A baseball is thrown straight downward with an initial speed of 40 ft=s from the top of the Washington Monu- ment (555 ft high). How long does it take to reach the ground, and with what speed does the baseball strike the ground?

Answer:

B

Explanation:

Vectors have both magnitude and direction.

A scalar quantity is something that has magnitude only.

Explanation:

There are two types of physical quantities i.e. vector quantities and scalar quantities.

The type of quantities that have both magnitude as well as direction are called vectors. While, the quantities having only magnitude are called scalars.

For example, if we say the ball is moving with a speed of 5 m/s towards north. It shows both direction and magnitude. Here, 5 m/s shows magnitude of speed and North shows the direction.

Hence, the correct option is (b) "magnitude and direction".

Answer:

11 kg of gas are emmited to the atmosphere

Explanation:

The problem consists in calculating the total mass of the gases produced in the combustion reaction of the carcoal.

In this case we consider the simple combustion reaction:

C+O2 -> CO2

In relation 1:1 molar with the oxygen.

In this case, the limitant reactant is the carbon, so we make the calculations based in the 3kg of carbon combusted.

First we convert the mass of carbon into mol of carbon

m,C=3kg=3000 g of C

Then we have the mol

n,C=3000g/(12 g/mol)=249.77 mol C

This number also corresponds to the number of oxygen mol reacting with the charcoal. Therefore we calculate the mass of oxygen.

m,O2=249.77 mol *(32 g/mol) = 7990 g =7.99 kg, aproximately 8 kg

Then the total mass of gas is the mass of charcoal plus the mass of oxygen

m,total=3kg + 8kg=11kg

Answer:

[tex]v_f=8.4m/s[/tex]    

Explanation:

We need to use the conservation of momentum Law, the total momentum is the same before and after the kid leaves the skate.

In the axis X:

[tex](m_{kid}+m_{skate})*v_{o}=m_{kid}*v_f[/tex]     (1)

[tex]v_f=(m_{kid}+m_{skate})*v_{o}/m_{kid}=(44+2.2)*8/44=8.4m/s[/tex]  

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

Explanation:

It is given that,

Mass of the aircraft 1, m₁ = 1400 kg

Mass of aircraft 2, m₂ = 1500 kg

Initially, the aircraft 2 is at rest, u₂ = 0

Initial speed of the aircraft 1, u₁ = 40 m/s

After the collision, the total speed of the system, V = 19.3 m/s

Time, t = 8.2 s

Since, two objects stick together it is a case of inelastic collision. The acceleration of the system is given by :

[tex]a=\dfrac{V}{t}[/tex]

[tex]a=\dfrac{19.3\ m/s}{8.2\ s}[/tex]

[tex]a=2.35\ m/s^2[/tex]

Distance covered by the system before stopping is given by :

[tex]x=\dfrac{1}{2}\times a\times t^2[/tex]

[tex]x=\dfrac{1}{2}\times 2.35\times (8.2)^2[/tex]

x = 79.007 meters

Hence, this is the required solution.

Answer:

[tex]V_B-V_A=-20736-0=-20736volt[/tex]

Explanation:

We have given charge on the particle [tex]q=-4\mu C=-4\times 10^{-6}C[/tex]

Mass of the charge particle [tex]m=3.2\times 10^{-6}kg[/tex]

From energy of conservation kinetic energy will be equal to potential energy

So at point A

[tex]\frac{1}{2}mv^2=qV[/tex]

At point a velocity is zero

So [tex]\frac{1}{2}(3.2\times10^{-6} )0^2=-4\times 10^{-6}V_a[/tex]

[tex]V_A=0volt[/tex]

At point B velocity will be 72 m/sec

So [tex]\frac{1}{2}\times 3.2\times 10^{-6}72^2=-4\times 10^{-6}V_b[/tex]

[tex]V_B=-20736volt[/tex]

So [tex]V_B-V_A=-20736-0=-20736volt[/tex]

Final answer:

The potential difference VB - VA between points A and B is -648 Volts, indicating that point A is at a higher electric potential than point B.

Explanation:

To find the potential difference VB - VA between points A and B, we use energy conservation. The work done by the electric field on the particle is equal to the change in kinetic energy of the particle. Since the particle starts from rest, its initial kinetic energy is 0, and its final kinetic energy is given by ½ mv2.

Therefore, the work done, which is equal to the potential energy change, is Work = ½ mv2 = q(VB - VA), where q is the charge of the particle. We can rearrange this to find the potential difference, VB - VA = ½ mv2/q. By plugging in m = 3.2 x 10-6 kg, v = 72 m/s, and q = -4.0 μC, we can calculate the potential difference.

First, convert the charge from microcoulombs to coulombs: -4.0 μC = -4.0 x 10-6 C. Then, plug the values into the formula: VB - VA = (0.5 * 3.2 x 10-6 kg * (72 m/s)2) / (-4.0 x 10-6 C). After calculating this expression, we get VB - VA = -648 Volts, which means VA is higher than VB by 648 Volts.

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation    x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find  x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

The acceleration of the object is the change in the velocity of the object within given time interval.

The acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.

Given that the position of an object is x = at3 - bt2 + ct, where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s.

The position of the object at time t = 4.1 s is calculated as given below.

[tex]x = at^3-bt^2+ct\\[/tex]

[tex]x = 4.1\times (4.1)^3 - 2.2 \times (4.1)^2 + 1.7\times 4.1[/tex]

[tex]x = 252.56\;\rm m[/tex]

The velocity v of the object at time t = 4.1 s is given below.

[tex]v = \dfrac{x}{t}[/tex]

[tex]v = \dfrac {252.56}{4.1}[/tex]

[tex]v = 62.67\;\rm m/s[/tex]

The acceleration of the object at time t =4.1 s is given below.

[tex]a = \dfrac {v}{t}[/tex]

[tex]a = \dfrac {62.67}{4.1}[/tex]

[tex]a = 15.26\;\rm m/s^2[/tex]

Hence we can conclude that the acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.

To know more about the acceleration, follow the link given below.

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Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

    fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

    fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between  the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.  

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

      μ kinetic <μ static

  In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

Answer:

33.4°C .

Explanation:

mass of bullet, m = 15 g = 0.015 kg

velocity of bullet, v = 1502 m/s

mass of wax, M = 2.5 kg

Initial temperature of wax, T1 = 31°C

Let T2 be the final temperature of wax.

Specific heat of wax, c = 0.7 cal/g°C = 0.7 x 1000 x 4 J/kg°C = 2800 J/kg°C

The kinetic energy of the bullet is converted into heat energy which is used to heat the wax.

[tex]\frac{1}{2}mv^{2}= M \times c \times \left ( T_{2}-T_{1} \right )[/tex]

[tex]0.5\times 0.015\times 1502 \times 1502 = 2.5 \times 2800 \times\left ( T_{2}-31 \right )[/tex]

[tex]2.42 =\left ( T_{2}-31 \right )[/tex]

[tex]T_{2}=33.4^{o}C[/tex]

thus, the final temperature of wax is 33.4°C .

Answer:

Part a)

[tex]\omega = 5 ft/s[/tex] counterclockwise

Part b)

[tex]\omega = 2.5 ft/s[/tex] counterclockwise

Part c)

[tex]\omega = 2.5 ft/s[/tex] clockwise

Explanation:

Let the cart has radius R = 1 ft

so here we have speed of the center of the cart is

[tex]v_c = 2.5 ft/s[/tex]

let the angular speed is given as

[tex]\omega[/tex] counter clockwise

Part a)

if the top most point of the rim has same speed as that of speed of the center but it is towards left

so we have

[tex]v = v_c + r\omega[/tex]

[tex]-2.5 = 2.5 + 1(\omega)[/tex]

so we have

[tex]\omega = -5 ft/s[/tex]

Part b)

if the speed of the top point on the rim is zero

[tex]v = v_c + 1(\omega)[/tex]

[tex]0 = 2.5 + \omega[/tex]

[tex]\omega = -2.5 ft/s[/tex]

Part c)

if the speed at the top position on the rim is 5 ft/s

[tex]5 ft/s = 2.5 ft/s + 1(\omega)[/tex]

[tex]\omega = 2.5 ft/s[/tex]

Explanation:

It is given that,

Position of the squirrel, [tex]y_o=24\ ft[/tex]

Initial speed of the squirrel, u = 8 ft/s

To find,

If the squirrel climbs down the tree in 2 sec, does it reach the ground before the nut

The position of squirrel as a function of time t is given by :

[tex]h(t)=-16t^2+8t+24[/tex]

Position at t = 2 seconds will be :

[tex]h(2)=-16(2)^2+8(2)+24[/tex]

h(2) = -24 m

At t = 2 s, the nut will hit the ground first than the squirrel.

The squirrel reaches the ground before the nut.

To determine whether the squirrel reaches the ground before the nut, we need to compare the time it takes for each to reach the ground. The equation for the height of the squirrel is given as h(t) = -16t^2 + 8t + 24. The squirrel climbs down the tree in 2 seconds, so we can plug in t = 2 into the equation to find the height of the squirrel at that time. h(2) = -64 + 16 + 24 = -24 ft.

Now, we need to find the time it takes for the nut to reach the ground. The height of the nut, h(t), is given by the same equation. We need to find the time when h(t) = 0. Setting the equation equal to zero and solving for t, we get -16t^2 + 8t + 24 = 0. Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), we get t ≈ 2.73 seconds or t ≈ -0.48 seconds. Since time cannot be negative, we disregard the negative solution. Therefore, the nut takes approximately 2.73 seconds to reach the ground.

Comparing the times, we can see that the squirrel reaches the ground before the nut. The squirrel takes 2 seconds to climb down the tree, while the nut takes approximately 2.73 seconds to fall. Therefore, the squirrel reaches the ground before the nut.

Answer:

Explanation:

(a)

Gold has an atomic mass of 197 g/mol means 197 g of gold contains 6.023×10^{-23} atoms.

therefore number of atoms in 17.4 g is

[tex]n= \frac{6.23\times10^{23}\times17.4}{197}[/tex]

[tex]5.32\times10^{22}[/tex]

Number of protons in each atom is 79.

therefore Total number of protons is 5.32×10^{22}×79 =4.2×10^24. protons

Their total positive charge is 4.2×10^{24}×1.6×10^{-19) =6.72×10^5 C.

(b)

equal number of protons should be there to neutralize the material therefore number of electrons is 4.2×10^(24.)

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

Answer:

d = .076 m

Explanation:

The time for frog A can be calculated from  equation of motion

[tex]v_f = v_o + at[/tex]

where v_f is final velocity, a is acceleration due to gravity

so from given data we have

[tex]-0.551= 0.551 + (-9.8)(t)[/tex]

t = 0.112 sec

Now we will use that time for frog B

[tex]v_f = v_o + at[/tex]

[tex]v_f = 1.23 + (-9.8)(0.112)[/tex]

[tex]v_f = 0.128 m/s [/tex](Note its positive)

For the displacement

[tex]s = v_o t + 0.5at^2[/tex]

[tex]s = (1.23)(0.112) + (.5)(-9.8)(0.112)^2[/tex]

d = .076 m

Answer: false, they hit the ground at the same time.

Explanation:

Answer:

The number of electrons an atom has relative to a neutral atom of the same element

Explanation:

the oxidation state is a number that indicates the ability to give or receive electrons. This number usually has a sign, which can be positive or negative.

If the oxidation state is a positive number, it means that this atom has the ability to receive such an amount of electrons. If the oxidation state is negative, the element can give that amount of electrons.

Answer:

Gravity is field force

Explanation:

because a gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body. Thus, a gravitational field is used to explain gravitational phenomena, and is measured in newtons per kilogram

Answer:

field force

Explanation:

Answer:

mouse speed is 5 m/s

Explanation:

given data

belt roll b = 3 m/s

head straight h = 4 m/s

to find out

mouse speed s

solution

we will apply here pythagorean theorem

that is

s = [tex]\sqrt{b^{2} +h^{2} }[/tex]   ...........................1

put here value in equation 1 as b = 3 and h = 4

so

s = [tex]\sqrt{b^{2} +h^{2} }[/tex]

s = [tex]\sqrt{3^{2} +4^{2} }[/tex]

s = [tex]\sqrt{9 + 16 }[/tex]

s = [tex]\sqrt{25 }[/tex]

s = 5

so mouse speed is 5 m/s

The mouse's speed relative to the factory floor is calculated using the Pythagorean theorem, taking into account both the speed of the mouse and the conveyor belt. The correct answer is 5 m/s, option E.

The speed of the mouse relative to the factory floor is calculated using the Pythagorean theorem given that the mouse's motion and the conveyor belt's motion form two sides of a right triangle. In this case, the speed of the mouse is 4 m/s and the speed of the conveyor belt is 3 m/s. The relative speed will therefore be the square root of (mousespeed)² + (conveyorbelt speed)². So, √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. Therefore, the mouse's speed relative to the factory floor is 5 m/s, option E.

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Answer:

46 meters

Explanation:

In the figure you can see law of cosines.

In your case

α = 22.5° + 42 ° = 64.5 °

b = 15 m (distance between your's tent and Joe's tent)

c = 42 m (distance between your's tent and Karl's tent)

a is the distance between Joe's tent and Karl's tent

Replacing in the first equation:

[tex]a^2 = 15^2 m^2 + 42^2 m^2 - 2 \times 15 \times 42 \times cos(64.5) [/tex]

[tex] a = \sqrt{2111.46 m^2} [/tex]

[tex] a = 46 m [/tex]

True, we see an emission spectrum from a neon sign due to the neon gas emitting light in the red-orange spectrum when electrically excited, with other gases emitting different signature colors.

The statement that one sees an emission spectrum from a neon sign is indeed true. A neon sign emits light because of the emission spectrum of neon gas inside the tube. When an electrical discharge excites neon atoms to a higher energy state, they emit light upon returning to their ground state. The characteristic red color of neon signs is due to the emission spectrum, with the most intense emission lines at 589 nm.

Signs that shine in colors other than red-orange contain different gases or mixtures of gases, which produce different emission spectra responsible for their signature colors. For instance, mercury vapor emits light with emission lines below 450 nm, resulting in a blue light, and sodium vapor emits light at 589 nm, creating an intense yellow light.

1)

Answer:

[tex]a = 18.68 m/s^2[/tex]

Part b)

[tex]a = 1.9 g[/tex]

Explanation:

Rate of the spinning of the dancer is given as

[tex]f = 2.6 rev/s[/tex]

angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(2.6) = 16.33 rad/s[/tex]

distance of the ear is given as

[tex]r = 7 cm = 0.07 m[/tex]

Part a)

Radial acceleration is given as

[tex]a = \omega^2 r[/tex]

[tex]a = (16.33)^2(0.07)[/tex]

[tex]a = 18.68 m/s^2[/tex]

Part b)

also we know that

[tex]g = 9.81 m/s[/tex]

so now we have

[tex]\frac{a}{g} = \frac{18.68}{9.81}[/tex]

[tex]a = 1.9 g[/tex]

2)

Answer:

Part a)

[tex]v = 226.2 m/s[/tex]

Part b)

[tex]a = 1.304 \times 10^3 g[/tex]

Explanation:

Length of the blades = 4.00 m

frequency of the blades = 540 rev/min

[tex]f = 540 \times \frac{1}{60} = 9 rev/s[/tex]

so angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(9) = 56.5 rad/s[/tex]

Part a)

Linear speed of the tip of the blade is given as

[tex]v = r\omega[/tex]

[tex]v = (4.00)(56.5)[/tex]

[tex]v = 226.2 m/s[/tex]

Part b)

Radial acceleration of the tip of the blade

[tex]a = \frac{v^2}{r}[/tex]

[tex]a = \frac{226.2^2}{4}[/tex]

[tex]a = 1.28 \times 10^4 m/s^2[/tex]

also we know

[tex]\frac{a}{g} = \frac{1.28 \times 10^4}{9.81}[/tex]

[tex]a = 1.304 \times 10^3 g[/tex]

Answer:

c)7 mph

Explanation:

Going downstream the speed of the boat is added to the speed of the water, on the way back it is opposite to it, then the speed of the water is substracted to that of the boat, and we have the total time of the trip, then we can write down a system of equations as follows:

x:=Speed of water that is being asked

t1:= time going downstream

t2:= time going upstream

(21+x)*t1=14  (1)

(21-x)*t2=14  (2)

t1+t2=1.5       (3)

then t1=1.5-t2, replacing this in (1) we have  (21+x)*(1.5-t2)=14, then t2=1.5-(14/(21+x))  (4). Doing the same in (2) we get t2=14/(21-x)  (5), this way we can use (4) and (5) and get that:

1.5-(14/(21+x))=14/(21-x), doing the operations we get that 588/(441-x^2)=1.5 or equivalently 1.5x^2-73.5=0 or x^2-49=0 or (x-7)(x+7)=0, then the possible answers are x=7 or x= -7, here we consider the positive answer, that in magnitude is the same as the other, because we already considered the direction of the water into the equations. Then the answer is c) 7mph.

Final answer:

The speed of the current is 3 miles per hour, obtained by setting up an equation based on the total time taken for an upstream and downstream trip and solving for the current's speed. So the correct option is a.

Explanation:

To find the speed of the current, we can set up an equation considering the speed of the boat in still water (which we know is 21 miles per hour) and the speed of the current, which we will call 'c'.

When the boat goes upstream, the effective speed is reduced by the speed of the current, so it's 21 - c miles per hour. Similarly, when the boat goes downstream, the effective speed is increased by the speed of the current, making it 21 + c miles per hour.

Let's calculate the time taken to travel upstream and downstream using the distance and the effective speed. For upstream, it is 14 / (21 - c) hours, and for downstream, it is 14 / (21 + c) hours. The total time for the trip is given as 1.5 hours, so we sum the upstream and downstream times to set up the equation:

14 / (21 - c) + 14 / (21 + c) = 1.5

By solving this equation for 'c', we find that the speed of the current is 3 miles per hour, which corresponds to option (a).

Answer:

Explanation:

a ) The velocity will never be zero . The velocity will be minimum at the highest point of projectile,  which will be equal to the horizontal component of the initial velocity.

b ) The velocity will be minimum when its kinetic energy will be minimum . Kinetic energy will be minimum when its potential energy will be maximum.

Its potential energy will be maximum at the highest point so velocity will be minimum at the highest point.

c ) Velocity will never be the same as initial velocity because constant force of gravitation is acting on the projectile all the time.

d ) At the moment when the projectile returns back and hits the ground, the speed becomes equal to the initial speed ( at t = 0 ) because its kinetic energy becomes the same as initial energy , the height becoming zero.  

In projectile motion on level ground, the velocity is never zero. The velocity is minimum at the apex of the trajectory and maximum at the launch and impact points. The velocity can never be the same as the initial velocity at a time other than t=0, but the speed can be the same as the initial speed at a time other than t=0 when the projectile lands.

a) In projectile motion on level ground with negligible air resistance, the velocity is never zero. The horizontal velocity remains constant throughout the motion, while the vertical velocity changes. However, the total velocity (magnitude of the velocity vector) is always non-zero.

b) The velocity is minimum when the projectile reaches its highest point, called the apex of the trajectory. The velocity is maximum at both the launch and impact points.

c) No, the velocity can never be the same as the initial velocity at a time other than t = 0. This is because the horizontal velocity remains constant during the motion, while the vertical velocity changes.

d) Yes, the speed (magnitude of the velocity vector) can be the same as the initial speed at a time other than t = 0. This occurs when the projectile lands, as the horizontal velocity is constant and the vertical velocity becomes opposite in direction but equal in magnitude to the initial vertical velocity.

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Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

Final answer:

Negative intrapleural pressure is caused by the natural elasticity of the lungs contracting and the thoracic wall expanding, with transpulmonary pressure determining lung size.

Explanation:

The pressure that results from the natural tendency of the lungs to decrease their size, due to elasticity, and the opposing tendency of the thoracic wall to pull outward and enlarge the lungs, is known as negative intrapleural pressure. This pressure is a result of two main forces: the inward pull due to the elastic recoil of the lung tissue and the surface tension of the alveolar fluid, and the outward pull from the pleural fluid and thoracic wall. The balance of these forces creates a negative pressure within the pleural cavity, which is crucial for the proper function of the lungs during breathing. The transpulmonary pressure, which is the difference between the intrapleural and the intra-alveolar pressures, determines the size of the lungs during the respiratory cycle.

Answer: Ok, if the position of the bicycle is negative, then you can think is in the -x range, lets call it -r.

And the velocity is positive, you know that the movement equation of something is x= v*t + x0

where v is te velocity, x0 the position and t the time.

so in our case x = v*t  - r.

so when v*t = r, for some time, the bicycle will be on your position.

So the bicycle was getting closer to you.

Answer:

y = 297.34 m

Explanation:

When coin is dropped from the balloon then due to inertia the initial speed of the coin will be same as the speed of the balloon

So here we have

[tex]v_i = 12.0 m/s[/tex]

initial height of the coin is

[tex]y_i = 290 m[/tex]

now at the maximum height final speed of the coin will be zero

so we will have

[tex]v_f^2 - v_i^2 = 2 a (\Delta y)[/tex]

[tex]0 - 12^2 = 2(-9.81)(y - 290)[/tex]

[tex]7.34 + 290 = y[/tex]

[tex]y = 297.34 m[/tex]

The maximum height reached by the coin, initially rising at 12.0 m/s from 290 m, is calculated by kinematic to be 297.35 m.

Initial velocity (u) of the coin is 12.0 m/s upward.

Initial height (h) is 290 m above the ground.

Acceleration due to gravity (g) is -9.8 m/s² (negative because it acts downward).

Calculate the maximum height:

Using the kinematic equation: v² = u² + 2a(s)

Where v = final velocity (0 m/s at the maximum height), u = initial velocity (12.0 m/s), a = acceleration (-9.8 m/s²), and s = displacement.

Set v = 0 and solve for s: 0 = (12.0 m/s)² + 2(-9.8 m/s²)s

0 = 144 - 19.6s

19.6s = 144

s = 144 / 19.6

s = 7.35 m

This displacement (s) is the distance the coin moves upward from the initial point.

Therefore, the maximum height above the ground is 290 m + 7.35 m = 297.35 m

Maximum height reached by the coin is 297.35 m.

Answer:

44257m

Explanation:

first we convert the initial velocity to m / s

Vo=initial velocity=18000Mph*(0.44m/s)/1mph=8046.7m/s

As the meteorite lost 90% of its speed, it means that it is moving with 10% of the initial speed

Vf=final velocity=(0.1)(8046.7)=804.67m/s

As the meteorite has a uniformly accelerated movement we can use the following equation

A=aceleration=(Vf-Vi)/t

A=(804.67-8046.7)/10=-724.2m/S^2

finally to calculate the displacement we use the following equation

(Vf^2-Vo^2)/2A=X=displacement

X=(804.67^2-8046.7^2)/(2*-724.2)=44257m

Final answer:

The final temperature of the gas, after expanding and being heated until the pressure returns to the initial pressure, is 3351°C.

Explanation:

The question involves thermodynamics and the behavior of an ideal gas when it expands and is then heated to reach the initial pressure. Initially, the gas occupies one-third of the tank's volume at a temperature of 935°C. When the partition is removed, the gas expands to fill the entire tank, which is three times its initial volume. Assuming an ideal gas and the process to be isobaric (constant pressure) during heating, we apply the ideal gas law and the concept of absolute temperature to determine the final temperature.

Initially, the gas temperature is 935°C, which is 1208K (since absolute temperature in Kelvin = temperature in Celsius + 273). Upon expansion, the volume triples without specifying pressure change, but we'll assume adiabatic free expansion for this initial step, meaning temperature remains unchanged. The gas is then heated to return to its initial pressure. Since the volume tripled and pressure returns to initial, using the relation P1V1/T1 = P2V2/T2, where P1=P2 (initial and final pressures are equal), V2=3V1 (final volume is three times the initial volume), and T1=1208K, we find that the final temperature (T2) can be obtained by rearranging the equation to T2 = T1(V2/V1). Thus, the final temperature is 3 times the initial temperature in Kelvin, or 3624K, which is 3351°C.

The hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."

You may devise an experiment to gauge the temperature of the air being blasted by the air conditioner both at night and during the day in order to verify this theory. This is a general description of the experiment.

Place a thermometer or temperature sensor close to the air conditioner's outlet, which is where the room's cool air is blown. Make sure the sensor is in the same spot for the measurements during the day and at night.

The exact relationship between the air temperature blasted by the air conditioner and the presence or absence of direct sunshine is what the experiment attempts to investigate. According to the theory, cooler air may be blasted by the air conditioner during the daytime when there is sunlight but warmer air during the night when there is no direct sunlight.

Hence, the hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."

To learn more about the hypothesis, here:

https://brainly.com/question/35154833

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