What is the change in air pressure over a given distance called?.

When two people each exert a force of 300N, pulling a car by a separate ropes in the east direction. the first person pulls at an angle of 20° N of E and the second person pulls at an angle of 20° S of E. then the work done on the car by each worker is 1578 J if the the car moves 0.5 m/s for 5.6s.

Given,

x component of Force F₁  = Fcos20° = 300cos20° = 281.9 N

y component of Force F₂  = Fcos20° = 300cos20° = 281.9 N

The actual force acting on the car is,

F₁(x) + F₂(x) = 281.9 N + 281.9 N = 563.8 N.

The distance travelled by the car,

d = v×t = = 0.5 × 5.6 = 2.8 m

The work W is force times distance

W = F.s = 563.8 N.× 2.8 m = 1578 J

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Answer: the answer to the question is B

Because of the fact that it keeps switching, from kinetic to potential, and again from potential to kinetic, as a result, while it is moving around in motion, it will lose energy.

:D

Best of Luck!

Answer:

Design 2

Explanation:

I had this same question

My answer:

"Design 2 is the well-designed one, because the air molecules are the most compact and could protect the individual better than 1,3,and no air bag."

Answer:

18 V

Explanation:

According to Ohm's Law, current equals voltage divided by resistance. (I=V/R)

Using this, we can rearrange the equation to have it solved for voltage.

V=IR

Then, you plug in the values for the current and the resistance.

V=(3.6)x(5.0)

V=18

Answer:

Explanation:

initial velocity u = 54 km/h = 15 m/s

final velocity v = 18 km/h = 5 m/s

distance s = 10 m

1. v^2 = u^2 + 2as

5^2 = 15^2 + 2a × 10

25 = 225 + 20a

25 - 225 = 20a

20 a = -200

a = -200/2

a = -100m/s^2

∴Deacceleration = -100m/s^2

2. v = u + at

5 = 15 -100t

5-15 = -100t

-10 = -100t

t = 100 / 10

∴t = 10 sec

Total distance covered by the car 10 m

To me, the hardest part of this whole thing is keeping the units straight.  We're starting out with information given to us in kilometers, hours, and meters, and we have to come up with answers in m/s² , seconds, and meters.

When I worked this problem, I jumped right in without thinking, and I immediately got bogged down when I had to go off to the side and convert some units.

Now I know better.  THIS time, before we get all tangled up trying to solve anything, let's get clever and change everything to m/s right now !

(54 km/hour) · (1,000m/km) · (1 hour/3600 sec)  =  15 meters/second

(18 km/hour) · (1,000 m/km) · (1 hour/3600 sec) = 5 meters/second

NOW I'll betcha it's gonna be about 70% faster and easier !

i). Acceleration = (change in speed / time for the change)

We know the distance, but not the time.  I know there's a formula for it, but I've learned so many formulas during my lifetime that I can't remember ALL of them.  So I just memorize some of them, and I work things out from the formulas that I know.  Here's how I do time:)

Average speed during the given slow-down = (1/2)·(15+5) = 10 m/s

Distance covered during the given slow-down = 10 m.

Time = (distance) / (average speed) = (10m) / (10 m/s) = 1 second

Acceleration = (change in speed) / (time for the change)

Acceleration = (5 m/s - 15 m/s) / (1 second)

Acceleration = (-10 m/s) / (1 second)

Acceleration = -10 m/s²   (or 'Deceleration' = 10 m/s² )

_____________________________________________

For parts ii). and iii)., there's a big shift in the question.

It only gave you the slow-down from 54 to 18 km/hr for the purpose of calculating the deceleration.  NOW, for the rest of the answers, it's talking about a complete stop ... 0 m/s .

____________________________________________

ii). Time required to stop = (initial speed) / (deceleration)

Time to stop from 54 km/hr = (15 m/s) / (10 m/s²)

Time to stop = 1.5 seconds

iii).  Distance covered = (average speed) · (time to stop)

Distance covered = (7.5 m/s) · (1.5 sec)

Distance covered = 11.25 meters

OR ... use the official formula:

Distance = (1/2) · (acceleration) · (time² )

Distance = (1/2) · (10 m/s²) · (1.5 sec)²

Distance = 11.25 meters           yay !

Answer:

infrasound

Explanation:

Answer:

using V= IR

I= 11Ampere

If the person attaches a nozzle to the end of the garden hose, the velocity of the water leaving the hose will increase. This is because the nozzle reduces the cross-sectional area of the hose, which increases the speed of the water. However, the flow rate of the water will decrease.

The time it takes to fill the bucket will depend on both the velocity and the flow rate of the water. While the velocity of the water leaving the hose will increase with the nozzle, the flow rate will decrease. The flow rate is the volume of water that flows through the hose in a certain amount of time, and it is measured in gallons per minute or liters per second.

Therefore, if the flow rate decreases due to the nozzle, the time it takes to fill the bucket will increase. This is because less water is flowing through the hose, so it will take longer to fill the bucket to the desired level.

In summary, while attaching a nozzle to the end of a garden hose can increase the velocity of the water leaving the hose, it will decrease the flow rate. As a result, the time it takes to fill the bucket will likely increase.

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1a) The total energy stored when all three weights are raised by 70 cm is 80.5 J.

1b) The height of the tank is 2 m.

2b) The potential energy is converted into kinetic energy when the water flows out of the taps 6.32 m/s.

3a) The mass of water that goes over the falls every second is 1 x 10⁶ kg.

3b) The gravitational potential energy of 1 kg of water at the top of the falls 1000 J.

3c) The speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy is 44.72 m/s.

3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.

3e) The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.

1a) To calculate the amount of energy stored in the three weights, we use the formula given below:

E = mgh

Where,E = Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

For the 4.5 kg weight:

E = 4.5 x 10 x 0.7 = 31.5 J

For each of the 3.5 kg weight:

E = 3.5 x 10 x 0.7 = 24.5 J

Thus, the total energy stored when all three weights are raised by 70 cm is:

31.5 J + 24.5 J + 24.5 J = 80.5 J

1b) To calculate how far the 4.5 kg weight must be lifted to store 45 J of energy, we use the formula:

E = mghh = E/mg = 45 / (4.5 x 10) = 1 m2a)

To calculate the gravitational potential energy stored in the water in the tank when it is full, we use the formula given below:

E = mgh

Where,E = Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

The mass of 1 litre of water is 1 kg and the tank can hold 200 litres of water. Therefore, the total mass of water in the tank is:

Mass = 200 kg

The height of the tank is 2 m.

Therefore, the gravitational potential energy stored in the water in the tank is:

E = mgh = 200 x 10 x 2 = 4000 J

Assumptions made in the answer:

We have assumed that the tank is full.

2b) To calculate the speed at which the water would come out of the bathroom and kitchen taps, we use the formula given below:

PE = KEPE = mghKE = 1/2mv²

Where,PE = Potential Energy (Joules)

KE = Kinetic Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

v = Velocity (m/s)

Assuming that the potential energy of the water in the tank is converted into kinetic energy when the water flows out of the taps, the potential energy stored in the water in the tank is given by:

PE = mgh = 200 x 10 x 2 = 4000 J

The potential energy is converted into kinetic energy when the water flows out of the taps.

Therefore, KE = 1/2mv²v² = 2KE/mv² = 2(4000)/200 = 40 m²/s²v = √(40) = 6.32 m/s (speed of the water coming out of the taps)

3a) To calculate the mass of water that goes over the falls every second, we use the formula given below:

Mass = Volume x Density

Where,Volume = 1000 m³/s, Density = 1000 kg/m³, Mass = 1000 x 1000 = 1000000 kg = 1 x 10⁶ kg

3b) To calculate the gravitational potential energy of 1 kg of water at the top of the falls, we use the formula:

E = mgh

Where,m = 1 kg, g = 10 N/kg, h = 100 m, E = 1 x 10 x 100 = 1000 J

3c) To calculate the speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy, we use the formula given below:

PE = KEP

E = mgh

KE = 1/2mv²

Where,PE = Potential Energy (Joules)

KE = Kinetic Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

v = Velocity (m/s)

Assuming that all the potential energy is converted into kinetic energy when the water reaches the bottom,

PE = KEKE = mghv² = 2mghv² = 2(1)(10)(100)v² = 2000v = √(2000) = 44.72 m/s

3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.

3e) To calculate the total energy transferred per second as the GPE stored in the water falling in one second is transferred to other energy stores, we use the formula given below:

Power = Energy / Time

Where,Power = 1 x 10⁶ x 10 x 100 = 1 x 10⁹ W = 1 GW (assuming that 1 m³ of water falls every second)3f)

The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.

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The equation to calculate the time it will take for the bar to reach a temperature of 95°C with given conditions is  t_final = (95\(e^{1/2x}\)  - 30) / 50.

To set up the differential equation and solve it, we can use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between the object's temperature and the surrounding temperature.

Let's denote T(t) as the temperature of the metal bar at time t, and Tsur as the surrounding temperature (which is the temperature of the boiling water).

According to the given information, the temperature of the metal bar increases 2°C in 2 seconds. This can be written as:

(dT/dt) = k(Tsur - T)

where dT/dt represents the rate of change of temperature, k is the constant of proportionality, and (Tsur - T) represents the temperature difference between the surrounding and the metal bar.

We can rearrange the equation as:

(dT/dt) + kT = kTsur

Now, let's solve this first-order linear ordinary differential equation.

The general form of a first-order linear differential equation is:

(dy/dx) + P(x)y = Q(x)

Comparing this with our equation, we can see that P(x) = k and Q(x) = kTsur.

The integrating factor is given by:

μ(x) = \(e^{\int\limits^ {} \, P(x)dx}\)

In our case, μ(x) = \(e^{\int\limits^{} kdx}\) = \(e^{kx}\)

Multiplying our differential equation by the integrating factor μ(x), we get:

\(e^{kx}\)(dT/dt) + k \(e^{kx}\)T = k \(e^{kx}\)Tsur

Now, the left side of the equation can be written as the derivative of the product of  \(e^{kx}\) and T with respect to t:

(d/dt)( \(e^{kx}\)T) = k \(e^{kx}\)Tsur

Integrating both sides with respect to t:

∫(d/dt)( \(e^{kx}\)T) dt = ∫k \(e^{kx}\)Tsur dt

\(e^{kx}\)T = kTsur t + C

where C is the constant of integration.

Now, let's solve for T:

T = (kTsur t + C) /  \(e^{kx}\)

To find the value of the constant C, we use the initial condition T(0) = 30°C:

30 = (kTsur * 0 + C) / \(e^{k * 0}\)

30 = C

Substituting C = 30 into the equation for T:

T = (kTsur t + 30) /  \(e^{kx}\)

Now, we know that when T = 95°C, t = t_final:

95 = (kTsur t_final + 30) /  \(e^{kx}\)

Solving for t_final:

t_final = (95 \(e^{kx}\) - 30) / (kTsur)

We can substitute the given values, k = 1/2 (since the temperature increases 2°C in 2 seconds), Tsur = 100°C (boiling water temperature), and solve for t_final when T = 95°C.

t_final = (95\(e^{1/2x}\) - 30) / (1/2 * 100)

Simplifying the equation further:

t_final = (95\(e^{1/2x}\) - 30) / 50

Now, we have the equation to calculate the time it will take for the bar to reach 95°C, given the initial temperature of 30°C and the temperature increase of 2°C in 2 seconds.

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Answer:

Massm=2.5kg

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mgh

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15U=25/10×10×=375j

Yes, melting lead into a liquid to form pellets is a physical change.

This is because the chemical composition of lead remains the same even after it has been melted and then solidified into pellets. In other words, the molecular structure of lead does not change during the melting process, but only the physical state of the material changes from a solid to a liquid and then back to a solid. This type of change is reversible and can be undone by cooling the lead pellets to their solid state. Therefore, melting lead to form pellets is an example of a physical change rather than a chemical change.

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--The complete Question is, Is lead melting into liquid to form pellets a physical change? --

The uterus lining is rebuilt by the end of the menstrual period. The rebuilding of the uterus lining starts after menstruation and the lining is typically completely rebuilt by day 14 of the menstrual cycle, which is when ovulation occurs and the uterus is preparing to potentially receive a fertilized egg.

The endometrium is the inner lining of the uterus, and it thickens every month to prepare for pregnancy. After menstruation, the endometrium grows and thickens to prepare for the implantation of a fertilized egg. The cells in the lining multiply and enlarge, and the glands in the lining begin to secrete mucus and other substances that help support the fertilized egg and promote its growth.

The rebuilding of the endometrium usually takes about two weeks after menstruation. This process is closely regulated by hormones such as estrogen and progesterone, which are produced by the ovaries and other parts of the body. These hormones help control the growth and development of the endometrium and other reproductive tissues.

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Answer:

PLS PLS PLS mark brainliest

Explanation:

The following are the environmental problems caused by the invasive algae which is known as Lyngbya.

A. It blocks the sunlight from reaching the water, thereby depriving the plants in the water from accessing it. So they won't be able to manufacture their foods and eventually die off.

B. The algae produce offensive odour like rotten egg , which drive people and other organisms away.

C. The algae prevent oxygen from entering the water, and prevent the resident organisms from making use of it, which leads to their death.

D. The algae also give off heat into the chilly spring waters.

In conclusion, the goals of the Crystal River project is to save the living organisms in the river from extinction and to make it a livable environment for them.

The exterior or housing, the audio reception and speaker system, the picture tube, and a complicated mass of electronics including cable.

Antennae input and output devices, a built-in antenna in most sets, a remote control receiver, and computer chips, make up the four main sets of parts that make up a television.

Typically, in addition to coaxial cable, HDMI, and other audio-video connectors, smart TVs also enable Ethernet, WiFi, USB, Bluetooth, and flash memory cards from digital cameras.

Therefore, the TV consists of exterior, audio reception, speaker system, and picture tube.

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An automobile and a golf cart traveling at the same speed collide head-on and the impact force is the same for both.

According to Newton's 3rd law of motion, when two bodies are interacted with each other and exert force on each other is known as action and reaction pairs of forces.

Force can be explained as a push or pull acting resulting in the interaction of two objects. According to this law, when the 1st body exerts a force on the 2nd body, the 1st body experiences a force with the same magnitude in the opposite direction.

Therefore, when an automobile and a golf cart traveling at the same speed collide head-on, then they exert equal force on each other but in opposite directions, therefore, the impact force is the same for both.  

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The distance over which the force acts on the box, given that 70 N is applied to the 10 Kg box is 3.125 m

We'll begin by obtaining the acceleration of the box. Details below:

Net force = mass × acceleration

40 = 10 × acceleration

Divide both sides by 10

Acceleration = 40 / 10

Acceleration = 4 m/s²

Finally, we shall determine the distance. Details below

v² = u² + 2as

5² = 0² + (2 × 4 × s)

25 = 0 + 8s

25 = 8s

Divide both side by 8

s = 25 / 8

s = 3.125 m

Thus, we can conclude that the distance covered is 3.125 m

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A wave transfers ENERGY from one place to another.

In a one-way ANOVA comparing the average speed, the treatments would be defined as the different levels or categories of the independent variable that are being compared.

In a one-way ANOVA (analysis of variance), the average speed is the dependent variable of interest, and the treatments refer to the different groups or levels of the independent variable. The independent variable in this case would be the factor that potentially influences or categorizes the average speed.

For example, if the study aims to compare the average speed of vehicles from different types of fuel (e.g., gasoline, diesel, electric), the treatments would be defined as the fuel types. Each fuel type represents a distinct level or category of the independent variable, and the average speed of vehicles within each fuel type would be compared.

The purpose of conducting a one-way ANOVA is to determine if there are significant differences in the average speed among the different treatments (in this case, fuel types). By analyzing the variance between the groups and within the groups, the ANOVA helps to assess whether the observed differences in average speed are statistically significant or simply due to random variability.

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In the inelastic collision of the carts, the two carts fuse into an object with double the mass of the original cart. The works due to friction and the frictional force exerted to stop the two carts over 20 meters is 980 Joules.

Generally, To calculate the work due to friction and the frictional force exerted to stop the two carts over 20 meters, we will need to know the coefficient of friction between the carts and the track, the mass of the two carts, and the acceleration due to gravity.

The equation for work due to friction is:

Work = Friction force * Distance

The equation for the friction force is:

Friction force = Coefficient of friction * Normal force

The normal force is equal to the mass of the two carts multiplied by the acceleration due to gravity:

Normal force = Mass * Gravity

We can substitute these equations into the equation for work to get:

Work = (Coefficient of friction * Mass * Gravity) * Distance

To calculate the work, we need to substitute in the values for the coefficient of friction, mass, gravity, and distance. Let's say the coefficient of friction is 0.5, the mass of the two carts is 10 kilograms, and the acceleration due to gravity is 9.8 meters per second squared. The distance the carts travel is 20 meters.

Substituting these values into the equation for work, we get:

Work = (0.5 * 10 * 9.8) * 20

Solving this equation gives us a final value for the work of: 980 Joules.

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4. iii) A critical angle, or "c," is the angle of incidence at which the cladding's refracted angle equals 90 degrees. According to c, it is 64.9 degrees.

Given data:

The refractive index of core, n1 = 1.50

Fractional refractive index change, ∆ = 4% or 0.04

We know that the relation between the refractive index of core and cladding is given by:

n1/n2 = √(1-∆)

For step index fiber, the refractive index of the cladding remains constant throughout the fiber. Therefore, the refractive index of cladding,

n2 = n1/√(1-∆)i) refractive index of cladding, n2 = 1.4286

Given data:

Refractive index of core, n1 = 1.50

Fractional refractive index change, ∆ = 4% or 0.04

For step index fiber, the refractive index of the cladding remains constant throughout the fiber.

Therefore, the refractive index of cladding,

n2 = n1/√(1-∆)

n2 = 1.50/√(1-0.04)

n2 = 1.4286

ii) Numerical aperture (NA) is defined as the sine of the maximum angle of incidence at which light is totally internally reflected by the fiber. It is given by:

NA = √(n12 - n22)

NA = √(1.50² - 1.4286²)

NA = 0.38

iii) Critical angle (θc) is defined as the angle of incidence at which the refracted angle in the cladding is 90 degrees. It is given by:θc = sin⁻¹(n2/n1)θc = sin⁻¹(1.4286/1.50)θc = 64.9 degrees

5. Given data:iii) Critical angle (θc) is defined as the angle of incidence at which the refracted angle in the cladding is 90 degrees. It is given by:θc = sin⁻¹(n2/n1)θc = sin⁻¹(1.4286/1.50)θc = 64.9 degrees

Thickness of the mica sheet, d = 8μm

Refractive index of the mica sheet, μ = 1.6

Shift of central bright band, δx = 7λ

We know that the condition for the bright fringe is given by:δx = μd(λ/Δy)

where, Δy = distance between the two interfering beams

Therefore, wavelength of light, λ = δxΔy/μd

Given data:

Thickness of the mica sheet, d = 8μm

Refractive index of the mica sheet, μ = 1.6

Shift of central bright band, δx = 7λ

We know that the condition for the bright fringe is given by:δx = μd(λ/Δy)

where, Δy = distance between the two interfering beams

Therefore, wavelength of light, λ = δxΔy/μdλ = 7λ/1λ = 5600 Å or 560 nm

Therefore, the wavelength of light used is 560 nm.

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The principle involved is the conservation of momentum, where the boat moves in the opposite direction to maintain total momentum zero.

The physics principle involved in the movement of the boat as the man jumps onto the jetty is the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant if no external forces act on it.

In this scenario, the boat and the man can be considered as an isolated system since there are no external forces acting on them. Initially, when the man is standing on the boat, the system is at rest, and the total momentum is zero.

When the man jumps off the boat and onto the jetty, he exerts a force on the boat in one direction. According to Newton's third law of motion, for every action, there is an equal and opposite reaction. As the man pushes off the boat, the boat experiences an equal and opposite force that propels it in the opposite direction.

Due to the conservation of momentum, the momentum gained by the boat in one direction is equal to the momentum lost by the man in the opposite direction. As a result, the boat moves away from the jetty, exhibiting a backward motion.

This principle can be mathematically expressed as:

Initial momentum of the system = Final momentum of the system

Since the initial momentum is zero, the final momentum of the system (including the man and the boat) must also be zero. The momentum gained by the boat ensures that the total momentum of the system remains conserved.

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Answer:

Exactly 27 m/s

Explanation:

P=mv

P(cannon)=2.7*1.7

P(cannon)=4.59 kg*m/s

P(cannon)=P(tshirt)

4.59=0.17*v

v=27 m/s

Answer:

39225J

Explanation:

Given parameters:

Mass of water  = 375grams of water

Change in temperature  = 25°C

Specific heat capacity of water  = 4.184J/g°C

Unknown:

Amount of heat absorbed  = ?

Solution:

To solve this problem, we use the expression below:

      H  = m c Ф  

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

  Insert the parameters and solve;

      H  = 375 x 4.184 x (25) = 39225J

So, the magnitude of the average force the wall exert on the bullet is 25,000 N. (See the solution steps in the attached image).

Hi ! I will help you to discuss about the relationship between impulse and momentum. However, now it's not just counting the amount of impuls and momentum. But we must look at the problem and relate it to the things we learned in previous chapters, like deceleration on straight motion. Here too, we will calculate the value of the force, so pay attention to the formulas that will be given.

Here, you will be given three formulas that you will use at each step.

Although you see the word "acceleration", here it can also apply to deceleration. The deceleration always have negative value of acceleration. Pay close attention to the formula:

\(\boxed{\sf{\bold{(v_t)^2 = (v_0)^2 + 2 \cdot a \cdot s}}} \)

With the following condition:

\( \boxed{\sf{\bold{a = \frac{v_t -v_0}{t}}}} \)

With the following condition :

Impulse is a change in the value of momentum. The impulse is influenced by the force and time factors, while the momentum is influenced by the mass and velocity factors. Consider the following formula.

\( \sf{I = \Delta p} \)

\( \sf{F \times t = m \cdot (v_t-v_0)} \)

\( \boxed{\sf{\bold{F = \frac{m \cdot (v_t-v_0)}{t}}}} \)

With the following condition :

[See at the picture that attached in this answer]

So, the magnitude of the average force the wall exert on the bullet is 25,000 N.

Cyclotron to make 1 milligram of the isotope fluorine-18 for use a a diagnostic tool with it PET scanner. It seems that the hospital should plan to make more F-18 the next morning, as the amount remaining at midnight is very small.

The half-life of F-18 is 1.8 hours, so after 1.8 hours, half of the original amount of F-18 will remain. After 3.6 hours, a quarter of the original amount will remain, and so on.

At 3:00 PM (9 hours after the F-18 was made), the hospital will have

= 1/2^(9/1.8)

= 1/2^5

= 1/32

= approximately 0.03125 milligrams of F-18 remaining.

At midnight (18 hours after the F-18 was made), the hospital will have

= 1/2^(18/1.8)

= 1/2^10

= 1/1024

= approximately 0.0009765625 milligrams of F-18 remaining.

While using cyclotron, based on these calculations, it seems that the hospital should plan to make more F-18 the next morning, as the amount remaining at midnight is very small.

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