We will have the following:
1. Windmill: From wnd energy to mechanical energy.
2. Flashlight: From chemical energy to ligth energy and thermal energy.
3. Firecracker: From chemical energy to thermal energy and kinetic energy.
4. Bycicle: From mechanical energy to kinetic energy.
5. Microwave: From electric energy to thermal energy,
your pet hamster sits on a record player that has constant angular speed . if the hamster moves to a point twice as far from the center, then its linear speed
Part 1 Assume that as a wave travels from one particular medium to another, its speed decreases. What happens to the wave's frequency?A. Its frequency increases.B. Its frequency decreases.C. Its frequency remains the same.D. It is impossible to determine without more data.Part 2Which best explains the correct answer to Part 1?A. A wave's frequency always remains the same. B. A wave's frequency is inversely proportional to its speed.C. Given that the wave's wavelength remains the same, if the wave's speed decreases, then the frequency must decrease.D. Given that a wave's speed is equal to the product of its wavelength and frequency, the wavelength must be known to determine the frequency.
Part 1
C. Its frequency remains the same.
Part 2
A. A wave's frequency always remains the same.
Explanation:The relationship between wavelength, frequency, and speed is given as:
[tex]v=\lambda f[/tex]When the frequency of a wave increases, its wavelength also increases.
The speed of a wave does not change as its frequency changes, and vice-versa.
Therefore, as the speed of the wave decreases, its frequency remains the same
A wave's frequency always remains the same regardless of the speed and wavelength
P and Q are points within a uniform electric field that are separated by a distance of 0.2 m as shown. The potential difference between P and Q is 75 V.How much work is required to move a +150 μC point charge from P to Q?0.023 J140 J0.011 J2800 J75 J
ANSWER:
0.011 J
STEP-BY-STEP EXPLANATION:
Given:
q = 150 μC
v = 75 V
r = 0.2
We calculate the work as the product between the charge and the potential difference, just like this:
[tex]\begin{gathered} W=q\cdot v \\ W=150\cdot10^{-6}\cdot75 \\ W=0.01125\text{ J} \\ W\cong0.011\text{ J} \end{gathered}[/tex]The work required is 0.011 J
Question 8 of 10 A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The brand of salt used B. The time it takes for the water to boil C. The kind of bottles used D. The amount of salt added to the water SUBMIT it's B
Answer:
Explanation:
A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The time it takes for the water to boil B. The amount of salt added to the water C. The kind of bottles used D. The brand of salt used
9.A stock car is moving at 25.0 m/s when the driverapplies the brakes. If it stops in 3.00 s, what is itsaverage acceleration?
The average acceleration a of an object that changes its speed from v_0 to v_f during a time interval t is:
[tex]a=\frac{v_f-v_0}{t}[/tex]Since the car is initially moving at 25.0m/s and it stops in 3.00 seconds, then its final speed is 0m/s.
Replace v_f=0, v_0=25.0m/s and t=3.00s to find the average acceleration of the car:
[tex]a=\frac{0\frac{m}{s}-25.0\frac{m}{s}}{3.00s}=-8.333\ldots\frac{m}{s^2}[/tex]Therefore, the average acceleration of the car when it stops from 25.0m/s in 3.00 seconds, is -8.33m/s^2.
A bicycle wheel, of radius 0.300 m and mass 1.47 kg (concentrated on the rim), is rotating at 4.00 rev/s. After 56.4 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? Nm
Given,
The radius of the wheel, r=0.300 m
The mass of the wheel, m=1.47 kg
The initial angular frequency of the wheel, f=4.00 rev/s
The time it takes for the wheel to come to a stop, t=56.4 s
The final angular velocity of the wheel, ω=0 rad/s
The initial angular velocity of the wheel is
[tex]\begin{gathered} \omega_0=2\pi f \\ =2\pi\times4.00 \\ =25.13\text{ rad/s} \end{gathered}[/tex]From the equation of motion,
[tex]\omega=\omega_0+\alpha t[/tex]Where α is the angular acceleration of the wheel.
On substituting the known values,
[tex]\begin{gathered} 0=25.13+\alpha\times56.4 \\ \Rightarrow\alpha=\frac{-25.13}{56.4} \\ =-0.45\text{ rad/s}^2 \end{gathered}[/tex]The moment of inertia of the wheel whose mass is concentrated on the rim is given by,
[tex]I=mr^2^{}[/tex]The torque applied by the friction is given by,
[tex]\begin{gathered} \tau=I\alpha \\ =mr^2\alpha \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=1.47\times0.300^2\times-0.45 \\ =-0.06\text{ Nm} \end{gathered}[/tex]The negative sign of the torque indicates that the torque is applied to the wheel in the direction opposite to the direction of rotation of the wheel.
Thus the magnitude average torque due to frictional force is 0.06 Nm
The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time between two events is measured by clocks at rest in a reference frame in which the two events: a. are separated by the distance a light signal can travel during the time interval
b. occur at the same time
c. satisfy none of the above
d. occur in Nashville occur at the same coordinates
The correct option is option (D) occur in Nashville occur at the same coordinates.
In general proper time is the time interval between two events as measured by an observer who sees the event occur at the same position.
In geometry, a coordinate system is a device that makes use of one or more numbers, or coordinates, to uniquely determine the location of the points or different geometric factors on a manifold which includes Euclidean area.
Math coordinates discover the vicinity of a factor on a graph or map. The ordered pair, (x, y) is the cope with of the factor. The Cartesian coordinate gadget is the graph used to locate the point. The graph is split up into four quadrants.
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ectile mo
A rifle is aimed horizontally at a target 35.0 m away. The bullet hits the target hits the
target 2.10 cm below the aiming point. a) What is the bullet's time of flight? b) What is
its muzzle velocity?
A 2.55kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0360m . The spring has force constant 870N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0160 m .)
The speed of the block when it has moved a distance of 0.0200 m from its initial position is 0.29 m/s.
The mass of the block is 2.55 Kg. Coefficient of kinetic friction is 0.45. Spring constant is 870N/m. The final compression of the spring is 0.0160 meters. The block is moved by distance of 0.0200 meters.
By considering the block-spring system, we can use Work energy theorem here,
According to work energy theorem,
Work done by conservative forces on the body + Work done by none conservative forces = Total change in kinetic energy of the body.
Work done by Conservative forces = 1/2Kx².
Where,
K is spring constant,
x is the compression in spring,
Work done by non-conservative forces = umgd
Where,
u is the coefficient of kinetic friction,
m is mass of the block,
d is the distance by which the block moves,
Here, we will take spring force as positive because the spring force and displacement both are in same direction,
Putting all the values,
1/2Kx² - umgd = 1/2mV² - 1/2mU²
1/2(870)(0.016)² - 0.45(2.55)(9.8)(0.02) = 1/2(2.55)V²
0.11136 - 0.22491= (1.275)V²
0.11355/1.275 = V²
V = 0.29 m/s.
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A wagon weighing 1000kg moves 50km/h on smooth rails. Later, a mass of 250 kg is placed in the wagon what is the velocity with which it moves?
The final velocity V is 40 km/h.
The final velocity V.
Mass of wagon m = 1000 kg
Initial velocity of wagon v = 50 km/h
Mass dropped velocity M = 250 kg
The final velocity =V
Initial system linear momentum [tex]P_{i}[/tex] = mv
=1000*50
=5*[tex]10^{4}[/tex] kg km/h
The system with the final velocity (wagon + mass dropped) be V .
Final system linear momentum [tex]P_{f}[/tex] = (m + M) V
=(1000+250)V
=1250 V
Basic principle of conservation of linear momentum,
[tex]P_{f}[/tex] = [tex]P_{i}[/tex]
1250V=5*[tex]10^{4}[/tex]
V=[tex]\frac{5*10^{4} }{1250}[/tex]
V= 40 km/h
So, the final velocity V is 40 km/h.
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A projectile is fired horizontally at an angle of 70° with the speed of 0.11km/s. Find the range and height. The final answer should be in meters.
In the given question projectile is fired horizontally at an angle of 70° with the speed of 0.11km/s ,(=110m/s).
to find the range and height
height=544.66m
time= 10.54 second
range=793.63m
vy=0
h=v^2sin^2θ/ 2g
v=0.11km/s =110m/s
θ=70°
range=v^2sin2θ/ g
horizontal velocity,
vx= 110cos70=37.622 m/s
vertical velocity,
vy=110sin70=103.36 m/s
using the equation of motion
s=ut+1/2 at^2
range, s=0, t=total time taken
range= 110^2 sin 2*70/9.8
range=793.63m
t= vsinθ/ g
=110 sin 70/9.8
time= 10.54 second
s=(110sin 70)*10.54 +1/2(-9.8)* (10.54^2)
s=1089-544.34
height=544.66m
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This is astronomy and give explanations
The scientific study of all celestial objects is known as astronomy. Galaxies, planets, stars, comets, meteors, and other events affecting the celestial bodies are all observed and studied by astronomers.
The words "Astro," which means "star," and "nomos," which means "rule," are derived from ancient Greek. Astronomy, when combined, refers to "star law." The natural sciences are the ones that it is the oldest.
Since there are so many celestial objects to study, it is practical to divide astronomy into various branches.
Here are some of the more well-known astronomical fields-
1. Planetary Astronomy
2. Solar Astronomy
3. Stellar Astronomy
4. Galactic Astronomy
5. Observational Astronomy; etc.
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Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will go the highest: if it is set at a 25-degree angle, or if it is set at a 35-degree angle?
Question 3 options:
25 degrees
35 degrees
The cannon ball will travel the highest distance when the angle of projection is 35 degrees.
What is the maximum height of a projectile?The maximum height reached by a projectile is calculated using the following formula.
H = u²sin²θ/2g
where;
u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 25 degrees;
H = (15² (sin 25)²) / (2 x 9.8)
H = 2.05 m
when the angle of projection is 35 degrees;
H = (15² (sin 35)²) / (2 x 9.8)
H = 3.78 m
Thus, the cannon ball will travel the highest distance when the angle of projection is 35 degrees.
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The table below shows part of the operating budget of a small dairy farm for the last year. The only expense not listed is maintenance. LAST YEARS OPERATING BUDGETExpense Fraction of BudgetFood 1/3Housing 1/3 Medical Care 1/4This year the managers of the farm will change the fraction for the budget for housing to ⅛ but will leave the fraction of the budget for good and medical care the same. Again, the remaining portion of the budget will be before maintenance expenses. What is the difference between the fraction of the budget for maintenance this year and last year?Show your workAnswer
Explanation
Step 1
let the fractions
[tex]\begin{gathered} Food\text{ }\frac{\text{1}}{3} \\ \text{ Housing }\frac{\text{1}}{3} \\ \text{Medical care }\frac{1}{4} \\ \text{ maintenance= M ( unknown)} \end{gathered}[/tex]so
a)This year the managers of the farm will change the fraction for the budget for housing to 1/8
so
[tex]\begin{gathered} Food\text{ }\frac{\text{1}}{3} \\ \text{ Housing }\frac{\text{1}}{8} \\ \text{Medical care }\frac{1}{4} \\ \text{ maintenance= M ( unknown)} \end{gathered}[/tex]b) hence,the total budget is
[tex]\begin{gathered} \text{past year= Operatingbudget= food+housidn +Medical care+maintenance} \\ \text{replace} \\ past\text{ year= }\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+m \\ past\text{ year=}\frac{11}{12}+m \end{gathered}[/tex]this year
[tex]\begin{gathered} \text{past year= Operatingbudget= food+housidn +Medical care+maintenance} \\ \text{replace} \\ thisyear\text{= }\frac{1}{3}+\frac{1}{8}+\frac{1}{4}+m \\ thisyear\text{=}\frac{17}{24}+m \end{gathered}[/tex]Step 2
find the fractions for maintenance on each year:
as we are using fraction of the total budget, we need to set the sum of the fractions equals 1, hence
a) past year
[tex]\begin{gathered} \text{total = 1, so} \\ past\text{ year=}\frac{11}{12}+m=1 \\ m=1-\frac{11}{12} \\ m=\text{ }\frac{1}{12} \end{gathered}[/tex]b) this year
[tex]\begin{gathered} \text{total = 1, so} \\ this\text{year=}\frac{17}{24}+m=1 \\ m=1-\frac{17}{24} \\ m=\text{ }\frac{7}{24} \end{gathered}[/tex]Step 3
now, find the difference
[tex]\begin{gathered} \text{difference =}\frac{7}{24}-\frac{1}{12} \\ \text{difference =}\frac{5}{24} \end{gathered}[/tex]therefore, the answer is 5/24
I hope this helps you
When you drop a 0.38 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 to- ward the earth’s surface. According to Newtons third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 x 10^24 kg, what is the magnitude of the Earth's acceleration toward the apple? Answer in units of m/s^2
The magnitude of the earth’s acceleration toward the apple with a mass of 0.38 kg is 6.2 × 10⁻²⁵ m/s².
What is gravity?Gravity is a fundamental force which is responsible for causing mutual attraction between all the things with mass or energy.
The gravitational force can be calculated by the formula:
F = m.a
where, F = force
m = mass of an object
a = acceleration of the object
Here, the mass of the apple, m = 0.34 kg
The acceleration of the apple towards the earth's surface (a) = 9.8 m/s²
The mass of the Earth, M = 5.98 × 10²⁴ kg
Now, by applying Newton's Law of the motion,
The magnitude of the force on the apple due to Earth is can be calculated as:
F = m × a
F = 0.34 × 9.8
F = 3.724 N
According to the Newton's third law of motion, the apple must exert an equal but opposite force on Earth.
So, the magnitude of the force on Earth due to the apple will be:
F = 3.724 N
Let the magnitude of the earth’s acceleration towards the apple be 'A' m/s².
Thus, the magnitude of the earth’s acceleration towards the apple will be given by:
F = MA
M = mass of the Earth
A = acceleration of the Earth
3.724 = [5.8 × 10²⁴] × A
A = 3.724 / [5.8 × 10²⁴]
A = 0.6227 × 10⁻²⁴
A = 6.2 × 10⁻²⁵ m/s²
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Archaeological samples are often dated by radiocarbon dating. The half-life of carbon-14 is 5,700 years.
After how many half-lives will the sample have only 1/64 as much carbon-14 as it originally contained?
How much time will have passed?
If the daughter product of carbon-14 is present in the sample when it forms (even before any radioactive decay happens), you cannot assume that every daughter you see is the result of carbon-14 decay. If you did make this assumption, would you overestimate or underestimate the age of the sample?
There are six half lives that have passed within 34,200 years. When we assume that we have some of the products present initially, then we overestimate the age of the sample.
What is the half life?The half life is the time taken to obtain only half of the number of atoms that were originally present in the radioactive material. We know that living things do contain the carbon- 14 isotope together with the carbon - 12 isotope in the substance. When the organism dies, the carbon - 14 which is radioactive begins to decay and its half life could be used to estimate the age of the sample as we know.
Given that after each half life, we have only half of the original amount that remains. We would have 1/64 of the original amount left after six half lives and this means that 34,200 years have passed.
If we the daughter product of carbon-14 is present in the sample when it forms (even before any radioactive decay happens), then we would over estimate the age of the sample.
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_______ is the study of the way all living things relate to each other in the world. The word comes from two Greek words which mean the study or science of the home.
Ecology is the study of the way all living things relate to each other in the world.
What is Ecology?This is referred to as the science which deals with the study of organisms and how they interact with their environment. It comprises of the study of the organisms present and their habitat.
Their habitat is the area where they live which comprises of many resources such as food, water etc which are needed for their growth and survival. The living things interact with it for beneficial reasons and is the reason why Ecology was chosen as the correct choice.
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Circular Motion
The Trajectory of a mass point is a helix:[tex]r(t)=\left[\begin{array}{ccc}rcos(wt)\\rsin(wt)\\ht\end{array}\right]\\[/tex]
a) Compute the velocity v(t)=r(t) and the acceleration a(t).
b) What is the angel between the velocity and the acceleration?
c) If h=0, what is the radial acceleration and the velocity of the mass point at r=2m and 40rpm
No idea how to answer this question, I have never seen anything like it.
Any help would be awesome :)
Answer and step by step explanation:
First of all, I'm assuming you have had calculus, or this is going to be very awkward. Then, I'm replacing w with the Greek letter omega, it's a pet peeve of mine, sorry.
Shockingly, the derivative of a vector is computed by taking the derivative of each component (by linearity). If you've never heard the word derivative yet, you can think of the x and y components as two harmonic motions out of phase by 90°, and the z component as having constant speed h (harmonic motion is what you see when a mass moves along a circle with constant angular velocity and you look at it from the side).
At this point, let's use the derivative method for question a:
[tex]v(t)=\dot{r}(t)=\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]; a(t)=\dot v(t)=\ddot{r}(t)=\left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right]\\[/tex]
Point b requires computing angles, which screams dot product to me.
[tex]\vec v(t) \cdot \vec a(t) =\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]\cdot \left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right] = + \omega^3r^2 sin (\omega t) cos (\omega t) -\omega^3r^2 cos (\omega t) sin (\omega t) = 0 = ||\vec v|| ||\vec a|| cos \theta \implies cos\theta = 0 \implies \theta=\frac \pi2[/tex]
Now, the implied part is granted by the fact that we are assuming neither the velocity nor the acceleration are both zero, so the only option is for the cosine being zero, that makes the two vector orthogonal.
Finally, for point c, let's just take the moduli of both velocity and acceleration [tex]||\vec v|| = \omega r; ||\vec a||=\omega^2 r[/tex] and let's convert the angular velocity in civiliz... err, IS units: [tex]40 rpm = 40\times \frac{2\pi}{60}rad/s = \frac 43 rad/s[/tex]
Let's replace and we get
[tex]v=\frac83= 2.6 m/s\\a= \frac{32}3 = 10.5 m/s^2[/tex]
If a piece of ribbon were tied to a stretched string carrying a transverse wave, then how is the ribbon observed to oscillate?a.)perpendicular to wave directionb.)both perpendicular to and parallel to wave direction
The ribbon was observed to oscillate perpendicular to the wave direction. Option A.
In transverse waves, particles move perpendicular to the direction of wave propagation. Examples of transverse waves are string vibrations and water waves. By moving the Slinky vertically up and down, you can create horizontal transverse waves. Her one type of mechanical wave is the transverse wave.
where the motion of the medium is perpendicular to the direction of energy propagation. A transverse wave is a wave whose vibration is perpendicular to the direction of wave propagation. This is in contrast to longitudinal waves, which propagate in the direction of oscillation. Water waves are an example of transverse waves.
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Consider a 1500kg rollercoaster moving at 4.5m/s at the top of a
30.0m hill. If instead, friction does 52kJ of work on the rollercoaster on its
trip down the hill, how fast will the rollercoaster be moving once it
reaches ground level?
The rollercoaster be moving once it reaches ground level with a speed of 23.21 m/s.
Applying work-energy theorem,
Work done by conservative forces on the body + work done by none conservative forces on the body = Total change in Kinetic Energy of the body.
Here, conservative forces are gravitational force and non-conservative force is frictional force,
MgH - Fr = 1/2MV² - 1/2MU²
Where,
M is mass of roller coaster,
g is acceleration due to gravity,
H is the height of the roller coaster,
Fr is the frictional force,
V is the final velocity of roller coaster,
U is the initial velocity of the roller coaster,
1500(9.8)(30) - 52000 = 1/2(1500)(V²-20.25)
(441000 -52000)2/1500 = V² - 20.25
518.67 = V² - 20.25
V² = 538.9
V = 23.21 m/s.
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When a force is exerted on a box, an equal and opposite force is exertedby the box. These forces are calledforces.O action-reactionO frictionalO centripetalO gravitational
Explanation
Newton's third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.
let's check a box on a table
The object represented by line E does not stop. Is this true or false?
The given graph is about velocity and time, the dependent and independent variable, respectively. As we can observe, line E crosses the zero level to the negative zone, which means the object didn't stop but changed its direction.
Therefore, the statement is true.For friction on an incline, what direction does Fp point?
The force of friction on an inclined plane, will act upward to oppose the downward motion of the object on the inclined plane.
What is force of friction?
The force of friction is the force that opposes the motion of an object.
Since the force of friction opposes the motion of an object, it acts upward along the plane for an object moving along an inclined plane.
As the object moves downwards, the force of friction will act upwards to oppose the downward motion of the object.
Mathematically, the magnitude of force of friction on an object moving along inclined plane is given as;
Ff = μmg
where;
μ is the coefficient of frictionm is the mass of the objectg is acceleration due to gravityThus, the force of friction on an inclined plane, will act upward to oppose the downward motion of the object on the inclined plane.
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The complete question is below:
For friction on an incline, what direction does force of friction (Fp) point?
A pilot wishes to fly 2500km due South. The air speed of the plane is 300km/h and there is a wind of 50km/h from the West. What heading should the pilot take and what is the time required to complete the flight?
Since there is a wind speed of 50 km/h to the west, this speed needs to be compensated by the heading of the plane.
If we decompose the plane speed, one component will be used to cancel out the wind speed:
To find the orientation theta, we can calculate the east component of the plane speed and equate it to the wind speed:
[tex]\begin{gathered} V_{\text{east}}=V_{\text{wind}} \\ V\cdot\sin (\theta)=50 \\ 300\cdot\sin (\theta)=50 \\ \sin (\theta)=\frac{50}{300}=\frac{1}{6} \\ \theta=\sin ^{-1}(\frac{1}{6}) \\ \theta=9.59\degree \end{gathered}[/tex]Now, let's calculate the south component:
[tex]\begin{gathered} V_{\text{south}}=V\cdot\cos (9.59\degree) \\ V_{\text{south}}=300\cdot0.986 \\ V_{\text{south}}=295.8\text{ km/h} \end{gathered}[/tex]Calculating the required time of flight, we have:
[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ 2500=295.8\cdot t \\ t=\frac{2500}{295.8} \\ t=8.45\text{ hours} \end{gathered}[/tex]Therefore the orientation is 9.59° east of south and the time required is 8.45 hours.
If the displacement, velocity and acceleration at an instant of a particledescribing S.H.M are respectively 7.5m, 7.5m/s, 7.5m/s?. Calculate themaximum velocity of the particle.
We are given the displacement, velocity, and acceleration of a particle that describes Simple Harmonic Motion. We are asked to determine the maximum velocity of the particle. To do that we can use the following equation for the maximum velocity of a particle describing SHM:
[tex]v_{máx}=-A\omega[/tex]Where:
[tex]\begin{gathered} A=\text{ amplitude} \\ \omega=\text{ angular frequency} \end{gathered}[/tex]To determine the values of amplitude and angular frequency we can use the expressions for displacement, velocity, and acceleration. The expression for displacement is:
[tex]x=A\cos (\omega t)[/tex]The expression for the velocity is:
[tex]v=-A\omega\sin (\omega t)[/tex]And the expression for acceleration is:
[tex]a=-A\omega^2\cos (\omega t)[/tex]Now, from the expression for the displacement we can solve for the amplitude, like this:
[tex]\frac{x}{\cos(\omega t)}=A[/tex]Now we can replace this in the expression got eh acceleration:
[tex]a=-\frac{x}{\cos(\omega t)}\omega^2\cos (\omega t)[/tex]Simplifying:
[tex]a=-x\omega^2[/tex]Now we can solve for the angular frequency:
[tex]-\frac{a}{x}=\omega^2[/tex]Taking square root to both sides:
[tex]\sqrt[]{-\frac{a}{x}}=\omega[/tex]replacing the values:
[tex]\sqrt[]{-\frac{-7.5\frac{m}{s^2}}{7.5m}}=\omega[/tex]Solving the operations:
[tex]1s^{-1}=\omega[/tex]Now we divide the formula for the displacement and the formula for the velocity, we get:
[tex]\frac{v}{x}=\frac{-A\omega\sin (\omega t)}{A\cos (\omega t)}[/tex]Simplifying we get:
[tex]\frac{v}{x}=-\omega\tan (\omega t)[/tex]Replacing the known values:
[tex]\frac{-7.5\frac{m}{s}}{7.5m}=-(1s^{-1})\tan (t)[/tex]Simplifying:
[tex]-1=-\tan (t)[/tex]Solving for "t":
[tex]t=\arctan (1)=0.78[/tex]Now we can replace these values in the formula for displacement to get the value of the amplitude:
[tex]x=A\cos (\omega t)[/tex]Solving for the amplitude:
[tex]\frac{x}{\cos (\omega t)}=A[/tex]Replacing the known values:
[tex]\frac{7.5m}{\cos (0.78)}=A[/tex]Solving the operations:
[tex]10.6m=A[/tex]Now we replace the values in the formula for the maximum velocity:
[tex]v_{\text{max}}=A\omega[/tex]Replacing:
[tex]\begin{gathered} v_{\max }=(10.6m)(1s^{-1}) \\ v_{\max }=10.6\text{ m/s} \end{gathered}[/tex]Therefore, the maximum velocity is 10.6 meters per second.
A 0.35-kg tennis racquet moving to the right at 20 m/s hits a 0.06-kg tennis ball that is moving to the left at 30 m/s. After the collision, the racquet continues to the right, but at a reduced speed of 10 m/s. What is the ball's velocity after the collision?
From conservation of momentum we have that:
[tex]0.35(20)-0.06(30)=0.35(10)+0.06v[/tex]Solving for v we have that:
[tex]\begin{gathered} 0.35(20)-0.06(30)=0.35(10)+0.06v \\ 7-1.8=3.5+0.06v \\ 0.06v=7-1.8-3.5 \\ 0.06v=1.7 \\ v=\frac{1.7}{0.06} \\ v=28.33 \end{gathered}[/tex]Therefore the velocity of the ball after the collision is 28.33 m/s
Which ideas describe the big crunch? (1 point)
O After the universe reaches its expansion limit, gravity will pull it all back together.
O The big bang actually never occurred, and the universe must have a different origin.
O The universe goes through infinite cycles of expansion and contraction.
O The universe is not only composed of the objects that we can see.
Answer:
a
Explanation:
after the universe reach the limit, gravity does its thing
WHAT IS CHROMATOGRAPHY ?
Answer:
Chromatography is a process for separating components of a mixture. To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.
Explanation:
hope it will be helpful
Answer:
Chromatography is a process for separating components of a mixture.
Explanation:
To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.
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In the experiment, the pressure of the gas is 1.2 x10^5 Pa at a temperature of 25.0°C.
When the cylinder is heated, the pressure reaches 2.1x10 Pa. Calculate the
temperature of the gas (in "C) at this pressure.
The temperature of the gas (in °C) at this pressure is 248.5 °C
Temperature is a bodily quantity that expresses the hotness of count or radiation. There are 3 kinds of temperature scales. Temperature is the degree of hotness or coldness of an item.
Temperature is a degree of the common kinetic energy of the debris in an object. whilst the temperature increases, the motion of those particles also increases. Temperature is measured with a thermometer or calorimeter.
Given;
P₁ = 1.2 x10⁵ Pa = 1.18430792
T₁ = 25.0°C = 298 K
P₂ = 2.1x10 Pa = 0.000207253886
T₂ = ?
using ideal gas equaion:-
PV = nRT
P₁/T₁ = P₂/T₂
T₂ = P₂T₁ /P₁
= 2.1x10⁵ x 298 / 1.2 x10⁵
= 521.5 K
= 248.5 °C
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Set the cannon to have an initial speed of 25 m/s at an angle of 30-degrees. Find how long it takes to hit the ground. Will the time it takes to get to the highest point be less than ½ the time, more than ½ the time, or ½ the time?
Question 7 options:
1/2 the time
more than 1/2 the time
less than 1/2 the time
The time taken by the projectile to reach the highest point would be one half of the total time of the projectile , therefore the correct answer is option D .
What is a projectile motion ?It is the motion of any object or body when it is ejected off the surface of the earth and follows any curving path while being affected by the gravitational pull of the planet .
As given in the problem Set the cannon to have an initial speed of 25 m / s at an angle of 30 degrees .
Thus, the right response is option D because it would take the projectile half as long to reach the highest position.
To learn more about projectile motion here,
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