Answer:
Explanation:
A 26-kg crate, initially at rest on a horizontal surface, requires a 72-N horizontal force to set it in motion.
Maximum Frictional force on stationary object F = μs R where μs is coefficient of static friction and R is reaction of ground which is equal to weight for object placed on horizontal surface .
F = μs x mg
Given F = 72 , m = 26
72 = 26 x 9.8 x μs
μs = .28
Kinetic friction : ---
Frictional force on moving object F = μk R where μk is coefficient of kinetic friction and R is reaction of ground which is equal to weight for object placed on horizontal surface .
F = μk x mg
Given F = 72 , m = 26
54 = 26 x 9.8 x μk
μk = .21
Why is it important for scientist to be skeptical
Answer:
I don't actually know, but it has to do with the research they id
Explanation:
to seem precise
can any one help me with this?
Which dog has the most kinetic energy? A. A dog of mass 10 kg running with speed 6 m/s B. A dog of mass 8 kg running with speed 4 m/s C. A dog of mass 12 kg running with speed 5 m/s D. A dog of mass 15 kg running with speed 3 m/s
Answer:
J stands for Joules.
A has the most KE
Explanation:
Dog A
En = 1/2 m * v^2m = 10 kgv = 6 m/sEn = 1/2 10 * 6^2En = 5 * 36En = 180 JDog B
En = 1/2 m * v^2m = 8 kgv = 4 m/sEn = 1/2 8 * 4^2En = 4 * 16En = 64 JDog C
En = 1/2 m * v^2m = 12 kgv = 5 m/sEn = 1/2 12 * 5^2En = 6 * 25En = 150 JDog D
En = 1/2 m * v^2m = 15 kgv = 3 m/sEn = 1/2 3 * 3^2En = 1.5 * 9En = 13.5 Jon the grid below sketch at least one complete cycle of a transverse wave with a 4.0 centimeter amplitude a freuqncy of 5.0 hertz
Draw the complete cycle of the wave by repeating the pattern of the peak, the equilibrium position, and the trough, with a distance of λ between each consecutive peak or trough. The number of cycles per second, or the frequency, should be 5.0 hertz.
What is Wave?
A wave is a disturbance that propagates through space and time, often transferring energy from one location to another without the physical transfer of matter. Waves can take many different forms, including sound waves, electromagnetic waves, and mechanical waves.
Draw a horizontal axis representing time, labeled in seconds or milliseconds.
Draw a vertical axis representing displacement or amplitude, labeled in centimeters or meters.
Choose a starting point for the wave, which represents the equilibrium position of the medium.
Draw the peak of the wave, which represents the maximum displacement of the medium from its equilibrium position. This should be 4.0 centimeters above the equilibrium position.
Draw the trough of the wave, which represents the minimum displacement of the medium from its equilibrium position. This should be 4.0 centimeters below the equilibrium position.
Determine the wavelength of the wave, which is the distance between two consecutive peaks or troughs. This can be calculated using the formula λ = v/f, where λ is the wavelength, v is the velocity of the wave, and f is the frequency. For a transverse wave on a string, the velocity is given by v = √(T/μ), where T is the tension in the string and μ is the linear mass density of the string.
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How long does take for a freely falling object to reach 4.0 m/s
Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
What does the cosmological principle allow cosmologists to assume?
A. Exactly what the specific fate of the universe will be and when
B. The galaxies are organized clusters of billions of stars, gas, dust, and matter in all other forms.
C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.
D. How the helium clouds impact human health and life expectancy.
Answer:
C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.
Explanation:
The Cosmological Principle assumes that the small portion of the universe that we can see is representative of the entire universe, even though we can only directly observe a tiny fraction of it. It's an assumption used by Cosmologists to simplify their models of the universe.
Calculate the net force for each of the following:
a.
20 N to the right and 30 N to the right:
Answer and Explanation:
A.
There would be a net force of 50 N going to the right.
Since both forces are going right, they add onto the force amount.
#teamtrees #PAW (Plant And Water)
Name:.
Date:
Homework: Writing an Argument About
Australia's Skin Cancer Rate
G
4
You now have enough evidence to explain why Australia's skin cancer rate is so high. Review your
argument from Chapter 2 and think about how you will convince the AHA that both Claim 2 and
Claim 3 are accurate. You will use your completed Reasoning Tool from Activity 3 to revise and add to
your argument.
Question: Why is the skin cancer rate in Australia so high?
Why was the geocentric model of our solar system widely accepted until the 16th century?
A.
The model was considered to work in tandem with the ancient Aztec calendar.
B.
Observations had led to the belief that the Sun, Moon, and stars revolved around Earth.
C.
The star catalogue matched the movement of constellations.
D.
No one had yet sailed entirely around Earth to disprove the geocentric model.
The correct answer is B. Observations had led to the belief that the Sun, Moon, and stars revolved around Earth.
The geocentric model, which placed Earth at the center of the universe with the Sun, Moon, and stars revolving around it, was widely accepted until the 16th century for several reasons. One of the primary reasons was that observations seemed to support this view. For example, it was observed that the stars appeared to move across the sky in a circular path, and it was assumed that this was due to their motion around Earth. Similarly, the Sun and Moon appeared to rise and set each day, which was consistent with their motion around Earth.
The geocentric model was consistent with the prevailing philosophical and religious beliefs of the time, which held that Earth was a special and unique place at the center of the universe. It was not until the development of the heliocentric model, which placed the Sun at the center of the solar system, and the refinement of astronomical observations, that the geocentric model was gradually abandoned.
The correct option is B. Observations had led to the belief that the Sun, Moon, and stars revolved around Earth.
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A sphere of radius r has surface area
A = 4r2and volume V = (4/3r^3) The radius of sphere 2 is triple the radius of sphere 1.
(a)
What is the ratio of the areas, A2/A1?
(b)
What is the ratio of the volumes, V2/V1?
Answer:
A2/A1 = 9
V2/V1 = 27
Explanation:
Area of Sphere 1; A1 = 4πr²
Volume of Sphere 1; V1 = (4/3)πr³
Area of Sphere 2; A2 = 4π(3r)² = 9*4πr²
Volume of Sphere 2; V2 = (4/3)π(3r)³ = 27*(4/3)πr³
Thus;
A2/A1 = 9*4πr²/4πr² = 9
V2/V1 = [27*(4/3)πr³]/[(4/3)πr³] = 27
5. If a jump rope is shaken fast enough to produce a wave with a wavelength of 0.5 m and
the crest of the wave passes a specific point 6 times per second, what is the velocity of the
wave?
Answer:
After solving the equation the wave is traveling at a velocity of 3 m/s.
What is velocity?Velocity is a vector quantity that describes the rate at which an object changes its position. It is a measure of both the speed and the direction of an object. Velocity is typically expressed in terms of meters per second (m/s). When an object is in uniform motion, its velocity is constant, meaning that it is not changing in magnitude or direction. If an object is accelerating, its velocity is changing over time, either in magnitude, direction, or both.
The velocity of the wave can be calculated using the following equation:
Velocity = Wavelength x Frequency
In this case, the wavelength is 0.5 m and the frequency is 6 Hz (6 cycles per second). So, the velocity of the wave is:
Velocity = 0.5 m x 6 Hz = 3 m/s
This means that the wave is traveling at a velocity of 3 m/s.
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An insect lands 0.1m from the centre of a turn table while the record is turning at 55 rev/min at what linear speed will the insect be carried
collision with the near stationary photograph
The linear speed will be the insect be 0.5759 meter/second carried collision with the near stationary photograph.
What is speed?
Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.
Given that an insect lands 0.1m from the center of the turn table.
Rotational speed of the turn table = 55 rev/min
= (55×2π/60) rad/second
= 5.759 rad/second.
Hence, the speed of the insect be = Rotational speed × length
= 5.759 rad/second × 0.1 M.
= 0.5759 meter/second.
Therefore, the speed of the insect be 0.5759 meter/second.
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Why must humans limit their exposure to X-rays and gamma rays?
A. The rays can change molecules and atoms in the body into ions.
B. Exposure can cause the human body to give off heat.
O C. Even a small amount of radiation can burn the skin.
D. These rays act similarly to microwaves on the human body.
Answer:
A. The rays can change molecules and atoms in the body into ions.
HELP PLEASE ITS URGENT!!!!
Answer:
1,000,000 seconds for it to travel 1 meter
Explanation:
Answer: think its 6.4
Explanation:
which items are matter?
Answer:
which items are matter?
battery , mobile phone
The system shown above consists of two identical blocks that are suspended using four cords, each of a different length. Which of the following claims are true about the magnitudes of the tensions in the cords? Select two answers.
Answer:
Option B and C are True
Note: The attachment below shows the force diagram
Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂;
w = T₁/cos 60° -----(1)
w = T₂/cos 30° ----(2)
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°
T₁ cos 30° = T₂ cos 60°
T₂/T₁ = cos 30°/cos 60°
T₂/T₁ =1.73
Therefore, option a is false since T₂ > T₁
Option B is true since T₁ cos 30° = T₂ cos 60°
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Assume your mass is 84 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb a flight of stairs 3.9 m high
Sulfur difluoride consists of a central sulfur atom and a fluorine atom on each side. Draw a model of the molecule formed by putting lines between the atoms to represent bonds so that each atom has the correct number of bonds.
Answer:
F - S - F
Explanation:
Answer:
F-S-F
Explanation:
PLATO ANSWER
An aeroplaneflying above groundnd490m with 100 meterpersecond how far on ground it will strike
The airplane will strike the ground at a horizontal distance of 490 meters.
To determine how far the airplane will strike on the ground, we need to consider the horizontal distance traveled by the airplane during its flight.
The horizontal distance traveled by an object can be calculated using the formula:
Distance = Speed × Time
In this case, the speed of the airplane is given as 100 meters per second and the time it takes to cover the distance of 490 meters is unknown. Let's denote the time as t.
Distance = 100 m/s × t
Now, to find the value of time, we can rearrange the equation as follows:
t = Distance / Speed
t = 490 m / 100 m/s
t = 4.9 seconds
Therefore, it takes the airplane 4.9 seconds to cover a horizontal distance of 490 meters.
Now, to calculate the distance on the ground where the airplane will strike, we can use the formula:
Distance = Speed × Time
Distance = 100 m/s × 4.9 s
Distance = 490 meters
It's important to note that this calculation assumes a constant speed and a straight flight path. In reality, various factors such as wind conditions, changes in speed, and maneuvering can affect the actual distance traveled by the airplane.
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Two students are canoeing on a river. While heading
upstream, they accidentally drop an empty bottle overboard. They
then continue paddling for 2 hours, reaching a point 2.5 km
farther upstream. At this point they realize that the bottle is
missing and, driven by ecological awareness, they turn around
and head downstream. They catch up with and retrieve the bottle
(which has been moving along with the current) 4.0 km downstream
from the turnaround point. (a) Assuming a constant paddling effort
throughout, how fast is the river flowing? (b) What would the canoe
speed in a still lake be for the same paddling effort?
The speed of the river flowing is 3.75 km/h, and the canoe speed in a still lake for the same paddling effort is 3.6 km/h (approximately).
The distance between the two points upstream and downstream is equal to the distance traveled by the empty bottle. Hence, the distance traveled by the bottle = distance traveled by canoeists = 4 km.Total distance traveled = distance upstream + distance downstreamTotal distance traveled = 2.5 + 4 = 6.5 kmTotal time taken = 2 hours upstream + (4/6) hours downstream (since they are covering 4 km in downstream with the current flowing downstream)Total time taken = 3.67 hours upstream + downstreamFrom the definition of speed, the speed upstream is given by:Speed upstream = distance/time upstreamSpeed upstream = 2.5/2Speed upstream = 1.25 km/hSimilarly, the speed downstream is given by:Speed downstream = distance/time downstream Speed downstream = 4/(4/6)Speed downstream = 6 km/hThe speed of the canoe in still water is given by the average of the upstream and downstream speeds:Speed in still water = (1.25 + 6)/2Speed in still water = 3.625 km/h or 3.6 km/h (approximately).
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a question was asked by a teacher to a student. She gave the student a jumbled word and told him to make words out of it. The jumbled word is gzeysktqix. Now you know what to do. see ya!
When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.
When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.
This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.
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You serve a tennis ball of mass 60g at a speed of 50
m/s, what is the impulse exerted on the ball? ( ball starts from rest )
Answer:
\(J = 3~Kg.m/s\)
Explanation:
Impulse and Momentum
The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.
The equation can be written as follows:
\(J =\Delta p = p_2-p_1\)
Where:
J = Impulse
p2 = Final Momentum
p2 = Initial Momentum
The momentum can be calculated as:
p = m.v
Where m is the mass of the object and v is the velocity.
The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:
\(\Delta p = 0.06Kg~50~m/s-0\)
\(\Delta p = 3~Kg.m/s\)
Thus the impulse is:
\(\marhbf{J = 3~Kg.m/s}\)
A train travels 8.81 m/s in a -51.0° direction.
The train accelerates for 2.23 s, changing its
velocity to 9.66 m/s in a 37.0° direction.
What is [delta]x?
The displacement of the train after 2.23 seconds is 25.4 m.
Resultant velocity of the trainThe resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
θ is the angle between the velocity = (90 - 51) + 37 = 76⁰R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
Displacement of the trainThe displacement is calculated as follows;
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
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A photon has 8.0 x 10-15 J of energy. Planck's constant
is 6.63 x 10-34 Jos.
What is the frequency of the photon?
O 1.21 x 1019 Hz
O 8.3 x 10-20 HZ
O 1.21 10-19 Hz
O 8.3 x 1020 HZ
Answer:
^I dont know how he got 12 but the answer is A i believe
Explanation:Smart like dat
The frequency of a photon is related to its energy and Planck's constant by the equation:
frequency = energy / Planck's constant.
frequency =\((8.0 x 10-15 J) / (6.63 x 10-34 J) = 1.21 x 10^19 Hz\)
Therefore, the correct answer is \(1.21 x 10^19 Hz\)
What is Planck's constant?Planck's constant is a fundamental constant in physics that is related to the behavior of particles at the atomic and subatomic level. It is denoted by the letter "h" and is approximately equal to 6.63 x 10^-34 Joule-seconds. Planck's constant is named after Max Planck, a German physicist who proposed its existence in 1900 to explain the behavior of radiation emitted by black bodies.
Planck's constant plays a central role in quantum mechanics, which is the branch of physics that describes the behavior of particles at the atomic and subatomic level. It appears in a number of important equations, including the equation for the energy of a photon (E = hf) and the equation for the uncertainty principle (ΔxΔp ≥ h/4π).
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A fire hose sends 2429 gallons of water per minute against a burning building. The water strikes the building at 66 m/s and does not bounce back. There are 3.785 L/gal Liters in a gallon, and the density of water is 1 kg/L.
a) What is the magnitude of the rate of change of momentum of the water? Answer in units of N.
b) The rate of change of momentum of the water 1. is negative. 2. cannot be determined. or 3. is positive?
c ) What force does the water exert on the building in 1.2 minutes? Answer in units of N.
Answer:
a) Δp = - 10113.2 Kg m / s, b) he rate of change is negative, c) F = 140.46N
Explanation:
a) For this part let's analyze a water particle, it has a velocity of 66 m / s and when it collides with the building its velocity changes to zero, so the change in moment is
Δp = mv_f - m v₀
Δp = -m v₀ (1)
the change of the moment in a second is
if 2429 gallons arrive in in minute (60s) in a second how many gallons arrive
c_agua = 2429/60
c_water = 40,483 gallon/ s
let's use the concept of density to find the mass
ρ = m / V
m = ρ V
let's reduce gallons to liters
c_water = 40,483 gal (3,785 l / 1 gal) = 153.23 l
m = 1 153.23
m 153.23 kg
we substitute in 1
Δp = - 153.23 66
Δp = - 10113.2 Kg m / s
b) From the previous result the rate of change is negative
c) Let's use the impulse ratio
I = f t = Δp
F t = Δp
F = Δp / t
F = \(\frac{10113.2}{1.2 \ 60}\)
F = 140.46 N
Consider the system of two blocks shown in Fig. P6.81, but with a different friction force on the 8.00 kg block. The blocks are released from rest. While the two blocks are moving, the tension in the light rope that connects them is 37.0 N. (a) During a 0.800 m downward displacement of the 6.00 kg block, how much work has been done on it by gravity? By the tension T in the rope? Use the work–energy theorem to find the speed of the 6.00 kg block after it has descended 0.800 m. (b) During the 0.800 m displacement of the 6.00 kg block, what is the total work done on the 8.00 kg block? During this motion how much work was done on the 8.00 kg block by the tension T in the cord? By the friction force exerted on the 8.00 kg block? (c) If the work–energy theorem is applied to the two blocks con- sidered together as a composite system, use the theorem to find the net work done on the system during the 0.800 m downward displacement of the 6.00 kg block. How much work was done on the system of two blocks by gravity? By friction? By the tension in the rope?
a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.
b) We cannot calculate the work done by the friction force.
c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.
(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:
Work_gravity = force_gravity * displacement * cos(theta),
where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).
The weight of the block is given by:
force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.
Plugging in the values, we get:
Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.
The work done on the 6.00 kg block by the tension in the rope is given by:
Work_tension = tension * displacement * cos(theta).
Plugging in the values, we get:
Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.
The negative sign indicates that the tension is in the opposite direction of the displacement.
Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:
Work_net = change_in_kinetic_energy.
Since the block starts from rest, its initial kinetic energy is zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.
Solving for velocity, we get:
velocity = sqrt(2 * Work_net / mass).
The net work done on the block is the sum of the work done by gravity and the tension:
Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.
Plugging in the values, we get:
velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.
Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.
(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:
Work_net = change_in_kinetic_energy.
Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.
The work done on the 8.00 kg block by the tension in the rope is given by:
Work_tension = tension * displacement * cos(theta).
Plugging in the values, we get:
Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.
The work done on the 8.00 kg block by the friction force can be calculated using the formula:
Work_friction = force_friction * displacement * cos(theta),
where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.
(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:
Work_net = change_in_kinetic_energy.
Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.
Simplifying, we get:
Work_net = 1/2 * 14.00 kg * velocity^2.
Using the value of velocity calculated in part (a), we get:
Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.
The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:
Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.
The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:
Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.
Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.
Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.
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A 0.530-kg cart moving at 0.572 m/s to the right collides elastically with a 0.25-kg cart initially at rest. The 0.25-kg cart then moves off rapidly and compresses a spring before the 0.530-kg cart can catch it again.
To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.First, let's find the velocity of the 0.530-kg cart after the collision. We can use the conservation of momentum:m1v1 + m2v2 = m1v1' + m2v2'where m1 and v1 are the mass and velocity of the 0.530-kg cart before the collision, m2 and v2 are the mass and velocity of the 0.25-kg cart before the collision, and v1' and v2' are the velocities of the carts after the collision.Plugging in the numbers, we get:(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s) = (0.530 kg)v1' + (0.25 kg)v2'Solving for v1', we get:v1' = [(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s)] / (0.530 kg + 0.25 kg) = 0.378 m/s to the rightSo the 0.530-kg cart moves off to the right at 0.378 m/s after the collision.Next, let's find the maximum compression of the spring. We can use the conservation of kinetic energy:(1/2)m2v2^2 = (1/2)kx^2where k is the spring constant and x is the maximum compression of the spring.We know the mass and velocity of the 0.25-kg cart before the collision (v2 = 0 m/s), so we can solve for k:k = 2(1/2)m2v2^2 / x^2 = m2v2^2 / x^2Plugging in the numbers, we get:k = (0.25 kg)(0 m/s)^2 / x^2 = 0This means that the spring constant is 0, which is not physically possible. Therefore, there must be an error in the problem statement or some missing information that would allow us to calculate the maximum compression of the spring.
White light (380nm-750 nm) strikes a diffraction grating (900 lines/mm) at normal incidence. What is the highest-order visible maximum that is formed?
Answer:
Highest order: m = 1
Explanation:
Formula for solving this is;
dsin θ_m = mλ
We want to find m, thus;
m = (dsin θ_m)/λ
We are told that White light spans from 380nm - 750 nm. Thus, at maximum, λ = 750 nm.
θ_m = 90°
d = (900 lines/mm) = 9000 × 10^(-7) lines/nm
Since we want to find m, the units nm has to cancel out in the equation.
Thus, we will write d in nm/lines as (10^(7))/9000 nm/lines
Thus;
m = ((10^(7))/9000 × sin 90)/750
m = 1.48
Approximately 1
A ball has the energy to move 30 m/s with the mass of 5. What is the energy
How do i solve this problem? Hint: The cannonball is being launched vertically upwards, therefore, there is no initial horizontal speed. The given initial speed will also be the initial vertical speed.
The initial velocity of the ball is given as 36.0 m/s.
The horizontal component of velocity of ball is given as,
\(v_x=v\cos \theta\)The ball is projected vertically, therefore, the angle made by ball is 90 degree.
Plug in the known values,
\(\begin{gathered} v_x=(36.0m/s)cos90^{\circ} \\ =(36.0\text{ m/s)(0)} \\ =0\text{ m/s} \end{gathered}\)Therefore, the initial horizontal velocity of ball is 0 m/s.
The vertical component of velocity of ball is given as,
\(v_y=v\sin \theta\)Plug in the known values,
\(\begin{gathered} v_y=(36.0m/s)\sin 90^{\circ} \\ =(36.0\text{ m/s)(1)} \\ =36.0\text{ m/s} \end{gathered}\)Therefore, the initial vertical velocity of the ball is 36.0 m/s.