Car A travelled 300k in 5 hours, what was its average speed?

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

Answer:

Reference point.

Explanation:

Motion can be defined as a change in location (position) with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

In this scenario, three runners compete in a 400m race on an oval-shaped running track. At a certain time, one runner is 15m away from the finish line. Thus, the term which best describes the finish line is a reference point.

A reference point can be defined as a fixed point that typically marks the position from which a body starts its motion or change its location.

Answer:

Explanation:

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Answer:

1400.38N

Explanation:

Step one

Given data

P1= 250N

D1= 0.02m

A1= πD1^2/4

substitute

[tex]A1= 3.142*0.02^2/4\\\\A1=3.142*10^-4[/tex]

D2= 0.15m

A1= πD2^2/4

[tex]A2= 3.142*0.15^2/4\\\\A2=1.76*10^-3[/tex]

Required

The load P2

Step two:

Applying the hydraulic expression for a non-compressible fluid

we know that

Pressure= force/are

P1/A1=P2/A2

[tex]250/3.142*10^-4= P2/1.76*10^-3[/tex]

cross multiply we have

P2= 1.76*10^-3*250/3.142*10^-4

P2=0.44/3.142*10^-4

P2=1400.38N

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

[tex]F[/tex] - External force exerted on the sled, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.

[tex]W[/tex] - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

[tex]W = m\cdot g[/tex] (1)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:

[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]W = 490.35\,N[/tex]

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]W[/tex] - Weight of the sled, measured in newtons.

If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:

[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]

[tex]F_{min,s} = 147.105\,N[/tex]

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)

Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.

If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:

[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]

[tex]F_{min,k} = 49.035\,N[/tex]

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Answer:

The distance between gas species are assumed to be large.

Explanation:

Hello!

In this case, since the ideal gas condition is a hypothetical state of gases at which they are at low pressure and high temperature, because it is assumed the molecules are neither attracted nor repelled by each other, we can infer that the correct statement is "the distance between gas species are assumed to be large" because the low pressure and high temperature ensure the molecules are far away from each other and therefore allowing the ideal equation to be used to model the case, otherwise a rigorous equation of state such as Peng-Robinson, Redlich-Kwong and other should be used to model it as gases actually undergo interaction between their molecules.

Best regards!

Answer:

The answer is "wedge​"

Explanation:

A wedge is an item that tightens to a meager edge. Pushing the wedge one direction makes a force in a sideways direction. It is normally made of metal or wood and is used for parting, tightening, lifting, or fixing, as in making sure about a hammer head onto its handle.

The wedge was used in ancient occasions to part logs and shakes; an ax is also a wedge, similar to the teeth on a saw. As far as its mechanical capacity, the screw might be considered as a wedge folded over a cylinder.

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      Numbers represent seconds

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

Answer:

[tex]\boxed {\boxed {\sf 14 \ meters \ per \ second}}[/tex]

Explanation:

Speed is the rate of motion and can be found by dividing the distance by the time.

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

The distance covered is 56 meters and the time taken is 4 seconds.

[tex]d= 56 \ m \\t= 4 \ s[/tex]

Substitute the values into the formula.

[tex]s=\frac{56 \ m}{4 \ s}[/tex]

Divide.

[tex]s= 14 \ m/s[/tex]

The speed of the car is 14 meters per second

Answer:

One bulb could go out and the strand will stay on.

Explanation:

In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.

On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.

So, the correct option is:

One bulb could go out and the strand will stay on.

Answer:

The answer is true, as gravity affects everything.

Answer:

0 because j×j =0

Explanation:

so a×b is 0

Answer:

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistors R1 ,R2 and R3 in series connected to a battery of V volt.

we can connect voltmeter across each resistor

V = V1 + V2 + V3

I1 = current across resistor R1

I1 = V1/R1

similarly

I2 = current across resistor R2

I2= V2/R2

I3 = current across resistor R3

I3 = V3/R3

we will see that

I1 = I2 = I3

also it will be equal to V/(R1 + R2+R3)

Answer:

0.37 m

Explanation:

Let the shoulder be the origin.

The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.

The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones  from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m

Substituting these into the equation for the center of mass, we have

y = (m₁y₁ + m₂y₂)/(m₁ + m₂)

y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)

y = (1.20 kgm + 0.65 kgm)/5.0 kg

y = 1.85 kgm/5.0 kg

y =  0.37 m

The distance of the center of mass from the shoulder is thus y = 0.37 m

Given :

A pinball machine launches a .045 kg ball at a speed of 9.2 m/s.

To Find :

The potential energy of the spring just before it launched the pinball.

Solution :

We know, their is no external force applied on system.

It means that kinetic energy will remains constant.

Initial Energy = Final Energy

[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\0 + P.E_i= \dfrac{mv^2}{2}+ 0\\\\P.E_i = \dfrac{0.045 \times 9.2^2}{2}\ J\\\\P.E_i = 1.9044\ J[/tex]

Therefore, the potential energy of the spring just before it launched the pinball os 1.9044 J.

Answer:

a. Weight of Object in Water = 20 N

b. Up thrust = 20 N

c. Weight of Water Displaced = 20 N

Explanation:

a.

The weight of the object remains same in the water as well. Because, the same force of gravity is acting there as well. Hence,

Weight of Object in Water = 20 N

b.

Since, the object floats on the water. Therefore, according to Archimedes' principle the up thrust force acting on the object must be equal to the weight of object:

Up thrust = Weight of object

Up thrust = 20 N

c.

From Archimedes' Principle, we know that the up thrust or the Buoyant force is equal to the weight of the water displaced by the object. therefore:

Weight of Water Displaced = Up thrust

Weight of Water Displaced = 20 N

Answer:

c

Explanation:

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

Answer:

inonic bond has greatest difference in electronegativiru

The type of bond that has elements with the greatest difference in electronegativity ionic bond. The correct option is is B.

Electronegativity is a chemical property that describes an atom's or functional group's tendency to attract electrons toward itself.

An atom's electronegativity is affected by both its atomic number and the distance between its valence electrons and the charged nuclei.

Electronegativity is a measurement of an atom's proclivity to attract electrons (or electron density) towards itself.

It governs the distribution of shared electrons between two atoms in a bond. The greater an atom's electronegativity, the more strongly it attracts electrons in its bonds.

Ionic bonds are formed by elements with large differences in electronegativity. Covalent bonds are formed between atoms of elements with similar electronegativity.

Thus, the correct option is B.

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4.80m/s

(d) Since v y >0 when the ball hits the wall, it has not reached the highest point yet.

Answer:

The value is  [tex]\alpha = 3.84 \ rad/s^2[/tex]

Explanation:

From the question we are told that

    The constant angular speed is  [tex]w = 11.9 \ rad/s[/tex]

     The time taken is  [tex]t = 3.10 \ s[/tex]

Generally the magnitude of the angular acceleration is  mathematically represented as  

          [tex]\alpha = \frac{w}{t}[/tex]

=>     [tex]\alpha = \frac{11.9}{ 3.10 }[/tex]

=>     [tex]\alpha = 3.84 \ rad/s^2[/tex]

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

W = Work (J)

F = Force / weight (N)

s = distance (m)

Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.

Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.

Work done is a vector quantity thus, characterised by a magnitude and direction.  The common unit of work done is joule.

Given that force applied on the weight = 10 N

displacement occurred = 40 m - 4 m = 36 m

Work done = F . ds

ds = 36 m and f = 10 N

Then W = 10 N × 36m

             = 360 J.

Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.

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Answer:

(a)When ignoring air resistance its accelerating increases steadily .

(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.

Hope this helped.

Answer: 9.3m/s

Explanation:

Your question isn't complete but let me help out:

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?

We would use Newton's law of motion to solve this which goes thus:

F = ma

f = m(v-u)/t

Cross multiply

ft = m(v-u)

where,

f = 55

t = 45/1000 = 0.045

m = 0.0060

u = -32

v = Unknown

Therefore,

55 × 45/1000 = 0.060(v - -32)

55 × 0.045 = 0.060(v + 32)

2.475 = 0.06(v + 32)

2.475 = 0.06v + 1.92

0.06v = 2.475 - 1.92

0.06v = 0.555

v = 0.555/0.06

v = 9.25m/s

v = 9.3m/s Approximately

Answer:

A.9.3 m/s

Explanation:

Answer:

Of course we feel bad or angry.

Explanation:

that it what i know....

Answer:

a. 6602.7 N b. 64.44 km/h

Explanation:

a. Find the force of friction required to keep it from  sliding down the hill.

The frictional force f equals the component of the weight of the car, W perpendicular to the inclined hill = Wcosθ times the coefficient of static friction, μ = 0.45.

Since f = μN = μWcosθ = μmgcosθ where m = mass of car = 1500 kg, g = acceleration due to gravity = 9.8 m/s² and θ = angle of incline of hill = 3.5°

So, f = μmgcosθ

= 0.45 × 1500 kg × 9.8 m/s²cos3.5°

= 6615cos3.5°

= 6602.7 N

b. Determine the speed of the car after it emerges from the mud (in km/h)

Since the car drops a vertical height of 14 m, its potential energy decreases by mgh and its kinetic energy increases by mgh where m =mass of car and h = height drop = 14 m. So its kinetic energy increase is ΔK = mgh = 1500 kg × 9.8 m/s² × 14 m = 205800 J

Since it has an initial velocity of u = 50 km/h = 50 km/h 1000m/3600 s = 13.89 m/s, its initial kinetic energy is K = 1/2mu² = 1/2 × 1500 kg × (13.89 m/s)² = 144699.08 J.

Its new kinetic energy after the drop is thus K' = K + ΔK = 144699.08 J + 205800 J = 350499.08 J

Let v be its velocity after the drop, since K' = 1/2mv²,

v = √(2K'/m) = √(2 × 350499.08 J/1500 kg) = √(700998.16 J/1500 kg) = √(467.332 J/kg) = 21.62 m/s

Now, from work kinetic energy principles, the kinetic energy change in the car is the work done on car by friction

So, ΔK" = -fd = -μmgd

Let v' be the velocity of the car after emerging from the mud and moving a distance d = 30.0 m.

So, 1/2m(v'² - v²) = -μmgd

v'² - v² = -2μgd

v'² = v² - 2μgd

Substituting the values of the variables, we have

v'² = (21.62 m/s)² - 2 × 0.25 × 9.8 m/s² × 30.0 m

v'² = 467.42 m²/s² - 147 m²/s²

v'² = 320.42 m²/s²

taking square root of both sides, we have

v' = √(320.42 m²/s²)

= 17.9 m/s

Converting v to km/h we have v' = 17.9 m/s × 3600 s/h × 1 km/1000 m = 64.44 km/h.

So, the car emerges from the mud with a speed of 64.44 km/h

Answer:

it comprises of the DNA/RNA bipolymer molecules