The work done by the spring force to ready the pen for writing is approximately 0.015 N· m or 0.015 J (joules).
To calculate the work done by the spring force to ready the pen for writing, we need to determine the total displacement of the spring when it is compressed.
The initial compression of the spring is given as 5.3 mm, and the additional compression to lock the pen into its writing position is 5.9 mm.
Therefore, the total compression of the spring is the sum of these two values:
Total compression = 5.3 mm + 5.9 mm = 11.2 mm = 0.0112 m
The work done by the spring force can be calculated using the formula:
Work = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring.
Plugging in the values:
Work = (1/2) * 233 N/m * (0.0112 m)^2
Work = 0.5 * 233 N/m * (0.00012544 m^2)
Work = 0.0147 N * m
Rounded to two significant figures, the work done by the spring force to ready the pen for writing is approximately 0.015 N·m or 0.015 J (joules).
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The concept of habituation is best exemplified by which of the following situations?
a. An infant recognizes her father's voice
b. A college student is no longer kept awake by her roommate's late night typing.
c. a kitten avoids a couch after being reprimanded for sitting on it
d. A rat learns to press a bar for food when a red light is flashed
e. A motorist drives at the speed limit when there is a police driver in sight on the highway
The concept of habituation is best exemplified by situation b, where a college student is no longer kept awake by her roommate's late night typing.
Habituation refers to the decrease in responsiveness to a stimulus after repeated exposure to it. This means that over time, the student becomes less and less responsive to the sound of typing, and it no longer disrupts her sleep.
Habituation is a basic form of learning and is thought to be an adaptive mechanism, allowing organisms to conserve energy by ignoring irrelevant stimuli. It is often observed in infants and young animals, but can occur in any organism that is capable of learning.
In situation a, an infant recognizes her father's voice, this is not an example of habituation, but rather an example of classical conditioning. In situation c, a kitten avoids a couch after being reprimanded for sitting on it, this is an example of classical conditioning or a form of punishment-based learning. In situation d, a rat learns to press a bar for food when a red light is flashed, this is an example of operant conditioning. And in situation e, a motorist drives at the speed limit when there is a police driver in sight on the highway, this is an example of social learning.
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you build a circuit with three incandescent bulbs in parallel one day in the phys 251 laboratory. bulbs 1 and 2 are identical, but bulb 3 has a larger resistance. bulb 1 burns out (becomes a break). we can say that:
If bulb 1 burns out in a parallel circuit with three incandescent bulbs, bulbs 2 and 3 will remain lit, but the overall resistance of the circuit will decrease, resulting in an increase in the total current flowing through the circuit.
In a parallel circuit, each bulb has its own path to the power supply, so if one bulb burns out, the current can still flow through the other bulbs. Bulbs 2 and 3 in this case will remain lit because they are still receiving the same voltage as before, but the overall resistance of the circuit will decrease due to the absence of bulb 1. This is because the resistance of the circuit is determined by the sum of the individual resistances of each bulb, and with one bulb removed, the total resistance of the circuit decreases.
Since the voltage across the circuit remains the same, the decrease in resistance results in an increase in the total current flowing through the circuit, which can cause the remaining bulbs to become brighter than they were before. It is important to note that the increase in current could potentially cause bulb 3 to burn out faster than it would have if all three bulbs were still functioning, as it is the bulb with the larger resistance and therefore may not be able to handle the increased current.
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During a snow day you decide to go sledding! You exert a force of 45N in order to pull a sled. You also perform 4723 J of work on the sled. Find the distance that you pulled the sled horizontally along the ground.
one example of these kinds of economic activities is ____
One very good example of kinds of economic activities simply is buying and selling goods.
How buying and selling of goods an an economic activitiesEconomics simply refers to the study of social science which human relationship in relationships with ends and and scarce means which have alternative uses.
However, economics from the description above also involves the buying and selling of goods or products. These selling of these goods is also a function of the quantity demanded.
In conclusion, we can now confirm and deduce from the explanation given above that we the quantity of goods demanded will determine its sales.
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So right now you think you're not moving? really? guess again. the earth is rotating
on its axis so if you were on the equator you would have traveled 40,000 km each day
(just a bit more than 460 meters each second or a bit over 1,000 mi/hr), but it's way
worse than that! we are traveling around the sun at a speed of 30,000 m/s. just in
case you were wondering, the mass of the earth is about 6.0 x 1024 kg.
a. what would be your momentum (let's say you have a mass of 100 kg) from the
rotation of the earth alone if you were on the equator?
b. what is the momentum of the earth traveling around the sun?
a) If you have a mass of 100 kg, your momentum from the rotation of the earth alone would be: 46,300 kg m/s ; b) The momentum of the earth traveling around the sun is 1.8 x 10²⁹ kg m/s.
a. To calculate your momentum from the rotation of the earth alone, we first need to find your velocity. As stated in the question, if you were on the equator, you would have traveled 40,000 km each day. To convert this to meters per second, we can use the fact that there are 86,400 seconds in a day:
40,000 km/day = 40,000,000 meters/day
40,000,000 meters/day / 86,400 seconds/day = 463 meters/second
So, if you have a mass of 100 kg, your momentum from the rotation of the earth alone would be:
momentum = mass x velocity
momentum = 100 kg x 463 m/s
momentum = 46,300 kg m/s
b. To calculate the momentum of the earth traveling around the sun, we first need to find the velocity of the earth. As stated in the question, the earth is traveling around the sun at a speed of 30,000 m/s.
The momentum of the earth can be calculated using its mass (6.0 x 10²⁴ kg) and velocity (30,000 m/s):
momentum = mass x velocity
momentum = 6.0 x 10²⁴ kg x 30,000 m/s
momentum = 1.8 x 10²⁹ kg m/s
So, the momentum of the earth traveling around the sun is 1.8 x 10²⁹ kg m/s.
In summary, if you were on the equator, you would have a momentum of 46,300 kg m/s from the rotation of the earth alone. The momentum of the earth traveling around the sun is much larger at 1.8 x 10²⁹ kg m/s.
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A car speedometer has a 4% uncertainty. what is the range of possible speeds when it reads 100 km/h? km/h (lowest) km/h (highest)
Answer:
104 highest and 96 the lowest
Explanation:
it has an uncertainness of 4%Meaning 104%on the highest and 96 on the lowest
Please I need help on this my teacher is not good at explaining! I attached a photo with the questions it also auto copied it here.
001 (part 1 of 3) 10.0 points
A ball of mass 0.3 kg, initially at rest, is
kicked directly toward a fence from a point
20 m away, as shown below.
The velocity of the ball as it leaves the
kicker's foot is 16 m/s at angle of 43° above.
the horizontal. The top of the fence is 3 m
high. The ball hits nothing while in flight and
air resistance is negligible.
The acceleration due to gravity is 9.8 m/s².
16 m/s
43°
T
3 m
Y
20 m
Determine the time it takes for the ball to
reach the plane of the fence.
Answer in units of s.
002 (part 2 of 3) 10.0 points
How far above the top of fence will the ball
pass? Consider the diameter of the ball to be
negligible.
Answer in units of m.
003 (part 3 of 3) 10.0 points
What is the vertical component of the velocity
when the ball reaches the plane of the fence?
Answer in units of m/s.
A ball of mass 0.3 kg, initially at rest, is kicked directly toward a fence from a point 16m away. The velocity of the ball as it leaves the kicker's foot is 16 m/s at an angle of 48 degrees above the horizontal. The top of the fence is 3 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8 meters per second square.
a) The time it takes for the ball to reach the plane of the fence is (t)= 1.49 s
b) The ball pass above the top of the fence will be (h)=3.81m
c) The vertical component of the velocity when the ball reaches the plane of the fence is [v(y)]= -2.71 m/s
What is velocity?The term velocity means that the object's covered distance within the given time. Example: A car goes 3m in 1 second the velocity of the car is 3m/s.
How can we calculate the parameters?a) To calculate the time we are using the formula,
x=vcosθt
Or, t = x/ vcosθ
Here we are given,
v= the initial velocity of the ball =16 m/s
θ = the initial angle = 48°
x= The distance of the fence = 16m
W have to calculate the time =t
Now we put the values in the above equation we get,
t = x/ vcosθ
Or, t= 16/16*cos48°
Or, t = 1.49 S
Now we know that, the time it takes for the ball to reach the plane of the fence is (t)=1.49 s
b) Now we have to calculate the height of the ball when it cross the fence. we are using the formula,
y= vsinθt-(1/2)gt²
t = the time we got earlier = 1.49 s
g= The acceleration due to gravity = 9.8 m/s².
Now we put the values in the above equation we get,
y= vsinθt-(1/2)gt²
Or, y = 16*sin48*1.49-(1/2)*9.8*(1.49)²
Or, y= 6.81 m
The ball pass above the top of the fence will be (h)=(y-y₁)=(6.81-3)=3.81m
c) To calculate the vertical component of the velocity when the ball reaches the plane of the fence, we are using the formula,
v(y)=vsinθ-gt
Or, v(y)=16*sin48-9.8*1.49
Or, v(y)= -2.71 m/s
From the calculation we assure, The vertical component of the velocity when the ball reaches the plane of the fence is [v(y)]= -2.71 m/s.
[Note: The component is negative cause the velocity is decreasing]
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A ball of mass 0.3 kg, initially at rest, is kicked directly toward a fence from a point 16m away. The velocity of the ball as it leaves the kicker's foot is 16 m/s at an angle of 48 degrees above the horizontal. The top of the fence is 3 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8 meters per second squared.
a) Determine the time it takes for the ball to reach the plane of the fence.
b) How far above the top of the fence will the ball pass? Consider the diameter of the ball to be negligible.
c) What is the vertical component of the velocity when the ball reaches the plane of the fence?
(c) A physics teacher sets up a demonstration to teach her Year 10 class ideas about the conservation of momentum. Figure 2 shows a straight, horizontal air track and two air track trolleys, S and T, which can move along it. Figure 2 Sharp pin o Cork S T has a mass of 3 x 10-1 kg and its velocity is 1.4 m/s to the left. S has a mass of 200 g and is stationary. When T collides with S they stick together. Calculate the velocity of the trolleys after the collision. Show your working clearly and give the unit and direction. [5 marks]
The velocity of the trolleys after the collision will be 0.84 m/sec. Both collide and stick together and move with the same velocity.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of S= 200 g =0.2 Kg
(u₁) is the initial velocity of S= 0 m/s
(m₂) is the mass T= 3 x 10⁻¹ kg
(u₂) is the initial velocity of T = 1.4 m/s
(v) is the velocity after collision =.?
According to the law of conservation of momentum;
Momentum before collision =Momentum after collision
\(\rm m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\ 3 \times 10^{-1} \times 1.4 + 0.2 \times 0 = v(0.3+0.2) \\\\ 0.5v=0.42 \\\\ v= 0.84 \ m/sec\)
Hence, the velocity of the trolleys after the collision will be 0.84 m/sec.
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What factors affect an objects kinetic energy?
Answer:
friction
air drag
every thing that opposes the motion affects kinetic energy
Explanation:
kinetic energy is a energy which is increase with increase in motion and potential energy is energy stored while the object is at rest
potential energy ∝ 1/(kinetic energy)
as kinetic energy increases potential energy decreases
What would the velocity of an object be if it was on a 1.75m rope and took 4 seconds to go around 2
revolutions
2.75 m/s
6.5 m/s
4 m/s
5.5 m/s
The velocity of the object is 5.5 m/s if it was on a 1.75 m rope and took 4 seconds to go around 2 revolutions. The formula used to calculate the velocity of an object in uniform circular motion is: v = 2πr / t Where, v is the velocity of an object in m/s, t is the time taken by the object in sπ = 3.14, r is the radius of the circle in meters
Now, we can calculate the velocity of the object using the given data: v = 2πr / tv = (2 × 3.14 × 1.75) / 4v = 5.5 m/s
Therefore, the velocity of the object is 5.5 m/s if it was on a 1.75 m rope and took 4 seconds to go around 2 revolutions. The velocity of an object in uniform circular motion can be calculated using the formula :v = 2πr / t Where, v is the velocity of an object in m/s, t is the time taken by the object in sπ = 3.14r is the radius of the circle in meters
Length of the rope = 1.75 m, Number of revolutions = 2, Time taken = 4 s
Now, we can substitute the given values in the formula and solve for v: v = 2πr / tv
= (2 × 3.14 × 1.75) / 4v
= 5.5 m/s
Therefore, the velocity of the object is 5.5 m/s if it was on a 1.75 m rope and took 4 seconds to go around 2 revolutions.
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What is the net force required to accelerate a 1.1 kg box at a rate of 2.8
m/s??
Answer:
Fnet = 13.86N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
\( Fnet = Fapp + Fg\)
Where;
Fnet is the net force Fapp is the applied forceFg is the force due to gravitationGiven the following data;
Mass = 1.1kg
Acceleration = 2.8m/s²
To find the applied force;
\( Fapp = ma\)
Substituting into the equation, we have;
\( Fapp = 1.1 * 2.8\)
Fapp = 3.08N
To find the gravitational force;
We know that acceleration due to gravity, g = 9.8m/s²
\( Fg = mg\)
Substituting into the equation;
\( Fg = 1.1 * 9.8\)
Fg = 10.78N
To find the net force;
\( Fnet = Fapp + Fg\)
Substituting into the equation, we have;
\( Fnet = 3.08 + 10.78\)
Fnet = 13.86N
Therefore, the net force required is 13.86N
Convert: 8 mm: ____________ cm
-1^10
centi=10 mm and mm=1/10=10^-1 cm
Describe the relationship between speed and thinking distance. Physics Paper 2
While there is no direct relationship between speed and thinking distance, higher speeds can result in longer thinking distances due to the increased reaction time needed by the driver.
The relationship between speed and thinking distance is not a direct one, as thinking distance is primarily influenced by the driver's reaction time rather than the actual speed of the vehicle. Thinking distance refers to the distance traveled by a vehicle during the driver's reaction time after perceiving a hazard.
However, there is an indirect relationship between speed and thinking distance in the sense that higher speeds generally result in longer thinking distances. When a vehicle is traveling at a higher speed, the driver needs more time to process information, make decisions, and react to potential hazards. Therefore, a higher speed can lead to a longer thinking distance.
It is important to note that thinking distance is just one component of the total stopping distance, which also includes braking distance. Braking distance is directly influenced by the speed of the vehicle. Higher speeds require longer braking distances to bring the vehicle to a stop.
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what goes back to its original shape after being stretched
A rubber band goes back to its original shape after being stretched.
Rubber bands are made of an elastic material that has the property of elasticity. When a rubber band is stretched, the polymer chains within the rubber band are extended, storing potential energy. Once the stretching force is released, the stored energy causes the rubber band to recoil and return to its original shape.
In summary, a rubber band is an example of a material that exhibits elastic behavior. Its ability to return to its original shape after being stretched is due to the elastic nature of the material, which allows the polymer chains to relax and regain their original configuration. This property makes rubber bands useful in various applications, such as holding objects together or providing tension in mechanical systems.
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6. Given cost=0 € (2): a) Determine sin28 b) Which quadrant does sin20 lie and what is the angle to the nearest tenth of a degree? Q
Since cost = 0 €, the value of sinθ will be 1. Recall that the Pythagorean identity for sine and cosine states that sin²θ + cos²θ = 1. So, sin²θ = 1 - cos²θ. Given cost=0 €,cosθ=0. Substituting cosθ = 0, we get;sin²θ = 1 - cos²θ. sin²θ = 1 - 0² = 1Therefore,sinθ = √1 = 1
This means that sin28 = 1 Since sin20 lies in the first quadrant (0° to 90°), it will have a positive value. To determine sin20, we can use a calculator or reference a trigonometric table. To the nearest tenth of a degree, sin20 is 0.3 and it lies in the first quadrant.
An identity that expresses the Pythagorean theorem in terms of trigonometric functions is known as the Pythagorean trigonometric identity, or simply the Pythagorean identity. It is one of the fundamental relations between the sine and cosine functions, along with the sum-of-angles formulas. The angle can be any real value, and the equation is s i n 2 + c o s 2 = 1. Given both the sine value and the quadrant in which the angle is located, we can use the Pythagorean identity to determine the angle of cosine.
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Explain how the distances between particles in a solid, liquid, and a gas help determine the desitys of each.
Answer:
The particles have more distance in them which impacts the density.
Explanation:
A solid's particles are much more closer together than that of a liquid's particles. Same thing goes for gas particles(they are much more far apart than liquid particles). This is why solids usually are more denser than liquids and gas, due to the way their particles are structured. Hope this helps!
Some minerals have special
properties. Which of these is a
special property that not all
minerals have?
A. crystalline structure
B. solid
C. radioactivity
D. nonliving
How long can a tow rope or chain be?
1. 20 feet 2. 15 feet
3. 5 feet
4. 10 feet
Answer:
"For towing, a tow chain should be of a length that keeps both vehicles within the maximum 4.5 meter distance, also tow chains an be any length 20 foot chains are often chosen"
Explanation:
- https://letstowthat.com
Also Quick note the feet of tow rope or chain varies on the situation but most longest or 20 feet.
How do you find the specific heat at constant volume of a gas?
To find the specific heat at constant volume of a gas, you need to measure the change in internal energy of the gas when it is heated at constant volume.
The specific heat at constant volume is defined as the amount of heat energy required to raise the temperature of a unit mass of the gas by one degree Celsius while keeping the volume constant.One way to measure the specific heat at constant volume is to perform an experiment in which the gas is heated in a container of fixed volume and the change in internal energy is measured. The heat energy supplied to the gas is equal to the change in internal energy plus the work done by the gas, which is zero in this case because the volume is held constant.The specific heat at constant volume can then be calculated using the equation:
C_v = ΔU / mΔT
where ,
ΔU is the change in internal energy,
m is the mass of the gas, and
ΔT is the change in temperature.
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If you want to have a high voltage change in a short amount of time, do you need a large or small current?
In order to have a high voltage change in a short amount of time, you would need a large current.
To clarify, a high voltage change in a short amount of time implies a rapid increase in the electric potential difference between two points. This can be achieved by having a large current flowing through the circuit, as current is directly proportional to the rate of flow of electric charge.
This is because voltage and current are related through Ohm's Law, which states that voltage equals current multiplied by resistance. Therefore, if you want to increase the voltage quickly, you need to increase the current.
However, it's important to note that working with high currents can be dangerous, so it's important to take proper precautions and use appropriate equipment.
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A juggler is throwing balls up in the air and catching them. An observer in the crowd makes the statement that the potential and kinetic energy of the ball is the same when the ball is halfway down from its highest point. In one paragraph, using your own words, explain using the relationship of mechanical energy, potential energy, and kinetic energy why the observer is correct.
Answer:
do you have the answer?
Explanation:
if so whats the answer?
which is a more scientific way of measuring earthquake? why?
Answer:Earthquakes are measured using seismographs, which monitor the seismic waves that travel through the Earth after an earthquake strikes. Scientists used the Richter Scale for many years but now largely follow the “moment magnitude scale,” which the U.S. Geological Survey says is a more accurate measure of size.
Dennis the Menace is shooting a rock with his slingshot with an initial height of 5 feet. The height in feet, of the rock above the ground is given by s(t)=-16t^2+44t+5, where t is time in seconds and t is >= 0. At what time will the rock be 15 ft above the ground?
Answer: 32
The average velocity of the rock would be 32 ft per second.
The rock will be 15ft above the ground 0.25 and 2.5secs after
The height of the rock above the ground is expressed according to the expression \(s(t)=-16t^2+44t+5\)
In order to get the time the rock will be 15ft above the ground, we will substitute s(t) = 15 ft into the formula above;
\(15= -16t^2+44t+5\\-16t^2+44t = 15 -5\\-16t^2+44t = 10\\-16t^2+44t- 10 = 0\\16t^2-44t+10=0\)
Factorize the expression to get the value of "t"
2(4t−1)(2t−5) = 0
4t - 1 = 0
4t = 1
t = 0.25secs
Similarly 2t - 5 =0
2t = 5
t = 5/2 = 2.5secs
This shows that the rock will be 15ft above the ground 0.25 and 2.5secs after
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A crate of mass 50kg is pushed along a floor with a force of 20N for a distance of
5m. Calculate the work done.
Answer:
F=50kg
D=5m
W=FD
therefore work done is 250Nm
The work done will be 100 J. It is obtained as the product of the force and displacement.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
m is the mass of the crate = 50 kg
F is the force of 20 N
d is the distance = 5 m
W is the work done.
\(\rm W =F \times d \\\\ \rm W =20 \times 5 \\\\ W= 100 \ J\)
Hence the work done will be 100 J.
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At lunch, the babysitter gives Tanya and Damien each a chocolate candy. Tanya puts her candy on her plate. Damien saves his candy in his pocket. After lunch, Damien discovers his candy has melted. Which statement accurately compares the atoms in the two candies?
The atoms in Damien’s candy have less energy.
The atoms in Damien’s candy are vibrating faster.
The atoms in Damien’s candy are unable to move.
Answer:
The atoms in Damien’s candy are vibrating faster.
When the atoms in something are going at rapid speeds they create heat.
Question 1 (5 points) What is the frictional force acting on a 100 kg box as it moves across the floor with an acceleration of 1.5 m/s2, if the applied force is 300N? -150 N Ob 150 m/s2 C 150 N od -300 N Review Answers
Answer:HJKGFHJYKIUGFJDTGYKUGFJDSX
Explanation:65TTY6Y Y YY. ;D DDS
HVC BJCHGNMGHJFYIKUHYGFTR6457TY8IUOHJKGYTDYR567YUOI8HJKBGJMVHDTR5E46T7Y TFGDGHYUFVYH T7UYGHUGFTY Y6TURF67URY
what are sunspots? is there a solid connection between sunspot numbers and climate change on earth?
Sunspots are dark, cooler regions that appear on the surface of the Sun due to the Sun's magnetic field becoming twisted and concentrated in certain areas. The extent of the connection is still a topic of scientific debate.
Sunspots typically occur in pairs or groups and can vary in size from a few hundred to tens of thousands of kilometers.There is evidence of a correlation between sunspot activity and climate change on Earth, although the extent of the connection is still a topic of scientific debate. During periods of high sunspot activity, the Sun emits more energy, including ultraviolet radiation, which can affect the Earth's atmosphere and climate. Some studies suggest that the increase in solar energy during high sunspot activity can lead to changes in global temperature and weather patterns, but other factors such as greenhouse gas emissions and natural climate variability also play significant roles in climate change.
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(A)derive the equation for circular motion,a=rw^2, where a is the centripetal acceleration w is the angular velocity and r the radius of the circle
(B) (i) the moon orbits the earth in a circle of radius 400000km. considering only these two objects, state what force act on the moon and explain how newton third law of motion applies to the system.
(ii) find the time for one complete revolution of the moon about the earth.
The equation for centripetal acceleration can be written as follows : a = v² / r = (rw)² / r = r × w² , the relationship between angular velocity and linear velocity : v = r × w
Evaluation :A.The following equation can be used to calculate the centripetal acceleration during a circular motion:
Consider a particle of mass m traveling at an angular velocity w in a circle of radius r. The particle's velocity is calculated as v = rw, where v is its linear velocity.
The net force exerted on the particle can be calculated using Newton's second law as follows:
F = m × a,
where a is the particle's acceleration.
The centripetal force is the net force exerted on the particle, which must be directed toward the circle's center.
As a result, the equation for centripetal acceleration can be written as follows:
a = v² / r = (rw)² / r
= r × w²
This equation is usually written as:
a = r × w².
What is the Third Law of Motion of Newton?According to Newton's third law of motion, there is an equal and opposite reaction to every action. The gravitational force exerted by the moon in this system is the same as that exerted by the Earth in this system. Newton's third law of motion, also known as the action-reaction pair of forces, describes this phenomenon.
B) (i) The equation below describes the gravitational force that is exerted on the moon:
F =G × (m_earth × m_moon) / r²
G is the gravitational constant, m_earth is the Earth's mass, m_moon is the moon's mass, and r is the distance between the Earth's and moon's centers.
(ii) We need to divide the circumference of the moon's circular path by its linear velocity in order to determine the time it takes for the moon to complete one complete revolution around the Earth:
C = 2 × pi × r
v = C / T
where C is the circumference of the moon's circular path, r is the circular path radius, T is the time it takes to complete one revolution, and v is the moon's linear velocity.
We obtain: by substituting the values:
C = 2 × pi × 400000 km = 800000 × pi km
v = C / T = (800000 × pi km) / T
We need to know the moon's angular velocity in order to calculate the time. These numbers indicate the relationship between angular velocity and linear velocity :
v = r × w
Substituting the values, we get:
T = C / (r × w)
= (800000 × pi km) / (400000 km × w)
= 2 × pi / w
Since the angular velocity w is not specified, it is not possible to determine how long it would take for the moon to complete one complete revolution around the Earth.
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Which is a property of every
mixture?
light will undergo total internal reflection only when it is
Total internal reflection occurs when a light ray hits a surface with a high refractive index at an angle larger than the critical angle. Thus, light will undergo total internal reflection only when it is refracted from a dense medium to a less dense medium at an angle greater than the critical angle.
The total internal reflection is when light waves are reflected back inside a medium instead of passing through it. The phenomenon occurs when the incident angle is greater than the critical angle that is required for the medium. Total internal reflection happens due to the changing of the speed of light when moving from one medium to another.
When a light ray goes from a more dense medium to a less dense one, it bends away from the normal, which is perpendicular to the surface. If the angle of incidence is too large, the ray never goes into the other medium but instead is reflected off the surface of the denser medium back into the original medium. This is referred to as total internal reflection.
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