What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m
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Related Questions
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.
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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.
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?
Answers
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
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
Answers
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.
Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 175 Hz when air temperature is 18.0°C
Answers
Answer:
Length = 0.4882 m
Explanation:
given data
fundamental frequency = 175 Hz
air temperature = 18.0°C
solution
we will apply here fundamental frequency formula that is
F = [tex]\frac{v}{4L}[/tex] ....................1
here v = [tex]331 \sqrt{1+\frac{T}{273}}[/tex]
here 331 m/s is speed of sound in air
so v = [tex]331 \sqrt{1+\frac{18}{273}}[/tex] = 341.74 m/s
now put value in equation 1 we get
F = [tex]\frac{v}{4L}[/tex]
[tex]175 = \frac{341.74}{4L}[/tex]
Length = 0.4882 m
Calculate the flow rate of blood (of density 0.846 g/cm3 ) in an aorta with a crosssectional area of 1.36 cm2 if the flow speed is 48.5 cm/s. Answer in units of g/s.
Answers
Answer:
55.80 g/s
Explanation:
From the question,
Flow rate = density×Area×velocity.
φ = ρ×A×V................... Equation 1
Where φ = flow rate of blood, ρ = density of blood, A = cross sectional area of blood, V = velocity of blood.
Given: ρ = 0.846 g/cm³, A = 1.36 cm², V = 48.5 cm/s.
Substitute these values into equation 1
φ = 0.846×1.36×48.5
φ = 55.80 g/s
Hence, the flow rate of the blood = 55.80 g/s
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?
Answers
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
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?
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Answer:
Explanation:
check this out and rate me
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.
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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
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
Answers
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.
Read more: https://brainly.com/question/23153766
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
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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
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
Answers
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
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
Answers
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
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to
Answers
Complete Question
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)
Answer:
The velocity is [tex]v = 3.79 *10^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]
The magnetic field is [tex]B = 0.38 \ T[/tex]
The force due to the electric field is mathematically represented as
[tex]F_e = E * q[/tex]
and
The force due to the magnetic field is mathematically represented as
[tex]F_b = q * v * B * sin(\theta )[/tex]
Now given that it is perpendicular , [tex]\theta = 90[/tex]
=> [tex]F_b = q * v * B * sin(90)[/tex]
=> [tex]F_b = q * v * B[/tex]
Now given that it is not deflected it means that
[tex]F_ e = F_b[/tex]
=> [tex]q * E = q * v * B[/tex]
=> [tex]v = \frac{E}{B }[/tex]
substituting values
[tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]
[tex]v = 3.79 *10^{5} \ m/s[/tex]
help me
Describe the different types of non contact forces.
Answers
Answer:
the correct answer is
All four known fundamental interactions are non-contact forces: Gravity, the force of attraction that exists among all bodies that have mass. ... Examples of this force include: electricity, magnetism, radio waves, microwaves, infrared, visible light, X-rays and gamma rays.
Explanation:
hope this helps you!!!!
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.
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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]
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?
Answers
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
A small glass bead charged to 8.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 940 μN.
Required:
What is the total charge on the rod?
Answers
Answer:
71nC is the total charge of the rod
Explanation:
See attached file
The total charge on the rod is equal to 3.3 × 10⁻⁸ C.
What is the force on a charge in an electric field?The force on the charge in a uniform electric field E is given by:
F = qE where q is charge in coulombs
The electric field due to the charge associated with the rod is given by:
[tex]E =\frac{kQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
Where r is the distance between the bead and the rod, L is the length of the glass rod and Q is the charge on the rod.
The force experienced by the bead charged is,
[tex]F =\frac{kqQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
From the above equation, we can find the value of Q as:
[tex]Q =\frac{Fr\sqrt{r+\frac{L^2}{4} } }{kq}[/tex]
Given, the value of force, F = 940μN = 940 ×10⁻⁹ C
The length of glass rod, L = 10cm = 0.1 m and r = 4cm = 0.04 m
[tex]Q =\frac{(940\times 10^{-6}N)(0.04m)\sqrt{(0.04m)+\frac{(0.1m)^2}{4} } }{(8.99\times 10^9 N.m^2/C^2)(8\times 10^{-9}C)}[/tex]
[tex]Q= 0.5228\times 10^{-6}\times\sqrt{0.0041}[/tex]
[tex]Q = 3.3\times 10^{-8} C[/tex]
Therefore, the total charge on the rod is 3.3 × 10⁻⁸ C.
Learn more about the force on a charge, here:
https://brainly.com/question/22042360
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If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answers
Answer:
14.79 kgm/s
Explanation:
Data provided in the question
Let us assume the mass of baseball = m = 0.145 kg
The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s
Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s
Based on the above information, the change in momentum is
[tex]\Delta P = m(v_f -v_i)[/tex]
[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]
= 14.79 kgm/s
Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s
A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wall at a point having coordinates (1.2, 1.9), where the units are meters, what is the distance of the fly from the corner of the room?
Answers
Answer:
[tex]D=2.25 m[/tex]
Explanation:
We can use the equation of a distance between to points:
[tex]D=\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}[/tex]
Now, we know that the origin point (x₀,y₀) = (0,0) and the point (x₁,y₁) = (1.2,1.9)
[tex]D=\sqrt{(1.2-0)^{2}+(1.9-0)^{2}}[/tex]
[tex]D=2.25 m[/tex]
Therefore the distance of the fly will be D = 2.25 m from the corner.
I hope it helps you!
There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.77 ✕ 105 km away, would the light first arrive on Earth?
Answers
Answer:
Explanation:
speed of light = 3 x 10⁸ m /s .
distance between moon and the earth = 3.77 x 10⁵ x 10³m .
Time taken by light to cover the distance
= distance / speed
= 3.77 x 10⁸ / 3 x 10⁸
= 1.256 s
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.
Answers
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
An electron traveling north enters a region where the electric field is uniform and points east. Will the electron speeds up or slows down?
Answers
Answer:
electron has a constant velocity in the north direction and accelerates in the west direction, the electron is accelerating and its velocity on this axis increasing in the west direction.
Explanation:
In this exercise we are asked about the speed of the electron
The electric force is F = q E
the charge of the electron is q = -e = - 1.6 10-19 C
If we use Newton's second law
Y axis (North-south direction)
[tex]v_{y}[/tex] = cte
X axis (East-West direction)
F = m a
- e E = m a
a = - e / m E
whereby electron has a constant velocity in the north direction and accelerates in the west direction, the electron is accelerating and its velocity on this axis increasing in the west direction.
Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters
Answers
Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
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.
Answers
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
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?
Answers
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.
Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.
Answers
Answer:
right A, B, C, D
Explanation:
They ask which statements are true
A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping
B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation
C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.
D) Right. It is a resonant process the driving force keeps the oscillation of the system
Statements that are right as regards oscillation are:
A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.
B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.
C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.
D. An oscillation that is maintained by a driving force is called forced oscillation.
Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.Therefore, the options are correct.
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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.
Answers
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. "
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A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a stationary particle with a charge of –5 × 10–6 C. The radius of the orbit is:
Answers
Answer:
r = 0.22m
Explanation:
To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.
Then, you have:
[tex]F_c=F_e=ma_c[/tex] (1)
m: mass of the particle = 20g = 20*10-3 kg
ac: centripetal acceleration = ?
q: charge of the particle = 5*10^-6C
Fe: electric force between the charges
The electric force is given by:
[tex]F_e=k\frac{qq'}{r^2}[/tex] (2)
r: radius of the orbit
q': charge of the particle at the center of the orbit = -5*10^-6C
Furthermore, the centripetal acceleration is:
[tex]a_c=\frac{v^2}{r}[/tex] (3)
v: speed of the particle = 7m/s
You replace the expressions (2) and (3) in the equation (1) and solve for r:
[tex]k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}[/tex]
Finally, you replace the values of all parameters in the previous expression:
[tex]r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m[/tex]
The radius of the circular trajectory is 0.22m
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.
Answers
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.
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a playground merry-go-round of radius r = 2.20 m has a moment of inertia i = 245 kg · m2 and is rotating at 11.0 rev/min about a frictionless vertical axle. facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round?
Answers
Answer:
8.92 rpm
Explanation:
Given that
Radius of the merry go round, r = 2.2 m
Initial moment of inertia, I1 = 245 kgm²
Initial speed of rotation, w1 = 11 rpm
Mass of the child, m = 26 kg
To solve the problem, we use the law conservation of momentum
I1w1 = I2w2
I2 = mr² + I1
I2 = 245 + 26 * 2.2
I2 = 245 + 57.2
I2 = 302.2 kgm²
Now, applying the formula, we have
I1w1 = I2w2
245 * 11 = 302.2 * w2
w2 = 2695 / 302.2
w2 = 8.92 rpm
Thus, the new angular speed of the merry go round is 8.92 rpm