Kinetic energy is the energy an object has due to its Motion.
Kinetic energy is a characteristic of a moving particle. It is a type of energy that a matter or particle possesses due to its motion.
It is expressed:
[tex]K_E = \frac{1}{2}mv^2[/tex]
Where m is the mass of the particle and v is velocity.
Hence, Kinetic energy is the energy an object has due to its Motion.
Learn more: https://brainly.com/question/999862
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.
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
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]
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:
check this out and rate me
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.
Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensitive plate. To avoid triggering the pressure plate, he must replace the gold with something of equal mass. The density of gold is 19.3.103 kg/m3, and the volume of a sphere is V = 4/3 Ar3. Indy has a bag of sand with a density of 1,602 kg/m3.
(A) What volume of sand must he replace the gold sphere with? If the sand was a sphere, what radius would it have?
Answer:
Volume of Sand = 0.4 m³
Radius of Sand Sphere = 0.46 m
Explanation:
First we need to find the volume of gold sphere:
Vg = (4/3)πr³
where,
Vg = Volume of gold sphere = ?
r = radius of gold sphere = 2 cm = 0.02 m
Therefore,
Vg = (4/3)π(0.2 m)³
Vg = 0.0335 m³
Now, we find mass of the gold:
ρg = mg/Vg
where,
ρg = density of gold = 19300 kg/m³
mg = mass of gold = ?
Vg = Volume of gold sphere = 0.0335 m³
Therefore,
mg = (19300 kg/m³)(0.0335 m³)
mg = 646.75 kg
Now, the volume of sand required for equivalent mass of gold, will be given by:
ρs = mg/Vs
where,
ρs = density of sand = 1602 kg/m³
mg = mass of gold = 646.75 kg
Vs = Volume of sand = ?
Therefore,
1602 kg/m³ = 646.75 kg/Vs
Vs = (646.75 kg)/(1602 kg/m³)
Vs = 0.4 m³
Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:
Vs = (4/3)πr³
0.4 m³ = (4/3)πr³
r³ = 3(0.4 m³)/4π
r³ = 0.095 m³
r = ∛(0.095 m³)
r = 0.46 m
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
Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
(a) What is the longest wavelength for which there will be destructive interference at point Q?
(b) What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. for destructive interference
λmax= 240m
b. for constructive interference
λmax = 120m
Explanation:
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
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.
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
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.
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 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
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
An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is
Answer:
15amps
Explanation:
V=IR
I=V/R
I = 120/8
I = 15 amps
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.
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
The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.
Required:
At this average price, calculate the cost of:
a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.
Answer:
Cost = $ 1.48
Cost = $ 0.005
Cost = $ 0.38
Explanation:
given data
electrical transmission varies = $0.070/kWh to $0.258/kWh
average value = $0.110/kWh
solution
when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be
first we get here energy consumed that is express as
E = Pt .................1
here E is Energy Consumed and Power Delivered is P and t is time
so power is here 0.04 KW and t = 2 week = 336 hour
so
put value in 1 we get
E = 0.04 × 336
E = 13.44 KWh
so cost will be as
Cost = E × Unit Price .............2
put here value and we get
Cost = 13.44 × 0.11
Cost = $ 1.48
and
when you making a piece of dark toast in 3.00 min with a 970-W toaster
so energy consumed will be by equation 1 we get
E = Pt
power is = 0.97 KW and time = 3 min = 0.05 hour
put value in equation 1 for energy consume
E = 0.97 × 0.05 h
E = 0.0485 KWh
and we get cost by w\put value in equation 2 that will be
cost = E × Unit Price
cost = 0.0485 × 0.11
Cost = $ 0.005
and
when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer
from equation 1 we get energy consume
E = Pt
Power Delivered = 5.203 KW and time = 40 min = 0.67 hour
E = 5.203 × 0.67
E = 3.47 KWh
and
cost will by put value in equation 2
Cost = E × Unit Price
Cost = 3.47 × 0.11
Cost = $ 0.38
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.
Read more: https://brainly.com/question/23153766
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
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
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
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 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
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]
A uniform 2.0-kg rod that is 0.92 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 29 N/m and 66 N/m. Find the angle that the rod makes with the horizontal.
Answer:
11.7°
Explanation:
See attached file
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?
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
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
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:
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
If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
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
https://brainly.com/question/13770820
#SPJ3