During a baseball game, a batter hits a popup to a fielder 83 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 5.7 s, how
high does it rise?
Answer in units of m

Robert has done 50,000 Joules of work in pushing the car 50 meters with a force of 1000 N

Work is a measure of energy transfer that occurs when a force is applied to an object and causes it to move in the direction of the force.

To calculate the work done by Robert in pushing the car, we need to use the formula:

work = force x distance x cos(theta)

In this case, the force is 1000 N, the distance is 50 meters, and we assume that the force is applied parallel to the ground, so theta is 0 degrees. Therefore, cos(theta) is equal to 1.

Plugging in these values, we get:

work = 1000 N x 50 m x 1

work = 50,000 Joules

Therefore, Robert has done 50,000 Joules of work.

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Hi there!

We can use the following equation for constant velocity:

\(\large\boxed{d = vt}\)

d = displacement (m)

v = velocity (m/s)

t = time (s)

Plug in the givens:

\(d = 3 * 7.15 = \boxed{21.45 m}\)

a) The ratio of the slit separation to the slit width is 2.  

b) There are four bright fringes within the first side peak.

a) The ratio of the slit separation to the slit width is 2. This is because the second side peak of the diffraction envelope is located at an angle of

2λ/d, where λ = wavelength of light and d = slit width.

The diffraction minima coincide with the two-slit interference maxima, which are located at angles of λ/d.

Therefore, the ratio of the slit separation to the slit width is 2.

(b) There are four bright fringes within the first side peak. This is because the first side peak of the diffraction envelope is located at an angle of

λ/d, where λ = wavelength of light and d = slit width.

The diffraction minima coincide with the two-slit interference maxima, which are located at angles of λ/d.

Therefore, there are 4 bright fringes within the first side peak.

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To solve this problem, we need to determine the frictional force acting on the block with different magnitudes of the applied force.

First, we need to find the normal force on the block, which is equal to the weight of the block. The weight of the block is given by:

W = mg = 2.5 kg x 9.8 m/s^2 = 24.5 N

Next, we need to find the force of the applied vertical force, which is given in the problem as "is". We can use trigonometry to find the vertical component of the force:

Fv = is sinθ

where θ is the angle between the force and the horizontal surface. Since the problem does not give us the value of θ, we will assume it to be 0°, which means the force is purely horizontal.

(a) If the magnitude of the applied force is 8.0 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 0) = 9.8 N

(b) If the magnitude of the applied force is 10 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 10) = 5.8 N

(c) If the magnitude of the applied force is 12 N, then the frictional force can be calculated as:

Ff = μkFn = μk(mg - Fv) = 0.25(24.5 - 12) = 3.1 N

Therefore, the magnitude of the frictional force acting on the block is 9.8 N, 5.8 N, and 3.1 N, for applied forces of 8.0 N, 10 N, and 12 N, respectively.

(a) When the horizontal force is 8 N the frictional force is 11.8 N.

(b) when the applied force is 10 N; the frictional force is 13.8 N.

(c) when the applied force is 12 N; the frictional force is 15.8 N.

(a) The magnitude of the frictional force on the block when the horizontal force is 8 N is calculated as;

F - Ff = ma

where;

F - μmg = ma

6 - 0.4 x 2.5 x 9.8 = 2.5 a

2.5 a = -3.8

a = -3.8/2.5

a = -1.52 m/s²

when the applied force is 8 N;

8 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 11.8 N

(b) when the applied force is 10 N;

10 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 13.8 N

(c) when the applied force is 12 N;

12 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 15.8 N

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A vector field and a starting position inside the field define the locus as a "field line." We have electric field lines for the electric fields.

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Answer:

yor welcome

Explanation:

hi

Answer:

A. infrared

Explanation:

Heat lamps used to keep foods warm are infrared on the light spectrum

The  equation describes the sum of the vectors plotted below is: \(\vec{r} = 4 \vec{x}+2 \vec{y}\)

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

According to the final position of the vector as shown in the figure, The final x co-ordinate is 4 and  the final y co-ordinate is 2.

the  equation describes the sum of the vectors plotted below is: \(\vec{r} = 4 \vec{x}+2 \vec{y}\)

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equal in magnitude but opposite in direction

Answer:

The Moon rotates on its axis and revolves around the Earth as the Earth revolves around the Sun.

Answer:

a bale

Explanation:

a bale is a group of turtles

Answer:

A bale or nest

Explanation:

Answer:

?

Explanation:

Answer:

Since the object is floating, the buoyant force equals the weight of the water displaced:

W = d g v  where d is the density of water and v = V/2 where v is the volume of the water displaced   and v = V/2 where V is the volume of the object.

W = 1000 kg/m^3 * 9.8 m/s^2 * (3.5 /2 m^3)

W = 17150 kg m / s^2 = 17150 N

When an object floats with half its volume beneath the surface of the water. If the volume of the object is 3.5 m,then the weight of the object would be 17167.5 N if the density of water is 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.

The density is the reciprocal of the specific volume of any substance.

The mathematical formula for density is given below

ρ =m /V

where ρ is the density of the substance

m is the mass of the substance

V is the volume of the substance

As given in the problem an object floats with half its volume beneath the surface of the water. The volume of the object is 3.5 m

The density of water is 1000 kg/m3​

According to the Archimedes principle, the buoyancy force acting on  the floating object is equal to the weight of the fluid displaced by it

W =  ρ g V

Given the object floats with half its volume beneath the surface of the water means the volume of the water displaced is half of the volume of the object

V=3.5/2 m³

W = 1000×9.81×3.5/2

   = 17167.5 N

Thus, the weight of the object is 17167.5 N

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Diamond, graphite and fullerenes (substances that include nanotubes and 'buckyballs' , such as buckminsterfullerene) are three allotropes of pure carbon. In all three allotropes, the carbon atoms are joined by strong covalent bonds, but in such different arrangements that the properties of the allotropes are very different.

Because of their distinctive structural makeup, graphite and diamond are distinct materials. Both have extremely high melting points due to their giant covalent structures. Diamond is incredibly strong and hard due to the four covalent links that each carbon atom in the mineral has with other carbon atoms. However, because each carbon in graphite is connected to three other carbons, it forms in layers.

Even though graphite is used to make pencils, despite the fact that each carbon atom only has three bonds, the layers are actually highly strong and have delocalized "free" electrons between them. Graphite appears soft because these electrons act as a lubricant between layers, allowing them to slide over one another. Graphite conducts electricity due to the free electrons. Diamond does not carry electricity because it lacks these free electrons.

There are more than three allotropes of carbon. These include diamond, graphite, graphene, carbon nanotubes, fullerenes, and carbon nanobuds.

Diamond

In a three-dimensional array, four additional carbon atoms are covalently attached to each carbon atom in a diamond. In essence, a diamond is one enormous molecule.

Graphite

The carbon atoms in graphite are bonded together in sheets of connected hexagons that resemble chicken wire. In essence, each sheet is a single molecule.

Each carbon atom in a sheet establishes solid covalent bonds with three other carbon atoms. The only forces keeping the sheets packed together are the modest intramolecular forces.

Graphene

In the form of a single sheet of graphite that is only one atom thick, graphene is made up entirely of carbon.

Nanocarbon tubes

Graphene sheet wrapped into a cylindrical tube of carbon atoms is how a carbon nanotube looks. Each atom connected to three other atoms, and the tube is one atom thick.

C60 and buckminsterfullerene

A single sheet of carbon atoms that has been folded into a spherical is what makes up buckminsterfullerene. Three additional atoms are connected to each carbon atom. With a carbon atom at each of the 20 hexagonal and 12 pentagonal corners, sixty carbon atoms are arranged in the shape of a ball.

There are numerous other known carbon balls, such as C70, C76, C84, and C540. They are collectively referred to as "buckyballs" or "fullerenes" and have varying amounts of pentagons and hexagons.

Nanocarbon buds

An allotrope of carbon called carbon nanobuds has fullerene-like "buds" that are covalently bonded to the sidewalls of carbon nanotubes.

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Explanation:

Falt isn't a size. I guess you were trying to spell flat. No, the earth isn't flat. instead, it is round.

Answer:

Earth is round. But the map shows you flat.

Answer:

3 m/s

The northward component of the puck’s velocity is 3 m/s

Explanation:

Applying the impulse momentum equation;

Impulse = change in momentum

Ft = m(∆v)

∆v = Ft/m

F = force = 6.0 N due north

t = time = 1.5 s

m = mass = 3.0 kg

Substituting the values;

Change in velocity ∆v = (6 × 1.5)/3.0 = 9/3

∆v = 3 m/s due north

And since the initial northward component of the puck’s velocity is zero.

The final northward component of the puck’s velocity is;

v = 0 + 3 m/s

v = 3 m/s

The northward component of the puck’s velocity is 3 m/s

As soon as the ball hits its highest point, it accelerates in the same direction as its downward velocity.

If the ball's velocity and acceleration are moving in the same direction, the ball is moving faster.

The direction of the ball's acceleration is initially downward (in the opposite direction of its velocity) due to the force of gravity, assuming the ball was thrown uphill and is travelling upward. The speed of the ball decreases owing to gravity as it rises until it reaches its highest point, where it temporarily stops moving.

The direction of the ball's velocity is now downward, and the direction of its acceleration is also downward as it comes back down (in the same direction as its velocity). This happens as a result of the ball falling faster and faster as it approaches the ground due to gravity.

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Answer: heat is transferred to and from objects -- such as you and your home -- via three processes: conduction, radiation, and convection.

Explanation:

Hopes this helps!

all except the rubber boots.

The answers should be The power lines themselves and The metal tools the worker uses (the 1st and 4th choices).

(For anyone curious, the image I attached to this answer is the image given for this problem.)

Answer:

Explanation:

According to Newton's Law of Gravitation, the gravitational force between two objects is always mutual, meaning that the force that one object exerts on the other is equal in magnitude but opposite in direction.

This means that if planet A exerts a gravitational force on planet B, then planet B will also exert a gravitational force on planet A. The magnitude of this force will be equal to the magnitude of the force that planet A exerts on planet B, but the direction will be opposite.

For example, if planet A exerts a gravitational force on planet B that is pulling planet B towards planet A, then planet B will also exert a gravitational force on planet A that is pulling planet A towards planet B. The magnitude of these two forces will be equal, but the directions will be opposite.

hat he sais

Answer:w

Explanation:

Usually, we use sampling when a population is hard to study, for some reason.

Sampling is a technique commonly employed in research and statistics when it is impractical or impossible to study an entire population directly. It involves selecting a subset, or sample, from the population and using the information gathered from the sample to make inferences about the entire population. This is done with the assumption that the sample is representative of the population and that the findings from the sample can be generalized to the larger population.

There are several reasons why a population might be difficult to study comprehensively. One reason is the size of the population. For example, if the population of interest is the entire world or a country, it would be practically impossible to study each individual in the population due to logistical constraints and limited resources. In such cases, sampling allows researchers to gather information from a smaller, manageable subset of the population.

Another reason for using sampling is when the population is dispersed or geographically scattered. If the population is spread out across a wide area, it can be challenging and costly to reach and collect data from every individual. Sampling allows researchers to select representative individuals or clusters from different regions, making data collection more feasible.

Additionally, there are cases where the population is inaccessible or hard to reach due to privacy concerns or ethical considerations. For example, if the population consists of individuals with certain medical conditions or sensitive personal information, it may be challenging to obtain consent or access to the entire population. In such cases, researchers can use sampling methods to obtain data from a subset of individuals who are willing to participate and meet the necessary criteria.

In summary, sampling is a valuable tool when studying populations that are hard to access, too large, or dispersed. It allows researchers to gather relevant data from a representative subset of the population and make valid inferences about the larger population, despite the challenges posed by studying the population as a whole.

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The question probable may be:

Usually, we use                  when a population is hard to study, for some reason.

Answer:

A. That it is expanding.

Answer:

Force = 60.08 N

Explanation:

Given that

Diameter d = 30 mm  

Holding pressure = 85 %  of Atmospherics pressure

Solution

As we know that  here 1 atm = 10⁵ N/m²

and pressure is known as force per unit area

pressure = \(\frac{F}{A}\)   ................1

put here value and we will get

F = \(0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N\)

solve it we get

Force = 60.08 N

Convex mirror always produces virtual images is the correct option.

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Answer:

 ΔV = -2.1 L

Explanation:

To solve this exercise we can use the ideal gas equation for two points

         PV = nRT

          P₁V₁ = P₂ V₂

where point 1 is on the surface and point 2 is at the desired depth,

          V₂ = \(\frac{P_1}{P_2} \ V_1\)

let's calculate

           V₂ = ( \(\frac{1 atm}{2.5 atm}\) ) 3.5 L

            V₂ = 1.4 L

this is the new volume, the change in volume is

          ΔV = V₂ -V₁

          ΔV = 1.4-3.5

          ΔV = -2.1 L