Sarah wanted to know the velocity of an airplane traveling from Houston to Dallas. After recording the distance (in miles) and time (in hours) it took to travel to Dallas, what other measurement does Sarah need to determine the airplanes velocity?

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

The car was moving at a speed of 36.95 m/s when it began to accelerate.

To find the initial velocity of the car when it began to accelerate, we can use the equation for average velocity:

V = (U + V)/2

Where:

Given that the car changes velocity at a constant acceleration, we can use the kinematic equation:

V = U + at

Where:

Substituting the known values, we have:

43.7 m/s = U + (2.5 m/s²)(2.7 s)

Simplifying the equation, we find:

U = 43.7 m/s - (2.5 m/s²)(2.7 s)

U = 43.7 m/s - 6.75 m/s

U = 36.95 m/s

Therefore, the car was moving at a speed of 36.95 m/s when it began to accelerate.

Let us say that:

1 = 1st player notation

2 = 2nd player notation (the opponent)

a. First let us establish the distance travelled by the 2nd player:

d2 = 13 m/s * (t + 1.5)

d2 = 13 t + 19.5

Then the distance of the 1st player:

d1 = v0 t + 0.5 a t^2         (v0 initial velocity = 0 since he started from rest)

d1 = 0.5 * 4 m/s^2 * t^2

d1 = 2 t^2

The two distances must be equal, d1 = d2:

2 t^2 = 13 t + 19.5

t^2 – 6.5 t = 9.75

Completing the square:

(t – 3.25)^2 = 9.75 + (- 3.25)^2

t – 3.25 = ±4.5

t = -1.25, 7.75

Since time cannot be negative, therefore:

t = 7.75 seconds

So he catches his opponent after 7.75 seconds.

b. Using the equation:

d1 = 2 t^2

d1 = 2 * (7.75)^2

d1 = 120.125 m

So he travelled about 120.125 meters when he catches up to his opponent.

It takes 3.25 seconds for the first player to catch the opponent. The first player has traveled 21.125 meters when they catch the opponent.

Given:

Opponent's speed (v₁) = 13 m/s

Acceleration of the first player (a) = 4.0 m/s²

Time (t) = 1.5 s

(a) We can use the following equation:

v = u + at

13  = 0 + 4.0 × t

t = 13 / 4.0

t = 3.25 s

Therefore, it takes 3.25 seconds for the first player to catch the opponent.

(b) we can use the equation:

s = ut + (1/2)at²

s = 0 × 3.25 + (1/2) × 4.0 × (3.25 )²

s = 0 + (1/2) × 4.0 × 10.5625

s = (1/2) × 4.0 × 10.5625

s = 21.125 m

Therefore, the first player has traveled 21.125 meters when they catch the opponent.

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The mass of the chain in two decimal places is 2.55 kg.

The tension at the end of the chain is due to the weight of the weight supported by the chain.

The mass of the chain is calculated by applying Newton's second law of motion as follows;

W = mg

where;

m = W / g

The mass of the chain is calculated as;

m = ( 25 N ) / ( 9.8 m/s² )

m = 2.55 kg

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The mass of the chain can be calculated using the formula Mass = tension / gravity. Given the values for tension and gravity, the mass of the chain is approximately 2.55 kg.

This is a problem related to the physics concept of force, mass, and acceleration due to gravity. Since we know that the tension in the chain isn't affected by the chain's length, we can assume that the tension is evenly distributed along the chain, with 25 N at each end.

The tension in the chain is equal to the weight of the chain. Therefore, we can use the formula for calculating the weight, which is Weight = mass x gravity. Since we know that the weight is equal to the tension, we can solve the formula for mass, which gives us Mass = tension / gravity. Substituting the given values, we have: Mass = 25 N / 9.8 m/s² = 2.55 kg.

Therefore, the mass of the chain is approximately 2.55 kg.

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Answer:

Instantaneous speed

Explanation:

It is given that, Henry was given this cool toy for his birthday. He can use it to tell how fast someone is going at that moment in time. The toy must calculate the person's instantaneous speed.

Instantaneous speed of an object is defined as the speed at a particular instant. Mathematically, it is given by :

[tex]v= \lim_{\Delta t \to 0}(\dfrac{\Delta x}{\Delta t})[/tex]

Instantaneous speed is a scalar quantity. It does not involve any direction.

So, Henry must calculate the person's instantaneous speed. Hence, the correct option is (d).

The equation of motion that we can use in this case is:

t = d / v

where t is time, d is distance, v is velocity

Therefore calculating for t:

t = 3.5 km / (25 km / h)

t = 0.14 h = 8.4 minutes

It takes approximately 8.4 minutes for a bird flying at 25 km/h to cover a distance of 3.5 km.

To determine how long it takes for a bird to fly 3.5 km at a speed of 25 km/h, we can use the formula for time, which is distance divided by speed. The calculation is as follows:

Time = Distance / Speed

Time = 3.5 km / 25 km/h = 0.14 h

To convert hours to minutes, multiply by 60:

Time = 0.14 h × 60 minutes/h = 8.4 minutes

Thus, it will take the bird approximately 8.4 minutes to fly 3.5km.

Answer:

X-Rays from space do not reach the ground.

Explanation:

To find the density of a cylinder, its volume is first determined using the formula V = πr²h. After correcting the radius and height according to the question, the density is calculated as mass divided by volume, approximately 6.65 g/cm³. Compared to standard material densities, the cylinder may be made of a type of stainless steel or a similar metal alloy.

To calculate the density of the cylinder, we first need to find its volume using the formula V = πr²h. Given that the length (height) is 3.23 cm and the diameter is 1.75 cm, the radius (r) would be half of the diameter, which is 0.875 cm. However, there seems to be a typo or mistake in the provided sample work as it uses a radius of 0.750 cm and a height of 5.25 cm, which do not match the question's parameters. Assuming the correct measurements, we would use:

V = π × (0.875 cm)² × 3.23 cm = 9.818 cm³

Now, to find the density, we divide the mass by the volume:

Density = Mass / Volume = 65.3 g / 9.818 cm³ = 6.65 g/cm³

Based on common material densities, this cylinder could potentially be made of stainless steel which typically has a density around 8 g/cm³; however, since our calculated density is slightly less, it could be a stainless steel alloy or a similar metal with a slightly lower density.

The force acting on a body is directly proportional to both mass and acceleration. Hence, the an increase in acceleration with a smaller mass does not make an increase  in force than that act on a larger mass with less acceleration. Hence, the net force will be the same for both masses.

Force is an external agent acting on a body to deform it or to change its state of motion or rest.  Force is a vector quantity thus, it is characterized by a magnitude and direction.

The force according to second law of motion is the product of mass and acceleration of the body. Hence, an increase in mass or acceleration or both  results in an increase in force.

Here, the force F1 is acting on the mass 3 Kg with an acceleration of 2a

and force F2 is acting on 6 Kg with an acceleration a.

Then F1  = 3 × 2 a = 6a

F2 = 6 × a = 6a.

Thus, net force of both are equal.

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Final answer:

To find the upward thrust force needed to accelerate the rocket, we use Newton's second law of motion. The upward thrust force needed is approximately 4989.2 N.

Explanation:

To find the upward thrust force needed to accelerate the rocket, we can use Newton's second law of motion, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 590 kg and the acceleration is the change in velocity divided by the time taken.

Using the formula acceleration = (final velocity - initial velocity) / time, we can calculate the acceleration:

acceleration = (28 m/s - 0 m/s) / 3.3 s =  8.48 m/s²

Now that we have the acceleration, we can calculate the upward thrust force:

force = mass x acceleration = 590 kg x 8.48 m/s² = 4989.2 N

Therefore, the upward thrust force needed to accelerate the rocket uniformly to an upward speed of 28 m/s in 3.3 s is approximately 4989.2 N.

To accelerate the 590-kg rocket uniformly to an upward speed of 28 m/s in 3.3 s, a thrust force of approximately 5010.2 N is required, calculated using Newton's second law.

To determine the upward thrust force needed to accelerate the rocket uniformly, we can use Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

First, calculate the acceleration (a) of the rocket using the kinematic equation:

v = u + at

where:

v is the final velocity (28 m/s),

u is the initial velocity (0 m/s, as the rocket is at rest),

a is the acceleration, and

t is the time (3.3 s).

Rearrange the equation to solve for acceleration:

a = (v - u) / t

Substitute the values:

a = (28 m/s - 0) / 3.3 s ≈ 8.48 m/s^2

Now, use Newton's second law to find the force (F):

F = ma

F = 590 kg × 8.48 m/s^2 ≈ 5010.2 N

The part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c, as the cell body, also known as the soma, is a central part of a eukaryotic cell.

The cell body includes the nucleus and other organelles like the mitochondria, endoplasmic reticulum, Golgi apparatus, and cytoplasm. The cell body is the main site for the metabolic and biochemical processes of the cell, and it is responsible for maintaining the integrity of the cell and regulating its activities. In the case of neurons, the cell body also contains the genetic information of the neuron in the form of DNA, which is used to synthesize the proteins and other molecules needed for the growth, repair, and maintenance of the neuron.

Hence, the part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c.

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True, the greater the mass of an object, the greater its force due to gravity.

The force due to gravity on object's is calculated by applying Newton's second law of motion as follows;

F = mg

where;

From the formula given above we can conclude that, the greater the mass of an object, the greater its force due to gravity.

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The statement addresses Isaac Newton's law of gravitation which suggests that the greater the mass of an object, the greater its force due to gravity. Gravitational attraction is a universal property of mass. Mass is constant while weight varies with the gravitational field strength.

The statement in question - 'The greater the mass of an object, the greater its force due to gravity' - is a fundamental principle in Physics, specifically in the study of gravity. Gravity is essentially a 'built-in' property of mass, meaning all masses in the universe interact via the force of gravitational attraction. The British physicist Sir Isaac Newton concluded that the gravitational attraction between two bodies is proportional to their masses and inversely proportional to the square of the distance between them. This is illustrated in Newton's law of gravitation, formulated as Fgravity = G × (M1 × M2) / R².

It's important to note also that the weight of an object, which is the force of gravity acting on it, varies with the strength of the gravitational field. Mass, however, is a constant property of an object and does not change regardless of the gravity acting on it. Mass is a measure of the amount of matter in an object, whereas weight is the gravitational pull acting on that matter.

Understanding these principles allows us calculate the masses of astronomical objects and understand a range of natural phenomena, from the motion of a falling apple to the orbits of planets.

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1. Some wooden rulers do not start with 0 at the edge, but have it set on a few millimeters because the cut of the woof alters the length of the first measures. Moreover, some objects are difficult to be aligned on the ruler, thus, in order to remove this difficulty, some wooden rulers do not start with 0 at the edge.

2. The micrometer has been badly bent which means that it will not measure properly and will not be perfectly parallel with the surface to be measured. This distortion will  change its precision. The device should be parallel with the surface to be measured in order to maintain its accuracy.

3. No, the parallax will not affect the precision of a measurement that we take as precision  is the closeness of the readings and the readings are already precise, thus taking the reading with the same parallax won't impair the precision but decreases the accuracy.

4. Length of box=18.1 cm, width= 19.2 cm and height=20.3 cm, then the a.volume=length×width×height

=[tex]18.1{\times}19.2{\times}20.3[/tex]

=[tex]7050cm^{3}[/tex]

b. The reading of the length is precise to 0.1 centimeters, while the volume's reading is precise to 10 cm³.

c. The stack of 12 of the boxes has height=[tex]12{\times}19.3[/tex]

=[tex]232cm[/tex]

d.The measure of one box is precise to 0.1 cm, while the measurement of the combined boxes is precise to 1 cm.

The net force acting on the refrigerator is 400 N to the right

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of refrigerator = m = 200 kg

magnitude of applied force = F = 400 N

magnitude of frictional force = f = 0 N

Asked:

net force = ΣF = ?

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = F - f[/tex]

[tex]\Sigma F = 400 - 0[/tex]

[tex]\boxed{\Sigma F = 400 \texttt{ N} }[/tex]

[tex]\texttt{ }[/tex]

We could also calculate the acceleration of the refrigerator as follows:

[tex]\Sigma F = ma[/tex]

[tex]a = \Sigma F \div m[/tex]

[tex]a = 400 \div 200[/tex]

[tex]\boxed{a = 2 \texttt{ m/s}^2 }[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

The answer is A 00.2

The time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.


Let's denote the speed of the slower plane as [tex]\(S\) (in mi/h)[/tex]. The faster plane, which is flying east, will have a speed of [tex]\(S + 30\) mi/h[/tex].

The relative speed between the two planes is the sum of their individual speeds:

[tex]\[ \text{Relative speed} = S + (S + 30) = 2S + 30 \, \text{mi/h} \][/tex]

Now, we know that the planes leave at noon, and we want to find the time it takes for them to be 1200 mi apart. Let t be the time in hours.

The distance traveled by the slower plane in t hours is [tex]\(S \cdot t\)[/tex], and the distance traveled by the faster plane is [tex]\((S + 30) \cdot t\)[/tex]. The sum of these distances is the total distance between the planes:

[tex]\[ S \cdot t + (S + 30) \cdot t = 1200 \][/tex]

Combine like terms:

[tex]\[ 2S \cdot t + 30 \cdot t = 1200 \][/tex]

Factor out t:

[tex]\[ t(2S + 30) = 1200 \][/tex]

Now, solve for t:

[tex]\[ t = \frac{1200}{2S + 30} \][/tex]

We know that when t is the time, it should be in hours, and we want to find at what time they will be 1200 mi apart. If they leave at noon, the time will be [tex]\(12:00 PM\) + \(t\) hours[/tex].

So, the time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.

Final answer:

The speed mentioned is approximately half the escape velocity from Earth and is not sufficient to escape Earth's gravitational field. The object will ascend to a maximum altitude before falling back down. A specific calculation would require further application of physics principles.

Explanation:

The question is related to the concept of escape velocity in physics, which is the speed needed to break free from a planet's gravitational pull without further propulsion. The escape velocity from the surface of the Earth is approximately 11 km/s. At a speed of 5534 m/s, which is approximately 5.5 km/s (half the escape velocity), the object will not have enough kinetic energy to completely escape Earth's gravity and hence will reach a maximum altitude before falling back to Earth. To calculate precisely how far from the center of the Earth you would get with the initial speed of 5534 m/s, you would need to use the equations for gravitational potential energy and kinetic energy, considering how much kinetic energy converts into potential energy as the object ascends against Earth's gravity.

The density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit, its density would be 1000 kg/m³.

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem we have to calculate the density of water in kilogram per centimeter cube,

1 cm³ = 1×10⁻⁶ m³

1 gram = 1×10⁻³ kg

The density of the water= mass of the water /volume of the water

                                 =1.00 g/cm3

                                 =1×10⁻³ kg /1×10⁻³ kg

                                 =1000 kg/m³

Thus, if the density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit its density would be 1000 kg/m³.

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The beanbag was thrown in a motion with an initial velocity of 10.78 m/s and reached a peak height of 5.929m.

The height and initial velocity of a thrown object can be calculated using kinematics, a branch of physics that describes motion of objects. Since the beanbag is thrown up and comes back to the same spot, the total time of its flight (2.2s) is split between the time going up and the time coming back down.

We can simplify calculations by assuming that the time for the journey upwards equals the time for the journey downwards, i.e., 1.1s each.

The highest point is reached when the velocity is zero. We can use the formula v= u + at to find the initial velocity (u), where v is the final velocity, a is acceleration due to gravity (-9.8m/s² due to the upward direction), and t is the time (1.1s). For the highest point, we can use the formula h = ut + 0.5at².

Calculating this, we find the initial velocity u = -9.8 × 1.1 = -10.78 m/s (it's negative because we took upward direction as negative). To calculate the highest point or maximum height, we find h = -10.78 × 1.1 + 0.5 ×  -9.8 × (1.1)² = -5.929 m (this is also negative due to upward direction). Hence, the beanbag was thrown up with an initial velocity of 10.78 m/s and reached a maximum height of 5.929 m.

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Final answer:

The Earth rotates at an angular velocity of 15 degrees per hour. Without specific flight path distances, we refer to Earth's rotation speed as a base reference for understanding angular velocity in a general sense.

Explanation:

The flight from Kampala to Singapore takes 7.0 hours. To calculate the plane's angular velocity relative to the Earth's surface, we need to understand that completing a flight between two points on Earth's surface involves moving over a certain segment of Earth's rotation. Since the Earth completes one full rotation (360 degrees) in 24 hours, we can calculate the angular velocity (in degrees per hour) by simply dividing the total degrees by the total hours.

Angular velocity = Total rotation / Time = 360 degrees / 24 hours = 15 degrees per hour.

However, this is the angular velocity of the Earth itself. A plane's angular velocity relative to Earth would depend on the distance traveled. But without specific details on the distance covered, we can only refer to the Earth's own rotation speed as a base reference, which is 15 degrees per hour. For a specific flight, beyond this basic understanding, actual calculation would require details about the exact flight path length over the Earth's surface.

Final answer:

The linear velocity of a tooth at the edge of the 14 inch diameter blade on a typical table saw can be calculated by converting the blade's rotational speed from revolutions per minute to radians per second and using the formula for linear velocity. The calculated value will give the speed of the tooth in miles per hour.

Explanation:

To calculate the linear velocity of a tooth at the edge of the blade, we need to first convert the blade's rotational speed from revolutions per minute to radians per second. We can do this by multiplying the revolutions per minute (rpm) by 2π (since there are 2π radians in one revolution) and dividing by 60 (to convert minutes to seconds). So, the angular velocity of the blade is:

ω = (3300 rpm) x (2π rad/rev) / (60 s/min)

Next, we can use the formula for linear velocity to find the speed of a point on the edge of the blade. The formula is:

v = ω x r

where v is the linear velocity, ω is the angular velocity, and r is the radius. In this case, the blade has a diameter of 14 inches, so the radius is half of that, which is 7 inches or 0.5833 feet. Converting this to miles, we get:

r = 0.5833 ft x (1 mile/5280 ft)

Finally, we can substitute the values into the formula:

v = (3300 rpm) x (2π rad/rev) / (60 s/min) x 0.5833 ft x (1 mile/5280 ft)

Calculating this equation gives the linear velocity of a tooth at the edge of the blade.