The amount of steering wheel movement needed to turn will ____________ the faster you go.

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.

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: 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:

Waning gibbous.

Explanation:

The same side that you see in one place of the continent at night, you will also see in other place on earth, regardless of the hemisphere. There is however a slight difference, but not ecognizable to the human eye thus Waning Gibbous will also be seen everywhere else in the night sky that night.

Final answer:

Your friend in Panama City will see the same waning gibbous moon phase as you see in New York, with only slight variances in the moon's position in the sky and visibility times due to the locations' relatively close longitude.

Explanation:

If you are in New York and you observe a waning gibbous moon, your friend in Panama City will see the same moon phase, provided the sky is clear. The moon goes through its phases at the same rate all over the world as it orbits Earth.

The difference in geographical location does not change the phase of the moon that is visible; it only changes the moon's position in the sky and the time it is visible.

Since New York and Panama City are relatively close in longitude, with Panama City being slightly to the west, the time difference for moonrise and moonset will not be significantly different. Therefore, both you and your friend should be able to witness the waning gibbous moon at approximately the same time during a clear night.

Answer:

110 N

Explanation:

When a force is applied on a body and body does not move, it means the body remains at rest.

In this condition, there is a contact force between the body and the floor which is called static friction.

Th static friction force is a self adjusting force and comes into play when the body is at rest.

Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.

So, the contact force between the chest and the floor is 100 N.

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:

1. Public Welfare, 2. Public Safety, 3. Public Works

Explanation:

I got them all right, if you go according to my answer, you will too

The government service to complete each sentence is

Government Services are services intended to serve all members of a community; it is usually provided by the government to people living within its jurisdiction. Examples are policing health care, and education.

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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: [tex]204.8\ miles[/tex]

Explanation:

Remember that:

[tex]V=\frac{d}{t}[/tex]

Where "V" is the speed, "d" is the distance and "t" is the time.

You are are driving home from school steadily at 65 miles per hour for 130 miles, then we can find the driving time at  65 miles per hour:

[tex]V_1=\frac{d_1}{t_1}\\\\65\ \frac{mi}{h}=\frac{130\ mi}{t}\\\\t_1=\frac{130\ mi}{65\ \frac{mi}{h}}\\\\t_1=2\ h[/tex]

You slow down to 55 miles per hour and you arrive home after driving 3 hours and 22 minutes, then we need to find the driving time at  55 miles per hour. But first you need to convert 22 minutes to hours:

[tex](22\ min)(\frac{1\ h}{60\ min})=0.36\ h[/tex]

Since the total time is:

[tex]t_{total}=t_1+t_2[/tex]

We can calculate [tex]t_2[/tex]:

[tex]t_2=t_{total}-t_1\\\\t_2=(3\ h+0.36\ h)-2\ h\\\\t_2=1.36\ h[/tex]

In order to calculate the distance from that point (where you slow down to 55 iles per hour) to your home, we need to solve for [tex]d_2[/tex] from the following formula and substitute values:

[tex]V_2=\frac{d_2}{t_2}\\\\V_2*t_2=d_2\\\\d_2=(55\ \frac{mi}{h})(1.36\ h)\\\\d_2=74.8\ mi[/tex]

Therefore, the distance between your hometown and your school is:

[tex]d_{total}=d_1+d_2\\\\d_{total}=130\ mi+74.8\ mi\\\\d_{total}=204.8\ mi[/tex]

Final answer:

To find the total distance to the student's hometown from school, first calculate the distance for each portion of the trip where speed is constant and then sum these distances. Use the formula distance = speed × time.

Explanation:

To solve the problem of finding how far the student's hometown is from school, we can use the simple formula for distance: distance = speed × time. However, given that the speeds vary over the journey, we must break the journey into parts where the speed is constant, calculate the distance for each part, and then sum those distances to find the total.

Firstly, we know the student drove 130 miles at 65mph. This part is straightforward. However, the overall time of travel is given as 3 hours and 22 minutes. To work seamlessly with speeds and distances, it's easier to convert this total travel time into hours, which is 3.367 hours (22 minutes is approximately 0.367 hours).

To find the distance traveled at 55mph, we subtract the time taken to travel the first 130 miles from the total time. The time taken to travel the first 130 miles at 65mph can be calculated as: time = distance / speed, which will give us the time in hours. Subtracting this time from the total time will give us the time spent traveling at 55mph, allowing us to calculate the second part of the journey's distance using the distance = speed × time formula.

Finally, to find the total distance to the student's hometown, we simply add the 130 miles to the distance traveled at 55mph.

Answer:1.2 m/s

Explanation:

mass of ball[tex]`(m_1)=2 kg[/tex]`

initial Velocity of [tex]m_1=3 m/s[/tex]`

mass of another ball [tex](m_2)=2 kg[/tex]`

initial velocity of [tex]m_2=0[/tex]`

Conserving momentum

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]`

[tex]2\times 3+2\times 0=(2+3)v[/tex]`

[tex]v=\frac{6}{5}=1.2 m/s[/tex]`

A normal distribution displays the highest data scores in the middle of the distribution.

A normal distribution curve, is used to represent symmetrical data set. A data set is a said to be symmetrical curve if the mean is equal to the median.

General properties of normal distribution curve include;

Thus, we can conclude that a normal distribution displays the highest data scores in the middle of the distribution.

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Answer:

The answer is the third option, C. A normal distribution displays the highest data scores in the middle of the distribution. Took the practice and this was correct.

Explanation:

Psychology 2022

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

Explanation:

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]

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.

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".

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.

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:

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:

The material stretched across the drum vibrates to produce the sound waves.

Explanation:

A drum is one of the oldest musical instrument made by man. It is made of a hollow body over which a material such as skin is stretched. When a drum is struck with a stick, the material vibrates up and down. This makes the air above the drum to contract and relax rhythmically resulting in soundwaves. The soundwaves travel through air to reach our ears where they are heard.

The quality of sound produced by  a drum is affected by its shape. A larger drum produces a lower pitched sound.

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:

It is the energy in/of a fluid due to applied pressure (force per area). For example, if there is a static fluid in an enclosed container, the energy of the system is only due to the pressure; if the fluid is moving along a flow, then the energy of the system is the kinetic energy as well as the pressure.

Explanation:

physics.stackexchange.com

The 'energy of pressure' in physics signifies the concept that pressure can be represented as energy per unit volume, particularly for fluids. Hydrostatic pressure reflects energy density due to a fluid's internal forces. Pressure multiplied by volume gives units of energy, aligning with the ideal gas law.

The term energy of pressure refers to a concept in physics where pressure is understood as a measure of energy per unit volume. This approach to describing pressure is particularly useful when discussing fluids, such as gases and liquids.

The traditional definition of pressure as force per unit area is related to energy by the work done (W = F.d); since work is a form of energy, we can derive an expression for pressure as an energy density (energy per unit volume).

In the context of fluids at rest, we talk about hydrostatic pressure, which is uniform in all directions within a fluid and equals the energy density due to the fluid's internal forces. The internal pressure, which is a property of the fluid, is indicative of the strength of intermolecular forces and is positive when these attractive forces dominate. This relates to the internal energy of a fluid, which is made up of potential and thermal energy.

When we consider the energy content of a gas, for example, we typically use the ideal gas law which relates pressure and volume to temperature. Through this relationship, we see that pressure multiplied by volume has units of energy (joules), and thus we view pressure as being indicative of the potential energy in a fluid per unit volume.

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:

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

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

Final answer:

To stop the train without crates sliding, we need to determine the stopping distance. By using the equation v^2 = u^2 + 2aS and considering the maximum static friction force, the stopping distance is found to be 225 / (0.64g) meters.

Explanation:

To determine the stopping distance of the train without causing the crates to slide over the floor, we need to find the maximum deceleration the train can have. Given that the coefficient of static friction is 0.32 and the initial speed of the train is 54 km/h, we can convert the speed to m/s by multiplying it by 5/18. Therefore, the initial speed of the train is 15 m/s. Using the equation v^2 = u^2 + 2aS, where v is the final velocity (0 m/s), u is the initial velocity (15 m/s), a is the acceleration, and S is the stopping distance, we can solve for the stopping distance.

Rearranging the equation, we have 0 = (15 m/s)^2 + 2aS. Since the train is stopping, the final velocity is 0. Plugging in the values, we get 0 = 225 + 2aS.

Now, we know that the coefficient of static friction (μ) is 0.32. The maximum static friction force (fs) can be calculated by multiplying the coefficient of static friction by the normal force (mg), where m is the mass of the crates and g is the acceleration due to gravity. However, since the crates are not sliding and the train is being stopped, the static friction force must be equal to or greater than the force required to stop the train (ma), where m is the mass of the crates and a is the deceleration. Therefore, we have fs ≥ ma.

Substituting the values, we get 0.32mg ≥ ma. Canceling out the mass, we have 0.32g ≥ a.

Now we have two equations: 0 = 225 + 2aS and 0.32g ≥ a. By substituting 0.32g for a in the first equation, we can solve for S.

Plugging in the values, we get 0 = 225 + 2(0.32g)S. Simplifying the equation, we get -225 = 0.64gS. Rearranging the equation to solve for S, we have S = -225 / (0.64g). Taking the negative sign out, as distance can't be negative, S = 225 / (0.64g). Therefore, the train can be stopped at a distance of 225 / (0.64g) meters without causing the crates to slide over the floor.

Answer:

Duration

Explanation:

According to my research on different job requirements and terminology, I can say that based on the information provided within the question the term being defined is called Duration. When speaking in terms of Job Assignments/Projects, Duration (like described in the question) is the amount of time that you have before that assignment or project needs to be finished.

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Final answer:

The term defined as the total number of work periods required to complete a schedule activity is known as 'duration' or 'work duration.' It is crucial for building an effective work schedule that considers member availability, project deadlines, and meeting times while also ensuring work-life balance.

Explanation:

The term defined as the total number of work periods required to complete a schedule activity is often referred to as duration or work duration. It is expressed in workdays or work weeks, and it does not account for holidays or other non-work periods. A clear understanding of duration is critical for constructing an effective work schedule or task schedule, and a timeline that takes into account the availability of team members, project deadlines, and meeting schedules.

Duration encompasses the work accomplished in the preceding periods, work currently being performed, and the work planned for the next periods. It also accounts for instances where workloads increase and might require adjustment to an employee's regular duty assignment. This could mean altering an employee's tour of duty to include hours when excess work needs to be completed over several days.

It is important to balance the hours worked during a workday with the need for rest, considering the average hours worked by full-time and part-time workers. Furthermore, maintaining productivity while being mindful of when meetings are scheduled and integrating work-related activities into the workday are all part of effective time management that reflects on an employee's job satisfaction and overall work-life balance.

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.

https://brainly.com/question/12134554.

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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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:

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.