We calculated that the pilot's apparent weight at the maximum point was 3 70.5 Newton. Additionally, the circular portion's lowest point is equal to 1598.5 Newton.
A stunt pilot performing a loop during an airshow has been described to us. the vertical circle's loop. During his performance, he gives us the radius of the circle as 3.08 into 10 to the power three m.
We are provided the weight of the pilot, which is 614. For one second, Newton maintains a speed of 2.2 into tend to the bar at a distance of 2 m. We need to know the pilot's apparent weight in newtons while he is at the highest point of the loop. Now, the formula for the pilot's mass can be expressed as being equal to the pilot's weight divided by the acceleration caused by gravity. adding the values together results in 614. too unexplored by exploration.
You get the mass of the pilot, which is 62.65 kgs Hill, when you add gravity. The apparent weight of the pilot will be equal to the normal force generated by the airplane seat. At this point, we may use the formula M to determine the apparent weight off the pilot at the highest point in the circular journey. N in this case stands for the pilot's apparent weight. We find the apparent weight of Mars to be 62.65 kgs multiplied by velocity, which is 2.2 and 2 10 to the part two m for a second, by adding the pilot's speed and the circular portion's radius, r. entire square.
The normal force of the seat on the pilot will be zero since the apparent weight of the pilot at his or her highest point throughout the circle in the case of weightlessness is Newton. With the numbers we have divided by Mars being She took 165 and shoes as replacements, we get the speed of the pilot at weightlessness state equal to 173.7 meters per second. As a result, its speed at this point will be equal to root of wait, multiplied by Baridis divided by mass.
Now your parent wait of the pilot at the lowest point can we given us Baron, well being equal to and we square r plus right, substituting the values, we have 62.65 Kgs to point to entertain the part to the whole square divided by yes, 614. Newton solving which we get the apparent weight of the pilot, the lowest point equal to 1598 0.5. Newton which is the apparent weight of the pilot at the lowest point of the circular what therefore Therefore we determined the apparent weight of the pilot at the highest point being equal to 3 70.5 Newton. And at the lowest point of the circular part being equal to 1598.5 Newton.
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the moderate temperatures of islands throughout the world has much to do with water's vast supply of internal energy. high evaporation rate. absorption of solar energy. high specific heat. poor conductivity.
High specific heat of the water. Option (c)
What is Specific heat?
The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's specific heat. Typically, calories or joules are used per gram and degree Celsius when referring to the units of specific heat.
The moderate temperature of islands has much to do with the water's high specific heat. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.
Hence, the water has high specific heat.
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While on the moon, Buzz Aldrin carried on his back a support system that would
weigh over 1,775 N on Earth. What did the backpack weigh on the moon? (gravity
on the moon is 1.62 m/s^2)
Answer:
W' = 287.55 N
Explanation:
We know that weight force is defined as follows:
W = m × g
So:
On Earth (W = 1775 N):
1775 = m × 9.8
m = 1775 / 9.8
m = 181.12 kg
On the Moon (g' = 1.62 m/s²):
W' = m × g'
W' = 181.12 × 1.62
W' = 293.42 N
Can anyone give me the step by step solution for this question?
Answer:
ans a
Explanation:
7) You push on box G that is next to box H, causing both boxes to slide along the floor, as shown in the figure.
The reaction force to your push is
Push
GH
A) the push of box G against you. B) the acceleration of box G.
C) the push of box H on box G. D) the push of box G on box H.
Answer:
c
Explanation:
because whenever you push g box into h box and then h displaces according to third law of motion h box will apply the force on g box so that's why
a 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30 m/s2 for 34 s, then runs out of fuel. ignore any air resistance effects.
A rocket with 200kgs is loaded extra 100kgs then the Rocket maximum altitude is 93005 m and total Time-of-Flight is 294 s
The Time-of-Flight concept (ToF) is a technique for determining distances between sensors and objects based on the amount of time that passes between a signal being sent and returning to the sensor after being reflected by an object. The Time-of-Flight principle can be employed with a variety of signals (also known as carriers), with sound and light being the most popular ones. The time it takes for photons to travel between two points—from the sensor's emitter to a target and back—in order to calculate distance for all Time-of-Flight (ToF) sensors.
Direct and indirect ToF both have unique advantages in particular situations. Each pixel in a scene can be simultaneously measured for distance and intensity.
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I WILL GIVE YOU BRAINIEST Write an objective summary of the Helen Keller text, From The Story of My Life.
Write a paragraph that explains what the text is about. Be sure to include the title and author's name.
When writing an objective summary:
Do not write a page-by-page recap. A summary should give the reader an idea of what the text is about without completely retelling the story.
Do not simply write a statement about the topic like - It is about a woman who is blind and deaf. It should include details.
The summary must be in your own words, but it has to be unbiased and not include opinions, experiences, or prior knowledge. Do not plagiarize.
Answer:
Helen Keller was born on June 27, 1880 in the small town of Tuscumbia, Alabama. When she was a year old, she was stricken with an illness that left her without sight or hearing. In the early years after her illness, it was difficult for her to communicate, even with her family; she lived her life entirely in the dark, often angry and frustrated with the fact that no one could understand her. Everything changed in March of 1887, when Helen's teacher, Anne Sullivan, came to live with the family in Alabama and turned Helen's world around.
Miss Sullivan taught Helen the names of objects by giving them to her and then spelling out the letters of their name in her hand. Helen learned to spell these words through imitation, without understanding what she was doing, but eventually had a breakthrough and realized that everything had a name, and that Miss Sullivan was teaching them to her. From this point on, Helen acquired language rapidly; she particularly enjoyed learning out in nature, where she and her teacher would take walks and she would ask questions about her surroundings. Soon after this, Helen learned how to read; Miss Sullivan taught her this by giving her strips of cardboard with raised letters on them, and then having her act out the sentence with objects. Soon, Helen could read entire books.
In May 1888, Helen went north to visit Boston with her mother and teacher. She spent some time studying at the Perkins Institute for the Blind, and quickly befriended the other blind girls who were her age. They spent a vacation at Brewster in Cape Cod, where Helen experienced the ocean for the first time. Following this, they spent nearly every winter up north.
Once she had learned to read, Helen was determined next to learn how to speak. Her teacher and many others believed it would be impossible for her to ever speak normally, but she resolved to reach that point. Miss Sullivan took her to the Horace Mann School in 1890 to begin learning with Miss Sarah Fuller, and Helen learned by feeling the position of Miss Fuller's lips and tongue when she spoke. The moment she spoke her first words, "It is warm," was a powerful memory for her: she was thrilled that she might be able to speak to her family and friends at last.
The winter of 1892 was a troubling time for Helen. Seemingly inspired by the beautiful fall foliage around her, she wrote a story called "The Frost King," and sent it up to her teacher at the Perkins Institute as a gift. It soon came out that Helen's story was quite like another in a published book, called "The Frost Fairies." Helen had been read the original story as a child, and the words had remained so ingrained in her mind that she'd unwittingly plagiarized them when she wrote her own story. This tainted Helen's relationship with her Perkins Institute teacher, Mr. Anagnos, and made her distrust her own mind and the originality of her thoughts for a long time.
In 1894 Helen attended the Wright-Humanson School for the Deaf in New York City, and began studying formal subjects like history, Latin, French, German, and arithmetic. In 1896, she began her studies at the Cambridge School for Young Ladies in Massachusetts, which would prepare her to eventually attend Radcliffe College, the women's college affiliated with Harvard University. This was her first time attending school with girls who could see or hear, rather than other students who were also deaf or blind. Though it was a challenge, she persevered; however, her mother eventually withdrew her from the Cambridge School to finish her Radcliffe preparation with a private tutor, because they did not agree with the Cambridge School principal's wish to lighten Helen's course load. She successfully qualified for Radcliffe in 1899, and entered college in the fall of 1900. Though college presented unique obstacles for Helen to overcome, she deeply appreciated her opportunity to attend.
Helen uses the final chapters of her memoir to discuss certain things that are particularly important to her, like her love of books, her favorite pastimes, and the friends she made who shaped her life. Two additional sections of the autobiography include Helen's personal letters written throughout her youth, as well as supplementary commentary by her editor, with a first-hand account by
or
The Story of My Life (1903) chronicles the early years of Helen Keller, a young woman who became both deaf and blind at a young age. The book explores the challenges she faced growing up as a child with disabilities, and introduces the amazing people who helped her along the way.
Explanation:
here
Isaac is practicing his volleyball skills by volleying a ball straight up and down, over and over again. His teammate
Marie notices that after one volley, the ball rises 3.6 m above Isaac's hands. What is the speed with which the ball left
Isaac's hand? (8.4 m/s) ANSWER FAST PLEASE!
The action of volleying a ball straight up and down comes under the free-fall motion of the ball. Free-fall is defined as the motion of a body which falls freely due to the gravitational pull of the earth.
The velocity (v) of the ball undergoing free-fall can be calculated using the free-fall formula,
[tex]v^{2}[/tex] = 2gh
where, g is the acceleration due to gravity and h is the height of the freely falling body.
Given that, the height to which ball rises, h = 3.6m. (g = 9.8 m/[tex]s^{2}[/tex])
Substituting h and g in the velocity formula,
[tex]v^{2}[/tex] = 2 × 9.8 m/[tex]s^{2}[/tex] × 3.6m
[tex]v^{2}[/tex] = 70.56 [tex]m^{2}/s^{2}[/tex]
The velocity of volley ball is,
v = [tex]\sqrt{70.56m^{2}/s^{2} }[/tex]
∴ v = 8.4 m/s
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two small, identical steel balls collide completely elastically. initially, ball 1 is moving with velocity ????1 directly toward ball 2, and ball 2 is stationary. what are the final velocities of ball 1 and ball 2, respectively?
When two small ball collide elastically in which first ball is moving with velocity, v with second ball which is in rest than after collision the second ball will move with velocity, v and first ball will come to rest.
What is elastic collision?An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.
An elastic collision is one in which the momentum, p and the kinetic energy, KE are both conserved.
[tex]p_{i} =p_{f}[/tex]
[tex]KE_{i} = KE_{f}[/tex]
Given in the question two identical steel ball that means have same mass and one is moving with velocity, v another is in the rest initially, after collision the momentum of first ball will transfer to second ball so second ball will move with velocity, v and first ball will come to rest.
So final velocity of first ball is zero and second ball is v.
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A 75-kg merry go round worker stands on the ride's platform 4.5 m from the center and her speed as she goes around the circle is 4.1 m/s. a) Draw a sketch of the scenario. b) Determine the force of friction necessary to keep her from falling off the platform. c) How far does she travel after 1 revolution? d) What is the period T (in seconds) after 1 revolution? Round answers to nearest tenth.
The friction that is necessary to keep her from falling off the merry go round is 280.16N, the distance travelled in one revolution is 28.28 meters and the time period T after in revolution is 6.89 seconds.
The mass of the worker is 75 kg, the radius of merry go round is 4.5m and the linear velocity of the merry go round is 4.1m.s.
a) The diagram is attached below.
b) The friction force while rotation will be equal to the centrifugal force, this is the reason why he will be able to go round,
So, we can write,
Friction force Fr = Centrifugal force
Fr = MV²/R
Where Fr is the friction force,
M is the mass of the worker,
V is the velocity of the merry go round,
r is the distance at which the worker is standing.
Putting all the values,
Fr = 75 x 4.1 x 4.1/4.5
Fr = 280.16N.
The friction required is 280.16N.
c) The distance she travelled by her after the first round will be equal to the circumference of the circle,
Distance = 2Πr
D = 2 x 22 x 4.5/7
D = 28.28m
d) time period T after one revolution,
T = Distance/velocity
T = 28.28/4.1
T = 6.89 seconds.
The time period is 6.89 seconds.
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PLEASE HELP ASAP, THANK YOU!
Why do you think Saturn and Jupiter have more moons than the other planets in our solar system?
This is more of an Astronomy question than physics but there’s no astronomy option
Answer: They are bigger
Explanation: They are the biggest planets in our Solar System and attract more to them so they would obviously have more moons than all of the other planets.
Answer:
Explanation:
Because of the huge mass of these planets.
a plane accelerates from 4 m/s/s for 28.7 s until it finally lifts on the ground. Assuming the plane started from the rest, what was the planes displacement before takeoff?
There were 1647 before liftoff. 38 m moving planes.
A Straight Line Motion is defined as?An object is considered to be in motion if its location throughout time changes in relation to its surroundings. It is the gradual change in an object's location. The only kind of motion that exists is linear motion.
Which three motion equations are there?The three equations are as follows: S = ut + 12at2, v = u + at, v2 = u + 2as.
The airplane's initial speed is U=0 m/s.
The plane accelerates at a 4 m/s rate.
with a time stamp of 28.7 seconds.
Distance traveled using the formula S=ut+a
⇒ S=0(28.7)+0.5(4) (4)
=1647.38 ~ 1647m
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What acceleration will result when a 44.7-N net force applied to a 7.7-kg object?
The acceleration when a 44.7 N net force is applied to 7.7 kg object is 5.8 m / s²
According to Newton's second law of motion,
F = m a
F = Force
m = Mass
a = Acceleration
m = 7.7 kg
F = 44.7 N
a = F / m
a = 44.7 / 7.7
a = 5.8 m / s²
Newton's second law of motion states that the force is equal to the rate of change of momentum. It can also be said as acceleration of an object is directly proportional to the net force acting on the object.
Therefore, the acceleration of the object is 5.8 m / s²
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the trees in a wooded area near a factory look like burnt toothpicks. what might be causing the problem?
A cliff jumper runs off a 15\,\text m15m15, start text, m, end text high cliff. Rocks extend 3.0\,\text m3.0m3, point, 0, start text, m, end text past the cliff bottom. We can ignore air resistance.
1.7 m/s is the horizontal velocity that is required to clear the rocks.
As the jumper runs off a 15 m high cliff.
And according to the question, starts with 0 velocities in the start direction thus he covers 15 m.
Given,
x = 15 m
a = g = 9.8 m, which is the acceleration due to gravity.
iv is the initial velocity which is 0, vo = 0
vf is the final velocity,
Using 2nd equations of motion,
x = iv × t + (1 ÷ 2) a × t²
15 = 0 + ((1 ÷ 2) × ( 9.8) × t²)
t = 1.75
Thus, the time is 1.75 seconds.
Using the same equation, horizontal velocity will be:
x = iv × t + (1 ÷ 2) a × t²
3 = vf × (1.75) + 1 ÷ 2 × (0) × (1.75)
vf = 1.7
Therefore, The horizontal velocity required to cover the rocks is 1.7 m/s.
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runs with constant velocity of 3.5m/s 2 seconds later bicyclist accelerates at 2.4m/s squared until he catches runner how long does it take to catch the runner and how far has the runner traveled
It takes 2.916s for the bicyclist to catch the runner and the runner had ran 17.2m with constant velocity of 3.5m/s
What is velocity?How quickly or slowly an object is moving can be determined by its velocity and speed. The need to determine which of two or more moving objects is moving faster arises frequently in our daily lives.
If both vehicles are traveling down the same road in the same direction, it is simple to determine which is moving more quickly.
To tell who is moving faster, though, is challenging if they are moving in the opposite direction. The idea of velocity is valuable in such circumstances
Lets say runner and bicyclist start at a point vi
The distance bicyclist cover to catch runner is [tex]S = (v_i)t = \frac{1}{2} at^2[/tex]
As the initial velocity is zero distance [tex]S = \frac{1}{2} at^2[/tex]
For the runner, distance he could run before bicyclist overpassed him is
S = vt
Now as athletes was 7m ahead so
= (athlete)S = (bicyclist)S
= (athlete)S = (bicyclist)S - 7
Substitute the values
= vt = [tex]\frac{1}{2} at^2[/tex]
= 3.5t = [tex]\frac{1}{2} (2.4)t^2[/tex]
= 3.5t = 1.2t²
= 3.5 = 1.2t
= t = 2.916s
Distance of runner = 3.5 × 2.916s
= 10.206 + 7m
= 17.2m
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According to Newton's first law of motion, what will an object in motion do when no external force acts on it?
come to a stop
move at the same velocity
speed up
change direction
Mark this and return
Answer:
Move at the same velocity
Explanation:
According to newton's 1st law of motion, an object at rest remains at rest or if in motion, it will remain in motion at constant velocity unless acted upon by a net external force
a block is given an initial velocity of 5.00 m/s up a frictionless incline of angle 20 degrees. the magnitude of the block’s acceleration is 2.5 m/????2 while it is on the ramp. how far up the incline does the block get before sliding down again?
The block will incline up with projectile motion until 0.085 m before sliding down.
We need to know about the projectile motion to solve this problem. The projectile motion is known as parabolic motion and the velocity is divided by 2 axes.
vox = vo cosA
voy = vo sinA
where vox is initial velocity of x axis, voy is initial velocity of y axis, vo is initial velocity and A is the angle.
From the question above, the given parameters are
vo = 5 m/s
A = 20⁰
By using the uniform motion, where vty = 0 m/s. Hence,
vty² = voy² - 2gh
0 = 5.sinA - 2 . 10 . h
20h = 5 . sin20⁰
h = 1/4 . 0.34
h = 0.085 m
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Clare de Iles is shopping. She walks 16m North to the end of an aisle. She then makes a right hand turn and walks 21m Eastward to the end of the aisle. Determine th emagnitude of Claire's resultant displacement. And the angle of her resultant.
The magnitude and direction of Clare's resultant displacement is 26.4 m and 37.3⁰ respectively.
What is the resultant displacement?The magnitude of Clare's resultant displacement is calculated as follows;
her initial position, a = 16 m north
Final position, b = 21 m east
Resultant displacement, d = √a² + b²
d = √(16² + 21²)
d = 26.4 m
The direction of the resultant displacement is calculated as follows;
tan θ = 16 / 21
where;
θ is the direction of the displacement obtained by forming right triangle from the initial and final positiontan θ = 0.762
θ = arc tan(0.762)
θ = 37.3⁰
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a 2000 kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. the spring is intended to stop the elevator, compressing 2.00 m as it does so. during the motion a safety clamp applies a constant 17,000 n friction force to the elevator.
A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s
B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²
In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:
F = mg
Where:
Mass of the elevator; m= 2000kg
Acceleration due to gravity; g = 9.8m/s
2000kg × 9.8m/s²= 19600N
F = 19600
Force of opposing friction clamp of gravity = 17000N
Net force on the elevator = force of gravity - Force of opposing friction clamp
Net force on the elevator = 19600 - 17000
Net force on the elevator = 2600 N
We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:
K.E = 1/2 mv²
Where
Mass of the elevator; m = 2000kg
Velocity of the elevator = 4.00m/s
K.E = (1/2)*2000kg*(4m/s)²
K.E = 16000J
The kinetic energy and energy gained will be absorbed by the spring across the next 2m
Therefore,
Energy; E = K.E + P.E
Where:
Kinetic energy K.E = 16000J
Potential Energy P.E = ?
P.E of spring = net force absorbed × distance at compression
Where:
Net force absorbed = 2600N
Distance at compression = 2.0m
P.E = 2600*2
P.E = 5200J
E = 16000J + 5200J
E = 21200J
Spring constant = k
To find k
Using:
E = (1/2)*k*(x)²
Where:
E = 21200J
k = ?
x = 2m
21200J = (1/2)*k*(2m)²
21200J*2 = (4m)k
K = 42400J/4m
K = 10600N/m
Therefore,
Acceleration at 1m compression = ?
Using:
F = K*X
Where
F is force provided by the spring = 10600N/m,
K = 10600 N/m
X = 1m
F = 10600N/m * 1m
F = 10600N (upward)
A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?
Using:
Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy
16000N + 2600N = (1/2)mv² + (1/2)k x²
18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)
18600 = 1000(v²) + 5300
18600 - 5300 = 1000(v²)
13300 = 1000(v²)
V² = 13.300
V =3.65m/s
B. The acceleration of the elevator is 1.00m below point where it first contacts a spring
Spring constant = net force on the elevator + resultant force
Where:
Spring constant = 10600N
Net force on the elevator = 2600N
Resultant force = ?
10600N = 2600N + resultant force
Resultant force = 10600N - 2600N
Resultant force = 8000N
Using the equation for Newton's 2nd law where F = ma,
a = F/m
Where:
Resultant force; F =8000N
Mass of the elevator; m =2000kg)
a = 8000 / 2000
a = 4m/s²
Here's the complete question:
In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.
1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?
2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?
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a parallel-plate capacitor initially has a potential difference of 200 v and is then disconnected from the charging battery. if the plate spacing is now quadrupled (without changing q), what is the new value of the voltage?
The new voltage value of parallel-plate capacitor is 800 V.
We need to know about the capacitance of the parallel-plate capacitor to solve this problem. The capacitance of the parallel-plate capacitor depends on the charge and voltage. It can be written as
C = Q/V
where C is capacitance, Q is charge and V is voltage.
Capacitance also can be calculated by the area and distance between capacitor plate
C = εo . A / d
where εo is vacuum permittivity (8.85 x 10¯¹² F/m), A is area of plate and d is distance.
From the question above, we know that
V1 = 200 V
d1 = d
d2 = 4 d
By substituting the given parameters, we can calculate the voltage
εo . A / d = Q/V
The voltage is proportional to the distance of plate. Hence, the ratio of voltage will be
V1 / V2 = d1 / d2
200 / V2 = d / 4d
V2 = 4 x 200
V = 800 N
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a winninmg team lifted a 3 kg trophy 2m up into the air what is the change in gpe of the trophy
The change in gpe ( gravitational potential energy ) of trophy of mass 3 kg that is lifted 2 m up in the air is 58.8 J
ΔPE = m g h
ΔPE = Change in potential energy
m = Mass
g = Acceleration due to gravity
h = Height
m = 3 kg
h = 2 m
g = 9.8 m / s²
ΔPE = 3 * 9.8 * 2
ΔPE = 58.8 J
Gravitational potential energy of an object is due to the presence of gravity. The higher the object is from the point of surface from which the gravity is exerted, higher will be its potential energy.
The change in gpe of the trophy is 58.8 J
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it takes 4.0 μj of work to move a 15 nc charge from point a to b. it takes -5.0 μj of work to move the charge from c to b. part a what is the potential difference vc−va ? express your answer in volts.
W is the work done by charge from A to B and from C to B then the potential difference (vc−va)=552.62V.
The difference between the energy levels that charge carriers have at two places in a circuit is known as the potential difference.
Voltage, commonly known as potential difference (p.d. ), is a unit of electrical potential. The charge carriers in a circuit pass through the electrical components, transferring energy to them. We gauge the potential difference using a voltmeter (or voltage).
V = I x, the potential difference formula
The quantity of current times the resistance equals the potential difference (also known as voltage), which is equal to that. When a charge travels between two locations in a circuit, a potential difference of one Volt is equivalent to one Joule of energy being used by that charge.
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arunner jack 12km north then turns and runs 16km in 3hours calculate his average velocity
A runner jack 12km north then turns and runs 16km in 3hours Therefore his average velocity is 1.3 km/h.
What is average velocity?
The change in position or displacement (x) divided by the time intervals (t) during which the displacement happens is the definition of average velocity. Based on the direction of the displacement, the average velocity might either be positive or negative. Meters per second (m/s or ms-1) is the SI measure for average velocity.
Given:
A runner jack 12km north then turns and runs 16km in 3hours .
Therefore average velocity= (16 - 12)/ 3 km/h
=1.3 km/h
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A ball of mass 1 kg is dropped from a height of 3 m. Assume no air resistance. After it bounces off the ground it only reaches a height of 2 m. What is the change in momentum of the before and after the bounce?.
energy at first = mg (3) J = 29.4J
Energy after bouncing: mg(2) = 19.6 J
Energy changed by the ball: 29.4J - 19.6J = 9.8J
Heat energy or sound energy are two ways that this energy is lost.
Work performed during the bounce equals 9.8J.
d. Momentum change = mvf-mvi
29.4 = 1/2mvi2
vf =7.668m/s
19.6J =1/2mvf2
vf=6.261m/s
Change of momentum is equal to 1kg x (-1.407m/s)/1kg = -1.407 kg/s (m(vf-vi)).
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what is the acceleration, in meters per square second, due to gravity on the surface of the moon? you will need to look up the mass and radius of the moon.
Answer:
F = m a = K M m / R^2 force on mass on moon
a = K M / R^2
a = 6.67E-11 * 7.36E22 / (1.74E6)^2
a = 6.67 * 7.36 / 1.74^2 * 10^-1 = 1.62 m/s^2 about 1/6 of earth
What do we call the illusion of movement that results from two or more stationary, adjacent lights blinking on and off in quick succession?.
In 1912 Wertheimer discovered the phi phenomenon, an optical illusion in which stationary objects shown in rapid succession, transcending the threshold at which they can be perceived separately, appear to move.
What is phi phenomenon ?When two neighbouring optical stimuli are presented in alternation with a relatively high frequency, an apparent motion is seen that is known as the phi phenomenon. At higher frequencies, beta movement is visible, but the stimuli themselves don't seem to move.
The concept of the phi phenomenon is crucial to the field of psychology known as Gestalt psychology. This field focuses on perception and is interested in comprehending how various components of a whole affect perception. The phi phenomenon tricks the brain into thinking that objects that aren't moving are.Learn more about Phi phenomenon here:
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water from a hose with a diameter leaves through a nozzle with a smaller diameter and fills a 22l bucket? whats the velocity of the water when its in the hose
Flow rate and velocity are physically connected but very different quantities. The flow rate of the river increases with increasing water velocity. However, river size also affects flow rate.
[v*π*(2/2)^2] *60 = 20*1000 [1 L = 1000 cm^3]
(22*60/7)*v = 20000,
or, v = 7*20000/22/60or
v = 106.06 cm/s = 1.06 m/s.
Hence, velocity of the water when its in the hose is 1.06 m/s.
Consider the rate at which a river flows to understand the difference. The flow rate of the river increases with increasing water velocity. However, river size also affects flow rate. For instance, the Amazon River in Brazil carries much more water than a swift alpine stream. The exact formula for the link between flow rate Q and velocity is
A is the cross-sectional area, and
V is the average velocity, therefore Q = A v.
This equation makes sense to me. According to the relationship, the size of a river, pipe, or other conduit, as well as the magnitude of the average velocity (hereinafter referred to as the speed), directly affect the flow rate. The cross-sectional area of a conduit increases with its size.
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if the parachutist comes to rest with constant acceleration over a distance of 0.660 mm , what force does the ground exert on her?
The force exerted by the parachutist on the ground is 471.24 N when she comes to rest at a distance of 0.660 m
Let t = Time taken
u = Initial velocity , v = Final velocity
s = Displacement, a = Acceleration
As given in the problem initial velocity u = 3.85 m/s
Final velocity v = 0 (rest)
s = 0.660 m
From the equation of motion v² = u² + 2as
we have a = v²-u² / 2s
⇒a = 0-3.85² / (2* 0.660)
⇒a = -11.22 m/s² (negative sign shows deceleration)
we know force F = mass m * acceleration a
Given mass of parachutist m = 42 kg
∴Force exerted by parachutist on ground
F = 42 * 11.22 = 471.24 N
If the distance is shorter, acceleration increases in magnitude and so does the force exerted by the parachutist.
The question is incomplete and the full question is probably " A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.660 m, what force does the ground exert on her? "
Know about when equations of kinematics can be applied
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What causes polar jet streams to form? (1 point)
A.rapid changes in wind direction
B.air masses moving away from each other
C.warm and cold air masses
D.high-pressure systems
Polar jet streams are formed because of the warm and cold air masses.
What is polar jet streams?
Polar front jet stream, also called polar front jet or mid-latitude jet stream stream, a belt of powerful upper level winds that sits at top the polar front.
The winds are strongest in the troposphere, which is the upper boundary of the troposphere, and move in a generally westerly direction in mid-latitudes.
The vertical wind shear which extends below the core of this jet stream is associated with horizontal temperature gradients that extend to the surface. As a consequence, this jet manifests itself as a front that marks the division between colder air over a deep layer and warmer air over a deep layer.
Polar jet streams are formed due to warm and cold air masses.
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Answer:
C
Explanation:
1. two ice skaters, megan and jason, push off from each other on frictionless ice. jason’s mass is twice that of megan. a. which skater, if either, experiences the greater impulse during the push? explain. b. which skater, if either, has the greater speed after the push-off? explain.
The impulse of the two skaters is the same but the speed of Jason is only half that of Megan.
What is the impulse?We define the impulse as the product of the force and the time. We know that two objects are taking of at the same time. Given the fact the they are taking off from the same kind of surface at the same time, the impulse of the two are the same.
We know that the speed of the object would depend on the mass of the object. The greater the mass, the lesser the speed of the object hence Jason would have a lesser speed than Mason.
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