The mass number of an atom of carbon (C) represents the total number of
A) protons in the atom.
B) neutrons and electrons in the atom.
C) protons and neutrons in the atom.
D) electrons in the atom.

Answer:

neutrons, more, physical, biological

Explanation:

Radioisotopes are those isotopes of an atom which due to excessive energies are unstable, this unstability results from combination of protons and neutrons which are unstable.

Thus to achieve stability, these isotopes loses their energy when they decay or disintegrate.

The time taken by the radioisotope to achieve about 50% stability after disintegration is known as Physical half-life.

The biological half-life refer to the time taken by the radioisotope for the elimination of about 50% of the radioactive substance.

Radioisotopes are unstable due to an excess number of neutrons, and as they decay, they become more stable. The physical half-life is the time for 50% to decay, while the biological half-life refers to the time till 50% leaves the body.

Radioisotopes are unstable forms of isotopes because they contain an excess number of neutrons. Radioisotopes lose nuclear energy as they decay or break down, thus becoming more stable. Radioisotopes are unstable forms of isotopes because they contain an excess number of neutrons. Radioisotopes lose nuclear energy as they decay or break down, thus becoming more stable. The physical half-life is the amount of time it takes for 50% of the radioisotope to become stable. The amount of time it takes for 50% of radioactive material to leave the body is referred to as the biological half-life.The physical half-life is the amount of time it takes for 50% of the radioisotope to become stable. The amount of time it takes for 50% of radioactive material to leave the body is referred to as the biological half-li.

Answer:

Displacement: 2.230 km    Average velocity: 1.274[tex]\frac{km}{h}[/tex]

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

First of all. We need to sum all the X components and all the Y componets.

∑[tex]Sx = Ax -0.916[/tex] ⇒  ∑[tex]Sx = [tex]3.52cos(49.7) - 0.916[/tex]

∑[tex]Sx = 1.361 km[/tex]

∑[tex]Sy = Ay - 0.918[/tex] ⇒ ∑[tex]Sy = 3.52sin(49.7) - 0.918[/tex]

∑[tex]Sy = 1.767[/tex]

The total displacement is calculated using Pythagoeran therorem:

[tex]S_{total} =\sqrt{Sx^{2}+ Sy^{2} }[/tex] ⇒

[tex]S_{total} = 2.230 km[/tex]

With displacement calculated, we can find the average speed as follows:

[tex]V = S/t[/tex]  ⇒  [tex]V = \frac{2.230}{1.750}[/tex]

[tex]V = 1.274\frac{km}{h}[/tex]

Answer: The primacy effect occurs when you're more likely to remember words at the beginning of a list. A suggested reason for the primacy effect is that the initial items presented are most effectively stored in long-term memory because of the greater amount of processing devoted to them.

So B would be your answer

To find the velocity of the cart when it reaches its greatest y coordinate, we can combine the x and y components of the velocity. The x-component can be found using Ux = v0x + ax*t, and the y-component can be found using Uy = v0y + ay*t. The total velocity, v, can then be found using v = sqrt(Ux^2 + Uy^2).

The velocity of the cart can be found by combining its x and y components. The x-component of the velocity, Ux, can be found using the equation Ux = v0x + ax*t. The y-component of the velocity, Uy, can be found using the equation Uy = v0y + ay*t. The total velocity, v, can be found using the equation v = sqrt(Ux^2 + Uy^2).

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When a car traveling in the negative x-direction (-x or westward direction) begins to slow down, it is decelerating. However, this deceleration while moving in the negative direction translates into positive acceleration. The example provided illustrates this with the change in velocity demonstrating a positive acceleration.

Considering a car traveling to the west (negative x-direction) that begins to slow down as it approaches a traffic light, the car is decelerating. However, because it is moving in the negative direction, the deceleration or slowing down translates into a positive acceleration. This epitomizes the fundamental rule that acceleration is in the same direction as the change in velocity.

To illustrate this, let's take the initial velocity of the car to be -25 m/s (moving to the west). As the car slows down, it ends at a speed less than that, say -5 m/s. The change in velocity is now given by final velocity - initial velocity, which equals -5 m/s - (-25 m/s) = 20 m/s.

This demonstrates a positive change in velocity. Hence, the slowing down of the car in a negative direction results in a positive acceleration.

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Final answer:

The percent uncertainty in the distance is 0.0593%, the uncertainty in the elapsed time is 1 second, the average speed is 391 m/s, and the uncertainty in the average speed is 0.231 m/s.

Explanation:

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

[tex]v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s[/tex]

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

[tex]x(t)=8.8 m/s *t[/tex]

[tex]y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}[/tex]

Now we calculate the time the hammer takes to get to the floor

[tex]17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s[/tex] or [tex]t=-2.38s[/tex]

Now, we keep the positive time result and calculate the horizontal distance travelled

[tex]x(1.47s)=8.8m/s*1.47s=12.94m[/tex]

Answer:

Δf = -82.57 Hz

Explanation:

For this problem we will use Doppler effect formula:

[tex]f = \frac{C+Vr}{C+Vs}*fo[/tex]    where:

C is the speed of sound

Vr is the velocity of the person = 0m/s

Vs is the velocity of the source (train): When approaching Vs = -41m/s and when receding Vs = 41m/s

fo = 342.5 Hz

When the train is approaching:

[tex]fa = \frac{C}{C+Vs}*fo = \frac{345}{345-41}*342.5=388.69Hz[/tex]

And when the train is receding:

[tex]fr = \frac{C}{C+Vs}*fo = \frac{345}{345+41}*342.5=306.12Hz[/tex]

So, the change of frequency will be:

Δf = fr - fa = 306.12 - 388.69 = -82.57 Hz

Final answer:

The volume of a python egg can be calculated using the formula V = 4/3πr³. By substituting the given values, the volume is found to be 16π cubic inches.

Explanation:

The volume V of a python egg can be calculated using the formula V = 4/3πr³. In this case, a = 2 inches, b = 2 inches, and c = 3 inches. To find the volume, substitute the given values into the formula:

V = (4/3)π(2)(2)(3) = (4/3)π(12) = 16π cubic inches.

Therefore, the volume of the python egg is 16π cubic inches.

Answer:

reaction time

Explanation:

According to my research on studies conducted by various neurologists, I can say that based on the information provided within the question this game best describes the reaction time task. This is the amount of time it takes for your brain to process the visual information and for your body to react accordingly. This is because the toy moles only pop their heads out for an extremely small amount of time in which you have to hit them with the hammer before they go back down.

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Answer:

Rocks will pass each other at 20.4m.

Explanation:

Alice drops the rock from [tex]y_{a,i}=40m[/tex] (y is in the vertical axis) at [tex]v_{a,i}=0[/tex].

Bob throws the rock from [tex]y_{b,i}=0m[/tex] at [tex]v_{b,i}}=20\frac{m}{s}[/tex].

[tex]g[/tex] is gravity's acceleration.

Position equation for alice's rock:

[tex]y_{a}(t)=y_{a,i} +v_{a,i}t - \frac{1}{2}gt^{2} =40m-\frac{1}{2}gt^{2}[/tex]

Position equation for Bob's rock:

[tex]y_{b}(t)=y_{b,i} + v_{b,i}t - \frac{1}{2}gt^{2} =20\frac{m}{s}t -\frac{1}{2}gt^{2}[/tex]

We we'll first find the time [tex]t_{0}[/tex] at which the rocks meet. For this, we will equalize Bob's and Alice's equations:

[tex]y_{a}(t_{0})=y_{b}(t_{0})[/tex]

[tex]40m-\frac{1}{2}gt_{0}^{2}=20\frac{m}{s}t_{0} -\frac{1}{2}gt_{0}^{2}[/tex]

⇒[tex]20\frac{m}{s}t_{0}=40m[/tex] ⇒ [tex]t_{0}=2s[/tex]

So, we can take [tex]t_{0}=2[/tex] and just put it in any of the two laws of motion to see at what height the rocks meet. We will take Alice's equation (using g=9.8m/s):

[tex]y_{a}(2) =40m-\frac{1}{2}g2^{2}=20.4m[/tex]

Rocks will pass each other at 20.4m.

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

[tex]F_{f} = -\mu N[/tex]

Since we're talking about the same car, the value of [tex]N[/tex] will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

Answer:

B

Explanation:

Right on edge 2021

Explanation:

The acceleration of an object is given by :

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

Dimension of velocity is, [tex][v]=[LT^{-1}][/tex]      

Dimension of time is, [tex][t]=[T][/tex]        

Dimension of acceleration is, [tex][a]=[LT^{-2}][/tex]    

Option 1.

[tex]\dfrac{v^2}{x}=[LT^{-2}][/tex]

Option 2.

[tex]xt^2=[LT^2][/tex]

Option 3.

[tex]\dfrac{x}{t^2}=[LT^{-2}][/tex]

Option 4.

[tex]\dfrac{v}{t}=[LT^{-2}][/tex]                    

So, from above calculations, it is clear that option (1),(3) and (4) have the dimensions of acceleration. Hence, this is the required solution.

Answer:

a)[tex]\dfrac{v^2}{x}[/tex]

c)[tex]\dfrac{x}{t^2}[/tex]

d)[tex]\dfrac{v}{t}[/tex]

Explanation:

Acceleration :

Acceleration is the rate of change of velocity of the particle.

The unit of acceleration is m/s².

In mathematical form,

[tex]a=\dfrac{dv}{dt}[/tex]

We will check the all option on base of unit

(a). [tex]\dfrac{v^2}{x}[/tex]

Where, v = velocity

x = position

The unit of velocity and position are m/s and m.

The dimension formula of velocity and position

[tex]v = LT^{-1}[/tex]

[tex]x=L[/tex]

Put the unit in the given equations

[tex]\dfrac{v^2}{x}=\dfrac{L^2T^{-2}}{L} =\dfrac{L}{T^2}[/tex]

(b). [tex]xt^2=L\times T^2[/tex]

(c). [tex] \dfrac{x}{t^2}=\dfrac{L}{T^2}[/tex]

(d). [tex]\dfrac{v}{t}=\dfrac{LT^{-1}}{T}=\dfrac{L}{T^2}[/tex]

Hence, This is the required solution.

Answer:

2,829 meters

Explanation:

X= Vo t + 1/2 a t^2

Being

X=1,5 m, t=0,18 s and a=9,81 m/s2

Then

Vo=7, 45 m/s

Then we use

V=a.t

Being

V=Vo, and a=9,81 m/s2

t=0,75 seconds to my window

We use again

X= 1/2 a t^2

Being

t=0,75 s and a=9,81 m/s2

X=2,82 m

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:

The acceleration you can get with that engine in your car is around 70,56 [tex](\frac{m}{s^{2} })[/tex] or 7,26 [tex](\frac{ft}{s^{2} } )[/tex] using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

[tex]F=m*a[/tex]

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

[tex]a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }[/tex]

[tex]a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }[/tex]

Note: N or Newton units are: [tex]\frac{kg * m}{s^{2} }[/tex]

[tex]a= 70,54 \frac{m}{s^{2} }[/tex]

Also in pounds you can compared

[tex]a= \frac{2400  lf }{3 306  lf}[/tex]

Note: lf in force units are: [tex]\frac{lf*ft}{s^{2} }[/tex]

[tex]a=7,26 \frac{ft}{s^{2} }[/tex]

Final answer:

The acceleration provided by a fighter airplane engine mounted on an average 1,500 kg car would be 70.56 m/s², calculated using Newton's second law of motion by dividing the force of 105,840 N by the mass of the car.

Explanation:

To calculate the acceleration that a fighter airplane engine could give to a car, we must use Newton's second law of motion, which states that the force exerted on an object equals the mass of the object multiplied by the acceleration of the object (F = m*a). In this case, the force given is 105,840 N from the airplane engine. Assuming an average car mass of about 1,500 kg (a reasonable estimate for a personal vehicle), the formula to find the acceleration (a) is as follows:

a = F / m

Substituting in the values provided:

a = 105,840 N / 1,500 kg

When the calculation is performed:

a = 70.56 m/s²

So, the acceleration that would result from mounting this aircraft engine into an average car would be 70.56 meters per second squared.

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:[tex]5k^2+6k[/tex]

Explanation:

Given

initial side of cube is k cm

New dimensions are

k+2 cm

k+3 cm

k cm

[tex]V_{initial}=k^3[/tex]

[tex]V_{Final}=\left ( k+3\right )\left ( k+2\right )\left ( k\right )[/tex]

Now  [tex]V_{final}-V{initial}=\left ( k+3\right )\left ( k+2\right )\left ( k\right )-k^3[/tex]

[tex]\Delta V=k^3+5k^2+6k-k^3=5k^2+6k[/tex]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

[tex]R_x = Rcos A[/tex]

[tex]R_y = Rsin A[/tex]

From above relation

Assuming base camp as the origin, location of 1st team is

[tex]R_1 = 37 km[/tex] away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

[tex]R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km[/tex]

[tex]R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km[/tex]

location of 2nd team is at

[tex]R_2 = 32 km[/tex], at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

[tex]R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km[/tex]

[tex]R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km[/tex]

Now position of 2nd team with respect to 1st team will be given by:

[tex]R_3 = R_2 - R_1[/tex]

[tex]R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j[/tex]

Using above values:

[tex]R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j[/tex]

[tex]R_3 = 51.49 i + 13.71 j[/tex]

distance of 2nd team from 1st team will be:

[tex]\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)[/tex]

[tex]\left | R_3 \right | = 53.28 km = 58.2 km[/tex]

Direction of 2nd team from 1st team will be:

[tex]Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}][/tex]

Direction = 14.90 deg North of east

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:

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:IU (international unit): An international unit (IU) is an internationally accepted amount of a substance.

                    This type of measure is used for the fat-soluble vitamins (such as vitamins AD and E) and certain hormones, enzymes, and biologicals (such as vaccines).

Explanation:

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]`

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:

1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R

Explanation:

We make use of relations between temperature scales with respect to degrees celsius:

[tex]1 K= 1^{\circ}C+273\\1^{\circ}F= (1^{\circ}C*1.8)+32\\1 R= (1^{\circ}C*1.8)+491.67[/tex]

This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.

So:

[tex]\frac{Q}{\Delta T(K)}=\frac{Q}{\Delta T(^\circ C)}=1600 \frac{kJ}{h} per K\\\frac{Q}{\Delta T(^\circ F)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per ^\circ F\\\frac{Q}{\Delta T(R)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per R[/tex]

Answer:

1.78 m upward

Explanation:

We can find the displacement of the volleyball by using the SUVAT equation:

[tex]v^2 - u^2 = 2ad[/tex]

where, assuming upward as positive direction:

u = 6.0 m/s is the initial velocity

v = 1.1 m/s is the final velocity

a = g = -9.8 m/s^2 is the acceleration of gravity

d is the displacement

Solving the equation for d, we find:

[tex]d=\frac{v^2-u^2}{2a}=\frac{1.1^2-6.0^2}{2(-9.8)}=1.78 m[/tex]

And since it is positive, the displacement is upward.

The integral concepts applied include kinematics and gravitational acceleration. Given initial velocity, final velocity, and knowing gravitational due to acceleration, we can determine the time of flight up to its highest point and the maximal height or displacement through kinematic equations. In this particular case, displacement happens to be approximately 2.275 meters upwards.

The subject of this question is Physics, specifically, it deals with the concept of kinematics. The first step to solve the problem is to use the final velocity equation in kinematics, v = u + at, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration (which would be gravity in this case, ‐9.8 m/s² since it's acting downward or against the direction of the initial velocity), and 't' time. From the given values, we know that v = 1.1 m/s (these are vector quantities, hence the direction is important, and in this case, both v and u are in the same direction, up the way), u = 6.0 m/s and a = -9.8 m/s². You should set up the equation as follows: 1.1 = 6 + (-9.8)t. By simplifying, we get t = (1.1 - 6) / -9.8 ≈ 0.5s. Next, to get the displacement, you can use another kinematic equation s = ut + (1/2)a*t² ('s' stands for displacement). Once you plug in the known values, you'll get s = 6*0.5 + 1/2*(-9.8)*(0.5)² ≈ 2.275 m.

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Answer:

A) adding hydrogens, decreasing the number of double bonds in the molecules.

Explanation:

The complete Question is:

An oil may be converted into a substance that is solid at room temperature by

A) adding hydrogens, decreasing the number of double bonds in the molecules.

B) removing water, causing a dehydration synthesis reaction to occur.

C) removing hydrogens, increasing the number of double bonds.

D) cooling it, so that double bonds form and the fats solidify.

Hydrogenation is a process that is commonly used in food industry to reduce the vegetable oil to solid or semi-soild state which is suitable for preserving. Oil is an unsaturated fatty acid having Carbon - Carbon double bonds. Hydrogenation i.e. addition of hydrogen in presence of catalysts to these oils reduce these double bonds to single bonds. This reduction of double bonds also change some physical properties of the oil, the most obvious of which is the melting point. The melting point increases and the oil is converted to a substance that is sold at room temperature.

Therefore, the correct answer to this question is given by option A.

Final answer:

An oil can be converted into a solid fat by hydrogenation, which involves adding hydrogen to unsaturated fatty acids to raise the melting point and create a solid at room temperature, commonly used in making margarine or shortening.

Explanation:

An oil can be converted into a solid fat through a process known as hydrogenation. During this chemical reaction, hydrogen is added to the unsaturated fatty acids present in the oil. This transformation is achieved using a catalyst such as nickel (Ni), platinum (Pt), or palladium (Pd), which facilitates the addition of hydrogen atoms to the oil's fat molecules. The result of these additions changes the molecular structure, thus converting unsaturated fatty acids into saturated fatty acids. This increase in saturation raises the melting point of the triglycerides found in the oil, often turning a liquid oil, like vegetable oil, into a semi-solid or solid form similar to margarine or shortening.

Saturated fats tend to be solid at room temperature due to their chemical structure, which lacks double bonds, allowing them to pack tightly together. Unsaturated fats, which have one or more double bonds in their hydrocarbon chains, are typically liquid at room temperature. The hydrogenation process effectively reduces or eliminates double bonds, increasing the degree of saturation and thereby raising the melting point of the fat. This is a common practice in the food industry to alter the texture and stability of edible fats for various applications.

The velocity of the object at time t0 = -5 can be found by taking the derivative of the position function, which results in v(t) = -22t. Substituting t = -5 into this velocity function yields v(-5) = 110 m/s.

The student asks for the velocity of an object at time t0 = -5 given the position function f(t) = −11t2. To find the velocity, we use the limit definition of the derivative. The derivative of the position function with respect to time gives the velocity function:

v(t) = f'(t) = d/dt (-11t2)

This results in:

v(t) = -22t

Now, we can find the velocity at t0 = -5 by plugging the value into the velocity function:

v(-5) = -22(-5) = 110 m/s

Therefore, the velocity of the object at t0 = -5 is 110 m/s.

Answer:

Option c that it means cm/year.

Explanation:

Nm means nanometer, 1x10^9m

Km/s^2 is acceleration

h.. it may be height

Answer:

D, km/s^2

Explanation:

This is a unit of speed because it is saying kilometers over seconds squared, and that is a unit of speed similar to miles per hour or mph or m/h.

Answer:

It takes 0.00127 seconds.

Explanation:

The equation its

[tex]y(x,t) = 4.7 \ mm  \ sin ( kx + 675 \frac{rad}{s} t + \phi)[/tex].

We want the time for ANY POINT, so, for convenience, lets take x=0

[tex]y(0,t) = 4.7 \ mm  \ sin ( k*0 + 675 \frac{rad}{s} t + \phi)[/tex].

[tex]y(0,t) = 4.7 \ mm  \ sin ( 675 \frac{rad}{s} t + \phi)[/tex].

Now, we want the positions

[tex]y(0,t_1)=2 \ mm[/tex]

[tex]y(0,t_2)=-2 \ mm[/tex]

so, for the first position:

[tex]y(0,t_1)=2 \ mm[/tex]

[tex]2 \ mm = 4.7 \ mm  \ sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].

[tex]\frac{2 \ mm} {4.7 \ mm }=  sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].

[tex] 0.42 =  sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].

[tex] sin^-1(0.42) =  675 \frac{rad}{s} t_1 + \phi)[/tex].

and for the second one:

[tex]y(0,t_2)=-2 \ mm[/tex]

[tex]-2 \ mm = 4.7 \ mm  \ sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].

[tex]\frac{-2 \ mm} {4.7 \ mm }=  sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].

[tex] -0.42 =  sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].

[tex] sin^-1(0.42) =  675 \frac{rad}{s} t_2 + \phi)[/tex].

Now, we can subtract both:

[tex]sin^-1(0.42) -sin^-1(-0.42) =  675 \frac{rad}{s} t_1 + \phi - 675 \frac{rad}{s} t_2 + \phi[/tex]

[tex]sin^-1(0.42) -sin^-1(-0.42) =  675 \frac{rad}{s} t_1 - 675 \frac{rad}{s} t_2[/tex]

[tex]sin^-1(0.42) -sin^-1(-0.42) =  675 \frac{rad}{s} (t_1 - t_2)[/tex]

[tex]\frac{sin^-1(0.42) -sin^-1(-0.42)}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]

[tex]\frac{0.43 - (-0.43)}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]

[tex]\frac{0.86}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]

[tex] 0.00127 s = (t_1 - t_2)[/tex]

The strings take 0.00127 seconds.

Final answer:

To find the time for a point on a sinusoidal wave to move from y = +2.0 mm to y = -2.0 mm, we calculate the time for half a cycle of the wave using the angular frequency.

Explanation:

The question involves determining the time it takes for a point on a string to move between two displacements in a sinusoidal wave motion. Given that the wave equation is y(x, t) = (4.7 mm)sin(kx + (675 rad/s)t + φ) and the displacements to consider are +2.0 mm and -2.0 mm, we can utilize the properties of the sine function and the given angular frequency to solve for the time interval. First, we find the phase angle (θ) corresponding to y = 2.0 mm by using the inverse sine function and then find the difference in time (t) to reach y = -2.0 mm, considering the periodic nature of the sine wave. Assuming we start from y = +2.0 mm at time t = 0, as the wave progresses to y = -2.0 mm, it completes half a cycle, so δt = π/ω, where ω = 675 rad/s is the angular frequency. Therefore, δt = π/675 s.