energy of the system, you take into account that if the zero of energy is at the equilibrium position, then the total mechanical energy is only the elastic potential energy of the spring.
You use the following formula:
[tex]E=U_e=\frac{1}{2}kA^2[/tex] (1)
k: spring constant = ?
A: amplitude of the oscillation = 7.50cm = 0.075m
The spring constant is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
[tex]k=4\pi^2f^2m[/tex] (2)
f: frequency of the oscillation = 1.95Hz
m: mass of the piece of iron = 1.53kg
You replace the expression (1) into the equation (2) and replace the values of all parameters:
[tex]E=\frac{1}{2}(4\pi^2f^2m)A^2=2\pi^2f^2mA^2\\\\E=2\pi^2(1.95Hz)^2(1.53kg)(0.075m)^2=0.645J[/tex]
The totoal mechanical energy of the system is 0.645J
Which statement describes a disadvantage of using natural gas as an energy source? It is expensive to use. It is hard to stop using. It is a renewable resource. It is scarce in some parts of the world.
Answer:
B : It is hard to stop using.
Explanation:
just took the quiz ! hope this helps with anyone who needs it !
Due to the dependency on natural gas as a fuel, it is hard to stop using.
What is natural gas?Natural gas is a fossil fuel which is obtained from the ground in association with petroleum.
Natural gas consists mainly of petroleum.
It is a non-renewable energy source.
Natural gas use contributes to global warming
However, due to the dependency on natural gas as a fuel, it is hard to stop using.
Learn more about natural gas at: https://brainly.com/question/815922
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?
Answer:
14.1 m/s
Explanation:
From the question,
μk = a/g...................... Equation 1
Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.
make a the subject of the equation
a = μk(g).................. Equation 2
Given: μk = 0.160, g = 9.8 m/s²
Substitute into equation 2
a = 0.16(9.8)
a = 1.568 m/s²
Using,
F = ma
Where F = force, m = mass.
Make m the subject of the equation
m = F/a................... Equation 3
m = 160/1.568
m = 102.04 kg.
Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.
Assuming the initial velocity of the skier to be zero
Fd+mgμ = 1/2mv²........................Equation 4
Where v = speed of the skier after crossing the patch, d = distance/width of the patch.
v = √2(Fd+mgμ)/m)................ Equation 5
Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16
Substitute these values into equation 5
v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]
v = √197.57
v = 14.1 m/s
v = 9.86 m/s
When a nucleus at rest spontaneously splits into fragments of mass m1 and m2, the ratio of the momentum of m1 to the momentum of m2 is
Answer:
p₁ = - p₂
the moment value of the two particles is the same, but its direction is opposite
Explanation:
When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved
initial instant. Before fission
p₀ = 0
since they indicate that the nucleus is at rest
final moment. After fission
[tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂
p₀ = p_{f}
0 = m₁ v₁ + m₂v₂
m₁ v₁ = -m₂ v₂
p₁ = - p₂
this indicates that the moment value of the two particles is the same, but its direction is opposite
Let us suppose the magnitude of the original Coulomb force between the two charged spheres is FF. In this scenario, a third sphere touches the grey sphere and the red sphere multiple times, being grounded each touch. If the grey sphere is touched twice, and the red sphere is touched three times, what is the magnitude of the Coulomb force between the spheres now
Answer:
F ’= 1/32 F
We see that the value of the force is the initial force over 32
Explanation:
In this problem the sphere that is touching the others is connected to ground, after each touch,
Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half
½ (Q / 2) = ¼ Q
For the red dial repeat the same scheme
with the first touch the charge is reduced to Q / 2
with the second touch e reduce to ½ (Q / 2) = ¼ Q
with the third toce it is reduced to ½ (¼ Q) = ⅛ Q
Now let's analyze what happens to the electric force
if the force is F for when the charge of each sphere is Q
F = k Q Q / r²
with the remaining charge strength is
F ’= k (¼ Q) (⅛ Q) / r²
F ’= 1/32 k Q Q / r²
F ’= 1/32 F
We see that the value of the force is the initial force over 32