How many measures are there until the next part enters in Kookaburra?
A. 2
B. 1
C. 3
D. 4

Answer:

At least [tex]0.128[/tex] (rounded up, to three significant figures as in speed,) assuming that the road is level and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

When an object is in a circular motion, the magnitude of the net force can be found in terms of the mass and speed of that object, as well as the radius of the circular path:

[tex]\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^{2}}{(\text{radius})} \end{aligned}[/tex].

Let [tex]m[/tex] denote the mass of this vehicle. Let [tex]v[/tex] denote the speed of this vehicle. Let [tex]r[/tex] denote the radius of this circular path. While in this circular motion, the magnitude of the net force will be on this vehicle will be:

[tex]\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^{2}}{(\text{radius})} = \frac{m\, v^{2}}{r} \end{aligned}[/tex].

Forces on this vehicle include:

If this road is level, the normal force from the road will balance the weight of this vehicle. The magnitude of the normal force will be equal to that of the weight of this vehicle, [tex]m\, g[/tex].

Let [tex]\mu_{\text{s}}[/tex] denote the static friction coefficient between the tire and the road. The static friction between the vehicle and the road cannot exceed [tex](\text{static friction coefficient}) \times (\text{normal force})[/tex].

Since [tex](\text{normal force}) = m\, g[/tex], the maximum possible value of static friction will be:

[tex]\begin{aligned} & (\text{static friction coefficient}) \times (\text{normal force}) \\ =\; & (\mu_{\text{s}})\, (m\, g) \\ =& \mu_{\text{s}}\, m\, g\end{aligned}[/tex].

Under the assumptions, the weight and normal force on this vehicle will be balanced. As a result, the net force on this vehicle will be equal to static friction and should also be no greater than [tex](\text{static friction coefficient}) \times (\text{normal force}) = \mu_{\text{s}}\, m\, g[/tex].

In other words:

[tex]\begin{aligned}(\text{net force}) &= (\text{static friction}) \\ &\le (\text{maximum static friction}) = \mu_{\text{s}}\, m\, g \end{aligned}[/tex].

Additionally, from circular motion:

[tex]\begin{aligned} (\text{net force}) &= \frac{m\, v^{2}}{r} \end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}\frac{m\, v^{2}}{r} \le \mu_{\text{s}}\, m\, g \end{aligned}[/tex].

Rearrange this inequality to separate the coefficient of static friction, [tex]\mu_{\text{s}}[/tex]:

[tex]\begin{aligned} \mu_{\text{s}} \ge \frac{v^{2}}{r\, g}\end{aligned}[/tex].

(Note that [tex]m[/tex] and [tex]g[/tex] are both greater than [tex]0[/tex].)

Substitute in [tex]v = 25.0\; {\rm m\cdot s^{-1}}[/tex], [tex]r = 500\; {\rm m}[/tex], and [tex]g \approx 9.81\; {\rm m\cdot s^{-2}}[/tex]:

[tex]\begin{aligned} \mu_{\text{s}} &\ge \frac{v^{2}}{r\, g} \\ & \approx \frac{(25.0\; {\rm m\cdot s^{-1}})^{2}}{(500\; {\rm m})\, (9.81\; {\rm m\cdot s^{-2}})} \\ &\approx \frac{25.0^{2}}{500\times 9.81} \\ &\approx 0.128\end{aligned}[/tex].

(Rounded up.)

The length of the rail is 3 m.

The length of the rail is determined by calculating the distance travelled by the skateboarder.

v² = u² + 2as

where;

10² = 8² + 2(6)s

100 = 64 + 12s

36 = 12s

s = 36/12

s = 3 m

Thus, the length of the rail is equal to the distance travelled by the skateboarder.

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The acceleration due to gravitational force on the moon and the Earth are different because the masses of the moon and the sun are different. Owing to the fact that gravity is a universal force, it is not possible to shield the gravitational field in a vacuum.

We know that there is a force that acts on any object that is found on the earth's surface. Now this force tends to act towards the center of the earth. We must realize that force is a vector and a such there must be the direction of a force at any given time that we are trying to observe the force.

We can then say that this force that acts on an object that has its direction as the center of the earth is what we call the gravitational force. This force does not only act on the earth but in the moon as well as act on other planets.

Having said this, let us bear in mind that the value of the acceleration due to gravity depends on the mass of the body. We know that the earth is larger than the moon hence the acceleration due to gravity is larger on the earth than on the moon.

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The magnitude of the net force on the object is 7.15 N.

The resultant velocity of the object is calculated as follows;

|v| = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)² ]

where;

|v| = √[(-3.75 - 4.04)² + (0 - -4.96)² + (4.44 - 4.44)² ]

|v| = 9.24 m/s

The net force experienced by the object is calculated as follows;

F = ma

F = (mv)/t

F = (2.97 kg x 9.24 m/s) / (3.84 s)

F = 7.15 N

Thus, we can conclude that the object will experience a net force of 7.15 N.

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The change to this process would shift the equilibrium to produce the maximum possible amount of SO3 is Decreasing the pressure. Option D.

Increasing the temperature shifts the equilibrium to the right with higher vapor concentrations, but keeping the system at this elevated temperature restores equilibrium. You can predict how given stress or change in conditions will affect equilibrium.

Sulfur trioxide formation is very slow at low temperatures. Increasing the temperature increases the speed at which equilibrium is reached. Decreasing the concentration of the reactants shifts the equilibrium to the left and produces less product. As the product concentration decreases the equilibrium shifts to the right producing more products.

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Answer: B

Explanation:

Answer:

uK=.9

Explanation:

friction=Fn(uK)

friction=135

Fn=150

135/150=.9

True because the vertical acceleration of a projectile is the acceleration due to gravity which is either 9.8 m/s/s or -9.8 m/s/s depending on perspective.

The distance travelled by the runner is 86.8 meters and the final velocity is 3.4 m/s.

Using kinematic equation we can find the distance travelled.

S = ut + 1/2 at^2

Where S = distance, u = initial velocity, t = time and a = acceleration.

The runner decelerates from a velocity of 9.00 m/s at a rate of 0.400 m/s^2. Initial velocity is 9.00 m/s and acceleration is -0.4 m/s^2 here. The distance travelled by the runner in the next 14 seconds is,

S = 9.00*14.0 + 1/2*(-0.4)*196

S = 86.8 m

The final velocity of the runner is using the formula v = u + at

v = 9.00+(-0.4)*14

v = 3.4 m/s

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iven:

he aangles of the stars;

Star A: p_A=0.67 arcsec

Star B: p_B=1.2 arcsec

Star C: p_C=0.35 arcsec

To find:xoplanation:

Order of the stars according to their distance.

he distacnce of a star in parsec is calculated using the formula,

Where p is the angles made by the stars in arcseconds.

On substituting the known values:

Star A,

Star B,

Star C,

inal answer:

Thus the closest star is star B and the farthest star is star C.

Thus the correct answer is option B.

The object with a mass of 2.3 kg that is the rock will have a weight of 22.5 N.

W = m * g

W = Weight

m = Mass

g = Acceleration due to gravity

g = 9.8 m / s²

For the book,

m = 1.1 kg

W = 1.1 * 9.8

W = 10.78 N

For the Box,

m = 4.5 kg

W = 4.5 * 9.8

W = 44.1 N

For the Rock,

m = 2.3 kg

W = 2.3 * 9.8

W = 22.5 N

Therefore, the object that has a weight of about 22.5 N is the rock

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The displacement between starting and endpoints is 13.89 units. Northeast.

A vector is an item that has both importance and direction. Geometrically, we can photo a vector as a directed line segment, whose duration is the importance of the vector, and with an arrow indicating the direction. The course of the vector is from its tail to its head. A displacement is a vector whose length is the shortest distance from the initial to the final function of a factor.

let 1 block = 1 unit

so, 14 block east means = 14î

      7 block north = 7 j

      2 block west = -2î

Net displacement in x- direction = 14î - 2î

                                                   = 12î

displacement towards y-direction = 7 j  

The resultant R = [tex]\sqrt{12^{2} +7^{2} }[/tex]

                          = [tex]\sqrt{193}[/tex]

                          =  13.89

The angle of displacement =   13.89/ 14î and  13.89/7 j  

 or

The angle of displacement = 12 tans 7/12

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The school of thought in which the instructor is most likely describing is functionalism and is denoted as option A.

This is referred to as a theory which is based on the doctrine of all aspects of a society and touches the mental states.

This type of theory also believes that institutions perform positive functions and so on. Darwin was a scientist who was involved in putting in positive efforts in the science field due to the skills which were developed and was solved many problems which is therefore the reason why Functionalism was chosen as the correct choice.

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Answer:

0.565

Explanation:

The distance is 34m , magnitude of the displacement is 6m and direction of displacement is towards west.

As per question, a dog walks 14 meters to the east and then 20 meters back to the west.

The distance moved for this motion is given below,

Distance = 14 + 20 m = 34m

The magnitude of the displacement and its direction are given below:

Displacement = 14 - 20 m = - 6m

negative sign indicates that the direction of displacement is towards west.

Magnitude of the displacement = 6m

Thus, the distance is 34m ,direction of displacement is towards west and  magnitude of the displacement is 6m.

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Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

Now for time:

Solving for t,

The total time the kangaroo takes in the air is 2.3s.

Given that the mass of each ball is m = 6.8 kg

The distance between them is d = 2r

Here, r is the radius of the ball.

The gravitational force of attraction is

We have to find the radius of the ball.

The gravitational force formula is

Here, the universal gravitational constant is

The radius will be

Thus, the radius of one of these balls is 0.114 m

The meters in the circuit that shows the amount of charge passing a point each second are A and C ( A )

In the circuit, the meter A and C are connected in series to a resistor and the meters B and D are connected parallel to a resistor. Ammeters should be connected in series and voltmeters should be connected in parallel because the resistance in ammeter is zero and the resistance in a voltmeter is infinite.

Ammeter is used to measure the current flowing through a circuit. It can also be said as amount of charge passing through the circuit. Voltmeter is used to measure the voltage across two points in a circuit.

Therefore, the meters in the circuit that shows the amount of charge passing a point each second are A and C ( A )

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Mary Cassatt has manipulated light and color in in the loge to emphasize the Division between male and female species.

The soft, blinking lighting fixtures of fireflies aren't just lovely—they'll additionally play a function in growing new species. A brand new observe suggests that the use of mild-up powers for courtship makes species break up off from every other at a faster tempo, presenting some of the clearest proof yet that the conflict to find mates shapes the range of life.

Males are extra colourful or ornamented than women in most, however no longer all, chook species. know-how this phenomenon calls for a primary draw close of the evolutionary forces that shape the conduct and morphology of people and species. Charles Darwin evolved plenty of the idea that allows explain this.

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The magnitude of the force in each case is;

a) 28 N

b) 14 N

c) -14 N

In physics, we know that force is a vector quantity. The force is a push or a pull and it must have direction thus we would see that the magnitude of the force would tend to differ in any direction that we would have to consider in the problem above. Hence, we shall now try to see the work that is done in each of the cases that involves the box as have been shown in the question that we can see above.

a) If the force have been applied horizontally, then the work that is done is the product of the mass,  the acceleration of the body

F = ma = 20 Kg * 1.4m/s^2 = 28 N

b) If the force is 30° above the horizontal then it has a vertical direction we have;

F = ma sin 30°

F = 20 Kg *   1.4m/s^2  * sin  30°  = 14 N

C) Now we have the negative vertical direction thus;

F = 20 Kg *   1.4m/s^2  * (-sin  30° ) = -14 N

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Given that as the sample of water is heated, the kinetic energy of its molecules increases.

Let's determine the sections of the heating curve which illustrates this process.

At point A, we can see the temperature of the water at 0 degrees, then as the temperature increases, the water molecules absorb more energy.

But as the phase changes, the temperature remains constant.

Since section A shows constant temerature, while section B also indicates constant temperature, we can sat section A followed by section B shows this process.

The sections which show as the energy overcome the attractions between molecules so that they begin to mover past one another can be said to be Section A followed by Section B

ANSWER:

D. A followed by B.

20 m/s is the final velocity of a boat that accelerates at 3 m/s² for twelve seconds if it started with a velocity of 4 m/s.

The speed at which that moves in a direction is known as its velocity. as the velocity of a car driving north on a road or even the pace at which a rocket takes off. Because the velocity vector was scalar, its integrand magnitude will always equal the motion's velocity.

Using the first equation of motion

V = U + at

where V = Final Velocity

U = Initial Velocity

a = Acceleration

t = Time

V = 4 + 3*12

V = 40 m/s.

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The coefficient of kinetic friction is equal to μ = F/N.

The coefficient of kinetic friction (fr), which is a numerical value, is obtained by dividing the resistive force of friction (Fr) by the normal or perpendicular force (N) pushing the objects together.

Static friction prevents the box from moving on its own; this friction must be overcome by an opposing force powerful enough to cause the box to move. Kinetic friction, sometimes referred to as dynamic friction, is a force that resists the relative movement of surfaces when they are in motion.

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( a ) R = u² sin 2θ / g

θ = 45°

R = 48 km = 48000 m

g = 9.8 m / s²

u² = ( 48000 * 9.8 ) /  sin 2 ( 45 )

u² = 470400 / 1

u = 685.86 m / s

( b ) H = u² sin²θ / 2g

H = 685.85² sin²45 / 2 * 9.8

H = 12098.15 m

( c ) The radius of Earth, the horizontal distance travelled by the shell and the distance from final position of cannon and center of Earth makes a right angled triangle.

According to Pythagoras theorem,

x² = 48² + ( 6.371 * 10³)²

x² = 2304 + ( 40.589641 * [tex]10^{6}[/tex] )

x² = ( 23.04 * 10² ) + ( 405896.41 * 10² )

x² = 405919.45 * 10²

x = 6371.18

At 48 km from the ship, the distance lower form the surface the shell lands  is,

d = 6371.18 - 6371

d = 0.18 km

Compared to the distance travelled of 48 km, the error of 0.18 km is insignificant

Therefore,

All choices offered are false.

The theme is heard first, followed by FIVE different variations of it.

===> 5

Explanation

o solve this , we need to use the formula:

so

Step 1

) Let

so, we need to convert from grams to kilograms ,to do that, divide the amount in grams by 1000

so

and

b) now, replace in the formula

so, the answer is 0.003999 Joules

hope this helps you

Answer:

The situation seems impossible because in order for the liquid to be tilted the glasses would also have to be tilted.

Explanation:

All of the glasses are in line and perfectly straight. There is no way that the liquid in the glasses would be tilted like how they are in the photo.

The speed  in meters per second if the string breaks suddenly is 9.6 m/s.

The angular speed is the speed of an object that is moving along a circular path. In this case, we have a puck of mass m = 0.095 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.69 m. The puck makes one revolution every t = 0.45 s.

Let us recall that in order to find the angular speed as is required by the question "how fast  in meters per second, does the puck move away?" we have to get the angle turned as well as the time taken.

We are told here that the angle turned is one revolution  or 2π radians and the time from the question is 0.45 s. Thus the speed is gotten from;

ω = θ/t

ω = 2π radians/0.45 s

ω = 13.96 rad/s

But v = rω

v = 13.96 rad/s * 0.69 m

v = 9.6 m/s

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The distance which this runner can cover in that time (8.5 seconds) if it starts from rest is equal to 39.7375 meters.

In order to determine the distance which this runner can cover in that time if it starts from rest, we would apply the second equation of motion. Mathematically, the second equation of motion is given by:

S = ut + ½at²

Where:  

Note: Initial velocity is equal to zero (0) since the runner starts from rest.

Substituting the given parameters into the formula, we have;

Distance, S = 0(8.5) + ½ × (1.1 × 8.5²)

Distance, S = ½ × (1.1 × 72.25)

Distance, S = 79.475/2

Distance, S = 39.7375 meters.

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Answer:

Fh=FN, where FN is larger as the box is being raised than when it was being held is correct.

Explanation:

Answer:

F ≈ 0,879 N

Explanation:

First, we calculate the value of the acceleration needed to stop the projectile.

That is:

V = V₀ + a.t

0 = 255 + a . (1,45)

a = - 255 / 1,45

a = -175,86 m/s²

Using the 2° Law of Newton:

∑F = m.a

F = m . a

F = 0,005 . (175,86)

F ≈ 0,879 N