Puck A and puck B are free to slide without friction on a horizontal air table; the mass of puck A has been measured to be 165. 0 grams , but the mass of puck B is unknown. The pucks are made of super-ball like material, so any collision between them should be elastic. An experiment is done with puck B at rest at the center of the air table, and with puck A sent at 55. 0 cm/s to make a glancing collision with puck B. After the collision, puck A is measured to have a speed of 29. 0 cm/s , and is observed to have been deflected by 27. 0 degrees from its original direction. What was the y component of puck B's momentum after the collision?

If you were standing 5 meters away from the cliff when you threw the keys, the total distance from the base of the cliff where you should look for the keys is 21.1 meters.

To find the distance from the base of the cliff where you should look for your keys, you can use the formula for the horizontal range of a projectile, which is:

range = (initial velocity * time of flight)^2 / (2 * acceleration due to gravity)

In this case, the initial velocity of the keys is 8.0 m/s (horizontally), the acceleration due to gravity is 9.8 m/s^2, and the time of flight can be calculated using the formula:

time of flight = (2 * initial velocity * sin(angle)) / acceleration due to gravit

Since the keys were thrown horizontally, the angle is 90 degrees, so the time of flight becomes:

time of flight = (2 * 8.0 m/s * sin(90)) / 9.8 m/s^2 = 1.63 seconds

Substituting these values into the formula for the horizontal range, we get:

range = (8.0 m/s * 1.63 s)^2 / (2 * 9.8 m/s^2) = 16.1 meters

This is the horizontal distance that the keys traveled during their time of flight. To find the distance from the base of the cliff where you should look for the keys, you need to add this distance to the horizontal distance you were from the cliff when you threw the keys.

For example, if you were standing 5 meters away from the cliff when you threw the keys, the total distance from the base of the cliff where you should look for the keys is 16.1 meters + 5 meters = 21.1 meters.

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Gravity is a fundamental, universal force that pulls objects toward Earth's center. It increases with mass and decreases with distance. Measured in Newtons, it affects all objects.

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The correct question would be as

Which statement describes gravity? Select three options. There is no defined unit of measurement for gravity.

Gravity is the force that pulls objects toward Earth’s center.

Objects that have a small mass will have no gravitational pull.

Gravitational pull between two objects increases as the mass of one increases.

Gravitational pull decreases when the distance between two objects increases

Answer:

Explanation:

A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.

The distance around the rectangular field is 128.88 m.

Perimeter is the total distance around an object.

To calculate the perimeter/ total distance around the rectangular field, we us the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the distance around the rectangular field is 128.88 m.

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Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, \(v_t=60\ m/s\)

Area of cross section, \(A=0.33\ m^2\)

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\(\dfrac{1}{2}\rho CAv_t^2=mg\)

Where

\(\rho\) is the density of air = 1.225 kg/m³

C is drag coefficient

So,

\(C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01\)

So, the drag coefficient is 1.01.

The quantity that must be measured in order to calculate the coefficient of friction is angle at which the surface is tilted

Since the block of mass m slides at a constant speed,

a = 0

If tilted at an angle θ,

∑ Fy = m ay

N - Wy = 0

N = m g cos θ

∑ Fx = m ax

Wx - f = 0

f = m g sin θ

f = μ N

f = Frictional force

μ = Co efficient of friction

N = Normal force

μ = f / N

μ = m g sin θ / m g cos θ

μ = sin θ / cos θ

μ = tan θ

μ ∝ θ

Therefore, to calculate the coefficient of friction, the quantity that needs to be measured is angle of tilt

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Explanation:

a) I = V / R

1.70 = 115 / R

R = 115 / 1.70

R = 67.647

R = 67.65 ohms

Therefore, equivalent resistance is 67.65 ohms

b) Equivalent resistance of circuit from above sum is 67.65 ohms

Given resistance of each bulb is 1.50 ohms

Number bulbs = Equivalent resistance / Resistance of each bulb

= 67.65 / 1.50

= 45

The angle θ is approximately 0.688°, and the angle θ' is approximately 1.988°.

To determine the angles θ and θ' in the given scenario, we can use Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the media involved. Snell's Law can be stated as follows:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:

n₁ is the refractive index of the medium of incidence (in this case, air with a refractive index close to 1),

θ₁ is the angle of incidence,

n₂ is the refractive index of the medium of refraction (in this case, linseed oil with a refractive index of 1.48), and

θ₂ is the angle of refraction.

Let's solve for the unknown angles θ and θ' using the given information.

Given:

Angle of incidence θ = 1°

Refractive index of linseed oil n₂ = 1.48

We need to find the angle of refraction θ₂.

Using Snell's Law, we have:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Since the refractive index of air (n₁) is approximately 1, we can simplify the equation to:

sin(θ₁) = n₂ * sin(θ₂)

Plugging in the values:

sin(1°) = 1.48 * sin(θ₂)

We can now solve for θ₂:

θ₂ = arcsin(sin(1°) / 1.48)

Calculating this value, we find:

θ₂ ≈ 0.688°

Now, let's determine the angle θ'.

Given:

Angle of refraction θ₂ = 0.688°

Refractive index of linseed oil n₂ = 1.48

We need to find the angle of incidence θ'.

Using Snell's Law, we have:

n₂ * sin(θ₂) = n₁ * sin(θ')

Since the refractive index of air (n₁) is approximately 1, we can simplify the equation to:

n₂ * sin(θ₂) = sin(θ')

Plugging in the values:

1.48 * sin(0.688°) = sin(θ')

Solving for θ', we find:

θ' = arcsin(1.48 * sin(0.688°))

Calculating this value, we get:

θ' ≈ 1.988°

Therefore, the angle θ is approximately 0.688°, and the angle θ' is approximately 1.988°.

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The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

The vertical height of fall of the child from the top of the incline is calculated as;

\(sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m\)

The gravitational potential energy of the child at the top of the incline is calculated as;

\(P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J\)

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

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The accuracy of a frequency counter relies heavily on the stability of its time base. The time base is the reference oscillator that provides the time intervals for counting the frequency of the input signal. Any instability or variation in the time base will directly impact the accuracy of frequency measurements. Factors such as temperature variations, aging of components, and noise can affect the stability of the time base. Therefore, maintaining a stable and precise time base is crucial for achieving accurate frequency measurements.

The factor that most significantly affects the accuracy of a frequency counter is time base stability. A stable time base ensures consistent and precise time intervals, leading to accurate frequency measurements. Variations or drifts in the time base can introduce errors and inaccuracies in frequency counting. While other factors play a role in the overall performance of a frequency counter, they have a lesser impact on accuracy compared to time base stability.

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Answer:

Netball is a ball sport for two teams of seven players it's rules are published in print and online by the international netball federation Games are played on a rectangular court divided into thirds with a raised goal at coach short end

Explanation:

It will help you

At standard temperature and pressure (STP), the temperature is 273.15 K and the pressure is 1 atmosphere (atm).

The average translational kinetic energy of an oxygen molecule at STP can be calculated using the formula:

KE = (3/2) * k * T

where KE is the average translational kinetic energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), and T is the temperature in kelvin.

Substituting the values for STP, we get:

KE = (3/2) * (1.38 x 10^-23 J/K) * 273.15 K

Simplifying this expression, we get:

KE = 6.21 x 10^-21 J

Therefore, the average translational kinetic energy of an oxygen molecule at STP is approximately 6.21 x 10^-21 J.

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All for the sattement A, B, and E is True and satement C, D, and F is False where you learned about astronomy from space.

A .The Compton, Chandra and Spitzer telescopes are all examples of space telescopes.

-This statement is True.

B. Most types of electromagnetic waves are not visible to ground-based telescopes.

-This statement is True.

C. Enough infrared energy gets to Earth's surface that infrared telescopes do not need to be put in orbit.

-This statement is False.

D. Cosmic rays are the absolute highest-energy type of electromagnetic wave.

-This statement is False.

E. The Hubble Space Telescope makes observations at all wavelengths.

-This statement is True.

F. Earth's atmosphere has no effect on visible light, so there is no need to place an optical telescope in space.

-This statement is False.

Astronomy is the scientific study of celestial objects and phenomena, including stars, planets, galaxies, and the universe as a whole. Astronomers use various tools and techniques, such as telescopes, spectroscopy, and astrophysics, to observe and understand the properties and behavior of these celestial objects.

Through the study of astronomy, scientists have discovered many fascinating things about the universe, such as the existence of black holes, the age and size of the universe, and the formation and evolution of stars and galaxies. Astronomy has also played a significant role in advancing our understanding of fundamental physics, including the laws of gravity and the behavior of matter and energy at extreme conditions.

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Answer:

His distance would be 80 or 13 something. I dont really understand this question. Just be sure to look into other answers pls.

Explanation:

Answer:

80 Percent

Explanation:

E=energy output/energy input×100

E=8000/10000×100

E=0.8×100

E=80 percent.

Mark brianliest if my answer suit your question

If the machine put, in 8000 J, and the user puts 10,000 J of work then the efficiency of the machine will be equal to 80%.

Work is a physics term used to describe the transfer of energy that takes place when an object changes above a distance because of an external force, at least some of which is given in the vector of the dislocation. The element of the force acting all along the path multiplied by the length of the path can be used to calculate work if the force is constant.

Mathematically, this idea is expressed as W = fd, where W is the effort and f is the force multiplied by d, the distance. Work is completed whenever the force is applied at an angle with about the displacement.

As per the data provided in the question,

Total input energy = 10,000 J

Total output energy = 8,000 J

Then, the efficiency (η) of the machine will be,

η = (output energy/input energy) × 100

η = 8000/10000 × 100

η = 80%.

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As the object is thrown up with the initial velocity of 20.0 m/s, and there is an air resistance force that would cause an acceleration of 3.00 m/s2, the speed of the object as a return to the ground is 10.8 m/s.

The kinematic equations are a set of equations that describe the motion of an object with constant acceleration.

When we have an initial velocity value, it is written as Vo, while for the final velocity, we simply write V or Vt. As an object moves through the air, air resistance slows the object’s speed.

The formula of the kinematic equation used for solving this case is

Vt = V0 + at (the gravity is 10 m/s2)

\(Vt = Vo + (-g-a)t\\\0 = 20 + (-10-3)t\\0 = 20-13t\\\13t = 20\\\t = \frac{20}{13}\)

After the time is known, now we can insert the value into the following formula :

\(Vt = Vo + (g-a)t\\Vt = 0 + 7.\frac{20}{13} \\Vt = 0 + \frac{140}{13} \\\Vt = 10.8 m/s\)

Thus, the speed of the object returns to the ground after being thrown up with an initial velocity of 20.0 m/s and acceleration of 3.00 m/s2, which is 10.8 m/s.

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Answer:

220 seconds

Explanation:

Answer:As an equation, the tangential velocity is the distance, 2π r, divided by the time, T. Thus, A point on the circle moves a distance 2π r in a time T. We can extend our equation by looking at a few ideas. These concepts include the angular speed, ω, and the frequency, f. The angular speed, ω, is a speed of rotation.

Explanation:

i know

It is velocity of a football just after being kicked. Velocity is the speed and direction of an object, and it is a vector quantity, meaning it has both magnitude and direction.

When a football is kicked, it has a certain velocity in a particular direction, which can be measured using a speedometer or other measuring device. Acceleration, on the other hand, refers to the rate of change of velocity, so it would only be relevant if the football's velocity was changing after being kicked (for example, if it were slowing down due to air resistance or gravity).

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Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

\(k = \frac{1}{2} m {v}^{2} \\ \)

m is the mass

v is the velocity

From the question we have

\(k = \frac{1}{2} \times 2.55 \times {3.81}^{2} \\ = 1.275 \times 14.5161 \\ = 18.5080275\)

We have the final answer as

Hope this helps you

Answer:

A) 1.21 kg

B) 1.26 kg

C) 3.13 kg

Explanation:

Let's say the mass of the cart is mc, the mass of each fan is mf, and the mass of the draggable mass units is M.

Newton's second law says the net force equals the mass times the acceleration.

∑F = ma

When there are three fans, the acceleration is 1.20 m/s².

∑F = ma

3 (2 N) = (mc + 3 mf) (1.20 m/s²)

5 kg = mc + 3 mf

When there are two fans, the acceleration is 1.07 m/s².

∑F = ma

2 (2 N) = (mc + 2 mf) (1.07 m/s²)

3.74 kg = mc + 2 mf

When there are three fans (one off) and two draggable mass units, the acceleration is 0.40 m/s².

∑F = ma

2 (2 N) = (mc + 3 mf + 2 M) (0.40 m/s²)

10 kg = mc + 2 mf + 2 M

Solving the system of equations, first subtract the second equation from the first:

mf = 1.26 kg

Now plug into either of the first two equation to find mc.

mc = 1.21 kg

Finally, plug both into the third equation to find M.

M = 3.13 kg

(A) The mass of the cart is 1.214 kg

(B) The mass of one fan is 1.262 kg

(C) The mass of one of the draggable mass units is 3.131 kg

The given parameters:

To find:

A. The mass of the cart.

B. The mass of a fan.

C. The mass of one of the draggable mass units

Applying the Gizmo observation:

Applying Newton's second law of motion:

\(F = ma\\\\where;\\\\m \ is \ the \ mass \ of \ the \ object s\\\\m = mass \ of \ cart\ (m_c)+ mass \ of \ fans \ (m_f) + \ draggable \ mass \ (m_d) \\\\\)

For 3 fans (all -on) and  zero draggable mass:

\(3(2) = 1.2(m_c + 3m_f)\\\\\frac{6}{1.2} = m_c + 3m_f\\\\5 = m_c + 3m_f \ \ -----(1)\)

For 3 fans (2 -on) and  zero draggable mass:

\(2(2) = 1.07(m_c + 2m_f)\\\\\frac{4}{1.07} = (m_c + 2m_f)\\\\3.738 = m_c + 2m_f \ \ ------(2)\)

For 3 fans (2 - on) and 2 draggable mass:

\(2(2) = 0.4(m_c + 2m_f + 2m_d)\\\\\frac{4}{0.4} = (m_c + 2m_f+ 2m_d)\\\\10 = m_c + 2m_f + 2m_d\ \ ------(3)\)

Find the mass of a fan by subtracting equation 2 from equation 1:

\(\ \ \ \ 5 \ \ \ \ \ \ = \ m_c + 3m_f\\-(3.738 \ = \ mc + 2m_f)\\ \\1.262 \ kg = m_f\)

Find the mass of the cart:

\(m_c = 5 - 3m_f\\\\m_c = 5 - 3(1.262)\\\\m_c = 1.214 \ kg\)

Find the mass of one of the draggable mass units:

\(2m_d = 10 - (2m_f + m_c)\\\\2m_d = 10 - (3.738)\\\\2m_d = 6.262 \\\\m_d = \frac{6.262}{2} \\\\m_d = 3.131 \ kg\)

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Answer:

newton's 2nd  law

Explanation:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

\(\\ \sf\longmapsto \dfrac{9}{5}(25)+32\)

\(\\ \sf\longmapsto 9(5)+32\)

\(\\ \sf\longmapsto 45+32\)

\(\\ \sf\longmapsto 77°F\)

\(\huge\purple{Hi!}\)

Recent research suggests that summer really does tend to be a time of reduced productivity. Our brains do, figuratively, wilt.

One of the key issues is motivation: when the weather is unpleasant, no one wants to go outside, but when the sun is shining, the air is warm, and the sky is blue, leisure calls.

on rainy days, men spent, on average, thirty more minutes at work than they did on comparatively sunny days.

bad weather made workers more productive, as measured by the time it took them to complete assigned tasks in a loan-application process.

When the weather improved, in contrast, productivity fell. The mere thought of pleasant alternatives made people concentrate less.

In summer, our thinking itself may simply become lazier.

Pleasant weather can often lead to a disconcerting lapse in thoughtfulness.

The better the weather, the easier it was to get the students to buy into a less-than-solid argument: on days that were sunny, clear, and warm, people were equally persuaded by both strong and weak arguments in favor of end-of-year comprehensive exams.

When the weather was rainy, cloudy, and cold, their critical faculties improved: in that condition, only the strong argument was persuasive.

Summer weather—especially the muggy kind—may also reduce both our attention and our energy levels.

In one study, high humidity lowered concentration and increased sleepiness among participants.

The weather also hurt their ability to think critically: the hotter it got, the less likely they were to question what they were told.

The shift toward mindlessness may be rooted in our emotions.

People get happier as days get longer and warmer in the approach to the summer solstice, and less happy as days get colder and shorter.

They also report higher life satisfaction on relatively pleasant days. The happiest season, then, is summer.

A good mood, generally speaking, has in turn been linked to the same type of heuristic, relatively mindless thinking.

A bad mood tends to stimulate more rigorous analytical thought. Weather-related mood effects can thus play out in our real-life decisions—even weighty ones.

There’s a limit to heat’s ability to boost our mood.

when temperatures reach the kind of summer highs that mark heat waves all over the world, the effect rapidly deteriorates.

on days when the temperature rose above ninety degrees, the negative impact on happiness levels was greater than the consequences of being widowed or divorced.

Our cognitive abilities seem to improve up to a certain temperature, and then, as the temperature continues to rise, quickly diminish.

the optimal temperature hovered around seventy-two degrees Fahrenheit, twenty-seven degrees Celsius, or roughly eighty-one degrees Fahrenheit.

Blistering heat does give us a perfectly good reason to eat ice cream: studies have shown again and again that blood glucose levels are tied to cognitive performance and willpower. A bite of something frozen and sweet, boosting depleted glucose stores, might be just what a brain needs as the temperature spikes.

The average velocity of the car in m/s for the first leg is 36.2 m/s and the average velocity (in m/s) for the total trip is 7 m/s

Speed is the distance travelled per time taken. The S.I unit is m/s. The average speed is the ratio of the total distance travelled to the to time taken. While velocity is the distance travelled in a specific direction per time taken

Given that a race car traveling northward on a straight, level track at a constant speed travels 0.760 km in 21.0 s. The return trip over the same track is made in 26.0 s.

(a) The average velocity of the car in m/s for the first leg of the run will be

Velocity = (0.760 x 1000)/ 21

Velocity = 760 / 21

Velocity = 36.2 m/s

(b) The average velocity (in m/s) for the total trip will be

36.2 - (0.760 x 1000)/ 26

36.2 - (760) / 26

36.2 - 29.2

7 m/s

Therefore, the average velocity of the car in m/s for the first leg is 36.2 m/s and the average velocity (in m/s) for the total trip is 7 m/s

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The scale is said to be reliable but not valid.

A person weighing 100 pounds is expected to return the same weight on mounting a scale, measuring 200 pounds means the scale is probably uncalibrated. However, when the person steps off and climbs back and the scale still shows 200 pounds this means the scale is reliable but not valid. The scale would have been rendered unreliable and invalid if it returns a different weight of the person. Hence, the scale is working but needs to be calibrated properly. It is also probable that the scale had been earlier set to 100 pounds which explains the reason for a doubled result.

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Answer:

Sediments can be carried from one place to another. The movement of sediments by wind, water, ice, or gravity is called erosion.

Answer:

Vx = 3.10 [m/s]

Vy = 11.59 [m/s]

Explanation:

To solve this problem we must decompose the velocity vector by means of the angle on the horizontal.

v = 12 [m/s]

Vx = 12*cos (75) = 3.10 [m/s]

Vy = 12*sin (75) = 11.59 [m/s]

To solve this problem, we can use the kinematic equations of motion. Since the bowling ball is fired horizontally, we know that the initial vertical velocity is zero. We can use the equation:

h = vi*t + (1/2)at^2

where h is the initial height of the cliff, vi is the initial vertical velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time it takes for the ball to hit the ground.

Solving for t, we get:

t = sqrt(2h / g) = sqrt(2192 / 9.8) = 8.88 s

Now that we know the time it takes for the ball to hit the ground, we can use the horizontal velocity to find the distance it travels:

d = v*t = 248 * 8.88 = 2203.24 m

Therefore, the bowling ball lands approximately 2203.24 meters away from the base of the cliff.

A golf ball would land at the same distance as the bowling ball. This is because the distance traveled by a projectile only depends on its initial velocity and the time it spends in the air. Since the golf ball and the bowling ball are fired with the same horizontal velocity, they will travel the same distance before hitting the ground, assuming they are both fired from the same height.

However, the golf ball would take longer to hit the ground because it has a lower initial vertical velocity than the bowling ball, so it would have a longer time of flight.

To find the velocity at impact, we can use the kinematic equation:

v^2 = vi^2 + 2ad

where v is the final velocity (which we want to find), vi is the initial vertical velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance traveled by the ball.

Using the distance found in part a, we get:

v^2 = 2*(-9.8)*2203.24

v = sqrt(2*(-9.8)*2203.24) = 196.67 m/s

The impact angle can be found using trigonometry. The angle theta can be found using the equation:

tan(theta) = opposite / adjacent

where the opposite side is the vertical distance the ball has fallen (which is 192 m) and the adjacent side is the horizontal distance the ball has traveled (which is 2203.24 m).

tan(theta) = 192 / 2203.24

theta = tan^-1(192 / 2203.24) = 5.02 degrees

Therefore, the velocity at impact with the ground is 196.67 m/s at an angle of 5.02 degrees.

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