Which statement describes a physical property related to the particles that
make up a material?
A. Carbon is flammable because its atoms combine easily with
oxygen atoms in the presence of heat.
B. Fluorine is highly reactive with other elements because its atoms
have 7 valence electrons.

C. Neon is not reactive with other elements, because its atoms have
8 valence electrons.
D. Gold is a relatively dense metal because its atoms have more
protons than atoms of many other metals.

Answers

Answer 1

Answer:

D

Explanation:

physical properties are properties that can be measured with instruments, such as a scale for weight.

The descriptions are chemical properties

Answer 2

Gold's atoms have more protons than those of many other metals, it is a particularly dense metal. Statement D explains the physical characteristics of the constituent parts of a substance.

What is density?

Density is defined as the mass per unit volume. It is an important parameter to understand the fluid and its properties. Its unit is kg/m³.

Gold is a relatively dense metal because its atoms have more protons than atoms of many other metals.

Statement D describes a physical property related to the particles that make up a material. Because density is the physical property of the material.

Hence option D is corect.

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Related Questions

A grapefruit falls from a tree and hits the ground 0.72 s later.
How far did the grapefruit drop?
What was its speed when it hit the ground?

Answers

Answer:

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

Explanation:

Free Fall Motion

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2.[/tex]

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:

[tex]\displaystyle y=\frac{9.8\cdot 0.72^2}{2}[/tex]

y = 2.54 m

The final speed is computed below:

[tex]vf=9.8\cdot 0.72[/tex]

vf = 7.06 m/s

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

What is the mass of 2.5 mol of Ca, which has a molar mass of 40 g/mol?

Answers

Answer:100 g of ca

Explanation:

Explain why when a firefighter rescues a dog that has fallen through ice on a
lake, they put their ladder on the ice first and then crawl out to the dog on the
ladder.

Answers

Answer:

This can spread their weight.

Explanation:

The ladder has a much larger surface area than the firefighter. Hence, his weight is spread out much more than usual, decreasing the pressure on the ice, preventing the ice from breaking/cracking.

Hope this helped!

The answer is: it can spread their weight

Select all the correct answers.
Which situations describe an elastic collision?
(A) Two glass marbles bounce off each other.
(B) Rodrick flops onto his sofa and sinks into the cushion.
(C) A tossed water balloon flattens when it lands on the grass.
D) A bowling ball knocks over five pins.

Answers

Correct me if I’m wrong, but I believe it’s A and D, though I’m not fully sure.
a and d i’m sure
good luck

Read the elapsed time on the stopwatch and answer the questions.
In which digit is there the least amount of confidence?
How many significant figures does this measurement have? I need help quick

Answers

Answer:

1.7

2.4

Explanation:

Answer:

1.7

2.4

Explanation:

a sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density?

Answers

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

[tex]Ro = m/V[/tex]

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

[tex]V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ][/tex]

And the mass m = 4 [gramm] = 0.004 [kg]

[tex]Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}][/tex]

Which could most likely describe the three surfaces?
Surface 1 is ice, Surface 2 is gravel, and Surface 3
is blacktop.
Surface 1 is gravel, Surface 2 is ice, and Surface 3
is blacktop.
Surface 1 is blacktop, Surface 2 is gravel, and
Surface 3 is ice.
Surface 1 is blacktop, Surface 2 is ice, and Surface
3 is gravel.

Answers

Answer:

Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Hope this helps! :]

Answer:

C. Surface 1 is blacktop, Surface 2 is gravel, and

Surface 3 is ice.

Explanation:

Someone please do this ! ASAP I’ll give brainliest!

Answers

Answer:

The first law, an object will not change its motion unless a force acts on it.

The second law, the force on an object is equal to its mass times its acceleration.

The third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

please help me.
During a football game, one of the players on the home team kicks the football that has a
mass of 0.6 kg so that the ball accelerates toward the opposing team at
23 m/s2. If no other forces act on the ball, how much force did the kicker apply to the
football?

Answers

Answer:

[tex]Force = 13.8N[/tex]

Explanation:

Given

[tex]Mass = 0.6kg[/tex]

[tex]Acceleration = 23m/s^2[/tex]

Required

Determine the applied force

From the question, we understand that no other force acts on the ball.

i.e. the only applied force on the ball is the force applied by the striker.

So, we apply Newton's second law to solve this question.

And this implies that:

[tex]Force = Mass * Acceleration[/tex]

[tex]Force = 0.6kg * 23m/s^2[/tex]

[tex]Force = 13.8N[/tex]

Hence, the applied force by the striker on the ball is 13.8N

Jorge conducted an experiment,and included the graph shown below as part of his lab report.

Jorges experiment involved which if the following?

-A Chemical change.

-A change in the chemical properties of a substance.

-A physical change.

-the formation of a new substance.

Answers

The correct answer is A physical change

Explanation:

Jorge's experiment shows water at different temperatures; in this experiment, it is expected at low temperatures such as -20°C water is in solid-state (ice), at medium temperatures such as 40°C water is in a liquid state (liquid water), and at high temperatures such as 120°C water is in gaseous state (water vapor). This implies during this experiment the changing factor is the physical state (solid, gas, or liquid), and this is a physical change because only the physical properties of water change but not its composition or identity. According to this, the correct answer is physical change.

Question 1) For resonance to occur, the object must experience a frequency equal to its what?


natural frequency


period


wavelength


amplitude

Question 2) What variable affects the natural frequency of an organ pipe?


pipe width


pipe length


pipe area


pipe use

Question 3) Resonance in air columns is used in which of these musical instruments?


violin


trumpet


drums


guitar

Question 4) Which term describes the number of crests that pass a point in a given amount of time?


wave speed


frequency


wavelength


period

Queston 5) Which of the following is the human ear able to detect?


dog whistle


sonar


ultrasound


musical note

Answers

Explanation:

Note that resonance can only occur when the natural frequency is greater than the damping rate, multiplied by the square root of 2. If the damping is too large, then resonance cannot occur.

1. An electric lamp a marked 240V, 6A. What
its resistance when it is operated at the correct
voltage?​

Answers

Answer:

The resistance of the lamp is 4Ω.

Explanation:

You have to apply voltage formula :

V = I × R

R = V ÷ I

R = 240 ÷ 60

R = 4 Ω

Earthquakes are essentially sound waves—called seismic waves—traveling through the earth. Because the earth is solid, it can support both longitudinal and transverse seismic waves. The speed of longitudinal waves, called PP waves, is 8000 m/sm/s. Transverse waves, called SS waves, travel at a slower 4500 m/sm/s. A seismograph records the two waves from a distant earthquake. The SS wave arrives 2.0 minmin after the PP wave. Assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.

Answers

Question

Earthquakes are essentially sound waves—called seismic waves—traveling through the earth. Because the earth is solid, it can support both longitudinal and transverse seismic waves. The speed of longitudinal waves, called P waves, is 8000 m/s Transverse waves, called S waves, travel at a slower 4500 m/s. A seismograph records the two waves from a distant earthquake. The S wave arrives 2.0 min after the PP wave.How far away is the Earthquake.  Assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.

Answer:

The distance is  [tex]d = 1.23 *10^{6} \ m[/tex]

Explanation:

From the question we are told that

   The speed of longitudinal seismic  wave is  [tex]v_p = 8000 \ m/s[/tex]

    The speed of Transverse seismic  wave is  [tex]v_s = 4500 \ m/s[/tex]

     The time difference between the arrival of longitudinal seismic with respect to Transverse waves is  [tex]\Delta t = 2.0\ min = 120\ seconds[/tex]

Generally the  time difference between the arrival of longitudinal seismic with respect to Transverse waves is mathematically represented as

     [tex]\Delta t = t_p - t_s[/tex]

=>  [tex]\Delta t =\frac{d}{v_p} -\frac{d}{v_s}[/tex]

=>   [tex]d = \frac{\Delta t}{ \frac{1}{v_p} - \frac{1}{v_s} }[/tex]

=>   [tex]d = \frac{120 }{ \frac{1}{8000} - \frac{1}{4500} }[/tex]

=>   [tex]d = 1.23 *10^{6} \ m[/tex]

Which force results from charged particles

Answers

Electrical force.
Hope that helped.

Answer:

electromagnetic force

Explanation:

PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!

compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 1 C, and at a distance of 1 m from each other.

Answers

Explanation:

Please mark me as the brainliest answer

Answer:

The electrostatic force is larger by a factor of 8.988E9 / 6.674E-11 = 1.35E20

Explanation:

William gave you all the ingredients, but not the answer.  Being both 1/r^2 forces (resulting from massless mediators in QFT), and with unit charges in the problem (1 coulomb, 1 Kg), separated by unit distance (1m), the only nontrivial numerical values in the problem are the constants of proportionality: Coulomb's constant (k) and Newton's gravitational constant (G).  So to "compare and constrast": the ratio of the forces is simply the ratio of these constants.  The electromagnatic force is 1.35 X 10^20 times stronger than the gravitational force.  Assuming positive charge on both objects (the problem is ambiguous on this), they are repelled, whereas the much weaker gravitational force is attractive.  (Gravity only has one kind of "charge" - it's unsigned - and the force is always attractive).  

The problem doesn't say the objects are pointlike: if they have some extent and are either conductive or made of some dielectric, then things get messy because the charge distribution on the object won't be uniform then, but save that for grad school. :-)

How much net force is needed to accelerate a 200 kg satellite 9.8 m/s2 ?

Answers

Answer:

1960 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 200 × 9.8

We have the final answer as

1960 N

Hope this helps you


can somone pls help me??!! i’m very stuck

Answers

Answer:

b

Explanation:

What will happen if the atom rearrange?
1. a new substance will form
2. start to boil
3. nothing will happen
choose the correct answer ​

Answers

Answer:

In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange amd form new bonds to make the products.

HELP
which two changes to a metal wire both increases resistance? the answer is B but why ?

Answers

Answer:

option C decreasing its thickness and increasing its temperature.

Explanation:

Resistance is directly proportional to length and temperature of the wire and inversely to area.

if you increase the temperature the resistance will increase.(resistance is directly proportional to temperature)

if you decrease its thickness (area) then the resistance will increase ( resistance is inversely proportional to area)

hope it helps:)

What is the difference
between interplanting and intercropping?

Answers

Answer:

Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.

Intercropping refers to large, often commercial, farming operations and is the procedure of planting alternating crop rows. Growing two or more crops in close proximity is known as intercropping.

What is interplanting?

Starting to grow two or more crops in close vicinity is known as intercropping.

The most common goal of intercropping is to increase yield on a given plot of land by utilizing resources that would otherwise go unused by a single crop.

Intercropping is the cultivation of more than one crop in the same space at the same time. Intercropping is the simultaneous planting of two or more species in a mixture, or the interplanting of one species during the growth of another.

The terms are frequently and regionally used interchangeably. Intercropping is the practice of planting alternating crop rows in large, often advertising, farming operations.

Thus, this is the difference between intercropping and interplanting.

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A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?

Answers

Answer:

The range of the projectile is 60 meters

Explanation:

To determine the range/distance of the projectile, the formula for velocity is used;

velocity = distance/time

where velocity is 15 m/s

time is 4 seconds

distance is unknown

From the formula above, distance is made the subject and thus

distance = velocity × time

distance = 15 × 4

distance = 60 m

The range of the projectile is 60 meters

A tortoise can run with a speed of 0.10 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm)!How long does the race take?What is the length of the race?

Answers

Answer: a. 126.21secs

b. 12.621 meters.

Explanation:

Given data:

Speed of tortoise = 0.1m/s.

Speed of hare = 2m/s.

Solution:

a. Distance traveled = Speed* Time

Speed of tortoise = 0.1 m/s

Speed of hare = 20*0.1 m/s = 2 m/s

2 minutes = 2* 60 s = 120 s

Let the time taken for the race be t seconds.

• Distance moved by tortoise

= (0.1 /s)* (t s)

= 0.1*t meter

•Hare has run for a time of (t - 120)s.

distance moved by hare

= Speed * Time

= (2 m/s)*(t- 120)s

= (2t - 240) meter.

Since hare is 20 cm (0.2 m) behind the tortoise, therefore

(0.1*t - 0.2) meter

= (2t - 240) meter

0.1*t - 0.2 = 2t - 240

Collect like terms

239.8 = 1.9t

Divide both sides by 1.9

t = 126.21secs

The race lasted for 126.21secs

b. Length of race

= Distance moved by tortoise

= 0.1*126.21 meter

= 12.621 meter

The length of the race is 12.621 meters.

Which two elements have similar properties and 8 electrons in their outermost shells?
A. Chlorine and bromine
B. Calcium and strontium
C. Nitrogen and phosphorus
D. Argon and krypton
** THE ANSWER IS D **

Answers

Answer:

D

Explanation:

their octet is complete they don't react with any thing

Argon and krypton have similar properties, and they have a complete octet, which means that they do not react with anyone. Hence, option D is correct.

What are inert gases?

The phrase "inert gas" is a bit misleading because, in some circumstances, these gases can really be reactive. As a result, these gases are typically referred to as noble gases in the context of chemistry and materials science.

The term "noble" has historically been used in chemistry (and earlier in alchemy) to characterize the resistance of some metals to chemical reaction, and the term "noble gas" denotes the same resistance.

Helium (He)Neon (Ne)Argon (Ar)Krypton (Kr)Xenon (Xe)Radon (Rn)

These are the inert gases which are mentioned in the periodic table.

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A space vehicle is coasting at a constant velocity of 22.4 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.206 m/s2 in the x direction. After 40.5 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.

Answers

Answer:

a. 23.9 m/s b. 69.58°

Explanation:

a. Since the space vehicle is moving at a constant velocity of 22.4 m/s in the y direction relative to the space station, its initial vertical velocity v = 22.4 m/s.

Also, since the space vehicle moves in the y direction, its initial horizontal velocity u = 0 m/s.

Since its acceleration a = 0.206 m/s² for time, t = 40.5 s, we find the final horizontal velocity u' from u' = u + at.

Substituting the values of the variables into the equation, we have

u' = u + at

u' = 0 m/s  + 0.206 m/s² × 40.5 s

= 8.343 m/s

≅ 8.34 m/s

The resultant velocity relative to the space station V = √(v² + u'²)

= √[(22.4m/s)² + (8.34 m/s)²]

= √[501.76 m²/s² + (69.56 m²/s²]

= √[571.32 m²/s²]

= 23.9 m/s

b. The direction of the vehicle's velocity relative to the space station is thus θ = tan⁻¹(v/u')

= tan⁻¹(22.4 m/s/8.34 m/s)

= tan⁻¹(2.6959)

= 69.58°

A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surface would the letters appear to be when viewed from directly above?

Answers

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest milliliter(mL). Which statement describes a change that could help improve the results of his experiment

Answers

The question is incomplete, the complete question is;

A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest millilitre (mL). Which statement describes a change that can help improve the results of his experiment?

A.) His measurements will be more precise if he takes measurements from an additional 100 flowers. B.) His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL. C.) His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL. D.) His measurements will be more accurate if he takes measurements from an additional 100 flowers.

Answer:

His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL.

Explanation:

In the measurements of volume using most graduated cylinders, the cylinders are calibrated to the nearest tenth owing to the uncertainty in the measurement of volume.

Hence if a cylinder has measures to the nearest milliliter(mL), then he can improve his experiment by using a graduated cylinder that measures to the nearest tenth of a mL

PLZ HELP!!! the moon umbriel orbits uranus (mass = 8.68 x 10^25 kg) at a distance of 2.66 x 10^8 m. what is umbriels orbital speed? (In hours)

Answers

Answer:

99.48

Explanation:

99.48

Suppose a 125 N force is applied to a lawnmower handle at an angle of 35° with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done? (hint: use cos to find the x of force vector)

Answers

Answer:

Workdone is 5734.06Nm.

Explanation:

Given the following data;

Force applied = 125N

Angle = 35°

Distance = 56m

To find the workdone by the lawnmower, we would first of all find the horizontal component of the force applied.

[tex] Horizontal force, Fx = mgCosd[/tex]

Where;

Fx represents the horizontal force. m is the mass of an object. g is the acceleration due to gravity. d is the angle of inclination (theta).

mg = weight = 125N

Substituting into the equation, we have;

[tex] Fx = 125 * Cos35[/tex]

[tex] Fx = 125 * 0.8192[/tex]

Fx = 102.39N

Workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

[tex] Workdone = 102.39 * 56[/tex]

Workdone = 5734.06Nm

Therefore, the work done by the lawnmower is 5734.06Nm.

During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m high. When the ball reaches the net,
(a) does the ball clear it and
(b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net,
(c) does the ball clear it and
(d) what now is the distance between the center of the ball and the top of the net?

Answers

Answer:

a

Yes it clears

b

 [tex]b= 0.19 \ m[/tex]

c

 No it does not clear

d

[tex]z= 0.86 \ m[/tex]

Explanation:

From the question we are told that

  The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]

  The height of the ball above the ground is  [tex]h = 2.3 7 \ m[/tex]

  The distance of the net is  [tex]d = 12 \ m[/tex]

   The height of the net is [tex]H = 0.9 \ m[/tex]

Generally the time taken for the ball to reach the net is mathematically represented as

     [tex]t = \frac{d}{v}[/tex]

=>  [tex]t = \frac{12}{23.6}[/tex]

=>  [tex]t = 0.508 \ s[/tex]

Generally the change in height of the ball after t is mathematically represented as

     [tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]

Here u is the initial velocity which is zero given that the ball was at rest initially

So

     [tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]

=>  [tex]\Delta h =1.28 \ m[/tex]

Generally the new height of the ball is mathematically evaluated as

      [tex]s= h-\Delta h[/tex]

=>   [tex]s = 2.37 - 1.28[/tex]

=>   [tex]s = 1.09 \ m[/tex]

From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]b = s - H[/tex]

=>   [tex]b = 1.09 - 0.90[/tex]

=>   [tex]b= 0.19 \ m[/tex]

Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal  , the velocity along the x-axis is  

      [tex]v_x = v cos(5)[/tex]

=>   [tex]v_x = 23.6 cos(5)[/tex]

=>   [tex]v_x = 23.5 \ m/s[/tex]

The velocity along the y-axis is  

      [tex]v_y = v sin(5)[/tex]

=>   [tex]v_y = 23.6 sin(5)[/tex]

=>   [tex]v_y = 2.06 \ m/s[/tex]      

Generally the time taken for the ball to reach the net is

      [tex]t = \frac{d}{v_x}[/tex]

=>   [tex]t = \frac{12}{23.5}[/tex]

=>   [tex]t =0.508 \ s[/tex]

Generally the change in height of the ball after t seconds is  

     [tex]c = v_yt + \frac{1}{2}gt^2[/tex]

=>  [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]

=>  [tex]c = 2.33[/tex]

Generally the new height of the ball after time t seconds is  

     [tex]e = h - c[/tex]

=>   [tex]e = 2.37 - 2.33[/tex]

=>   [tex]e = 0.04 \ m[/tex]

From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]z = H-e[/tex]

=>   [tex]z = 0.90 - 0.04[/tex]

=>   [tex]z= 0.86 \ m[/tex]

   

(a) Yes, the ball clears the net.

(b) The distance between the center of the ball and the top of the net is 0.203 m.

(c) No, the ball does not clear the net.

(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.

What is a Projectile motion?

When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.

Given,

The horizontal distance traveled by the ball is 12 m.

The height of the top of the net is 0.90 m.

The height of the horizontal launch of the ball is 2.37 m.

The time for the horizontal motion of the projectile that is the ball is,

[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]

The equation for the vertical motion of the projectile can be solved by substituting the above result.

[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]

Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.

The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]

The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]

For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,

[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]

Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,

[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]

The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.

[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]

Thus, the distance between the center of the ball and the top of the net is -2,32 m.

When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.

Learn more about projectile motion here:

https://brainly.com/question/11049671

can y’all please help me with this 3 part question?

Answers

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

[tex]v_{f} =v_{o} +g*t[/tex]

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

[tex]v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)[/tex]

44100 = 20*y

y = 2205 [m]

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