Part 3/3 find the thermal energy transferred. Answer in units of KJ

Part 3/3 Find The Thermal Energy Transferred. Answer In Units Of KJ

Answers

Answer 1

0.0We are asked to determine the volume of a gas given the moles, the temperature, and the pressure. To do that, we use the following formula:

[tex]PV=nRT[/tex]

Where:

[tex]\begin{gathered} P=pressure \\ V=\text{ volume} \\ n=\text{ number of moles} \\ R=\text{ universal gas constant} \\ T=\text{ temperature} \end{gathered}[/tex]

Now, we substitute the values for the final state. Since the process is isothermally this means that the final temperature is the same as the initial temperature. we get:

[tex](1.8atm)V_f=(2mol)(8.31415\frac{J}{Kmol})(243K)[/tex]

We need to convert the pressure from atmospheres to Pascals. To do that we use the following conversion factor:

[tex]1atm=101325Pa[/tex]

Multiplying by the conversion factor we get:

[tex]1.8atm\times\frac{101325Pa}{1atm}=182385Pa[/tex]

Now, we substitute the value in the formula:

[tex](182385Pa)V_f=(2mol)(8.3145\times\frac{J}{Kmol})(243K)[/tex]

Solving the operations:

[tex](182385Pa)V_f=4040.847J[/tex]

Now, we divide both sides by 182385Pa:

[tex]V_f=\frac{4040.847J}{182385Pa}[/tex]

Solving the operations:

[tex]V_f=0.022m^3[/tex]

Therefore, the final volume is 0.022 cubic meters.

Part B. We are asked to determine the work done. To do that we will use the formula for isothermic work:

[tex]W=nRT\ln(\frac{P_0}{P_f})[/tex]

Where:

[tex]P_0,P_f=\text{ initial and final pressure}[/tex]

Now, we plug in the values:

[tex]W=(2mol)(8.31451\frac{J}{Kmol})(243K)\ln(\frac{0.21atm}{1.8atm})[/tex]

Now, we solve the operations:

[tex]W=-8681.51J[/tex]

Therefore, the work done is -8681.51 Joule. To convert to kilojoules we divide by 1000:

[tex]W=-8681.51J\times\frac{1kJ}{1000J}=-8.68kJ[/tex]

Part 3. In an isothermic process the change in internal energy is zero, therefore, according to the first law of thermodynamics we have:

[tex]Q-W=0[/tex]

Therefore:

[tex]Q=W[/tex]

Therefore, the amount of heat is equal to the amount of work. Therefore, the thermal energy transferred is:

[tex]Q=-8.68kJ[/tex]


Related Questions

if the fundamental frequency of an 80 cm long guitar string is 450 Hz, what is the speed of the traveling waves?

Answers

Given that the frequency of the wave is, f = 450 Hz

and the length of the guitar string is, L = 80 cm = 0.8 m

We have to find the speed of the wave, v.

As the frequency is the fundamental frequency, so it will be the first harmonic.

The formula to find the wavelength is

[tex]\lambda=2L[/tex]

Substituting the values, we get

[tex]\begin{gathered} \lambda=2\times0.8 \\ =1.6\text{ m} \end{gathered}[/tex]

The formula to find the speed of the traveling wave is

[tex]v=\lambda\times f[/tex]

Substituting the values, we get

[tex]\begin{gathered} v=1.6\times450 \\ =720\text{ m/s} \end{gathered}[/tex]

Thus, the speed of the traveling wave is 720 m/s

If you swing an object on a string around in a circle, how can you feel the effects of the centripetal force?(1 point)


A in the mass of the object
B in the length of the string
C in the tension in the string
D in the speed of the object

Answers

The effects of the centripetal force is felt in the tension in the string.

option C is the correct answer.

What is centripetal force?

Centripetal force is the inward or radial force experienced by an object moving in a circular path.

Mathematically, the centripetal force experienced by an object moving in a circular path is given as;

F = ma

where;

m is the mass of the objecta is the centripetal acceleration

F = mv²/r

where;

m is the mass of the objectv is the velocity of the objectr is the radius of the circular path

The centripetal force on the object is measured in Newton and it is equal to the tension in the string.

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Answer:

C in the tension in the string

Explanation:

Please helpppppppppp - the diagram shows the trajectory of a ball that is thrown horizontally from a top of a building. The ball’s vertical and horizontal velocity vectors along with the resultant factors are also indicated if the ball takes three seconds to reach the ground how fast is it moving by the time it reached the ground

Answers

[tex]\begin{gathered} t=3\text{ s} \\ g=9.81m/s^2 \\ v_y=\text{ (}9.81m/s^2\text{)(3s)} \\ v_y=\text{ 29.43 m/s} \\ v_x=1.9\text{ m/s} \\ v=\sqrt{v^2_y+v^2_x} \\ v=\sqrt{(\text{ 29.43 m/s})^2+(1.9\text{ m/s})^2} \\ v=29.49\text{ m/s} \\ \text{The velocity of the ball when it hits the ground is }29.49\text{ m/s} \end{gathered}[/tex]

tion 14A 37 kg child holds a 5.9 kg package in a 67 kg boat at rest, and then the child throws the packagehorizontally from the boat at a velocity of +12.4 m/s. What is the velocity of the boat after this?Vunits

Answers

Answer:

-0.704 m/s

Explanation:

The conservation of momentum demands that since the child + boat + box system is isolated, the initial momentum must equal the final momentum.

Now since initially, everything is at rest, the initial momentum of the system is zero.

The final momentum of the system is

[tex](m_c+m_b)v_1+(m_{\text{box}})v_2[/tex]

where

m_c = mass of the child

m_b = mass of the boat

m_box mass of the boat

v1 = belcity of child + boat

v2 = veloctiy of the box.

Equating the initial momentum to the final momentum gives

[tex](m_c+m_b)v_1+(m_{\text{box}})v_2=0[/tex]

Now in our case

m_c = 37 kg

m_b = 67 kg

v1 = unknown

m_box = 5.9 kg

v2 = 12.4

Therefore, the above equation gives

[tex](37+67)v_1+(5.9)(12.4)=0[/tex]

solving for v1 gives

[tex]v_1=\frac{5.9\cdot12.4}{37+67}[/tex]

which evaluates to give (rounded to the nearest hundredth)

[tex]\boxed{v_1=-0.70m/s}[/tex]

which is our answer!

A/An _____ is described as a device that detects current differences and then opens the circuit preventing electrocution.circuit breakerfuseground fault interruptershort circuit

Answers

ANSWER

ground fault interrupter

EXPLANATION

A ground fault interrupter is an automatic switch that measures the difference of current between the "hot" and neutral wires. If this difference is greater than a fixed values (only a few milliamperes) the switchs opens and stops the circulation or current.

If a person is in contact with a wire in the circuit the current is diverted through the person's body to the ground. Therefore the ground fault interrupter will detect a difference of current. Hence this device is used to prevent electrocution.

A block with a mass m1 is hit by a force of magnitude F which causes the block to have an acceleration of magnitude a. If a second block of mass m2 is hit by the same force of magnitude F which causes the block to have an acceleration of magnitude 2a, then which of these could be the two masses? A) m1= 200kg ; m2= 100kgB) m1= 50kg ; m2= 25 kgC) m1= 100kg ; m2= 50kgD)m1= 10kg ; m2= 50kgE)Any of these

Answers

We are given that a force "F" accelerates an object of mass "m1". According to Newton's second law, this can be represented by the following equation:

[tex]F=m_1a[/tex]

Now, we are given that a second object of mass "m2" is accelerated by "2a" using the same force. Using Newton's second law we get:

[tex]F=m_2(2a)[/tex]

Now, we will divide both equations, we get:

[tex]\frac{F}{F}=\frac{m_1a}{m_2(2a)}[/tex]

Now, we simplify by canceling put the "F" and the "a":

[tex]1=\frac{m_1}{2m_2}[/tex]

Now, we multiply both sides by "2m2", we get:

[tex]2m_2=m_1[/tex]

Therefore, the first mass must be twice the second mass.

The options that meet this condition are:

[tex]m_1=200kg,m_2=100kg\text{ }[/tex][tex]m_1=50kg,m_2=25kg[/tex][tex]m_1=100kg,m_2=50kg[/tex]

9. The graph is a plot of the velocity versus time for an objectmoving in a straight line. The x position of the objectat t = 0 seconds is 0 meters. At what time after t = 0 secondsdoes the object again pass through its initial position?(a) 1 second(b) Between 1 and 2 seconds(c) 2 seconds(d) 3 seconds

Answers

Velocity:

The velocity in simple word, is defined as the change in the position of the particle with respect to the time. The velocity is considered as a vector quantity as it has both the magnitude and direction in it. It can be measured or calculated in meter per second.

At t = 2 seconds, the speed of an object is -10 m/s. As it passes through its initial position. Hence the correct answer is (2)

The postal service will not ship goods over 50 lbs without a special label. Donovan wants to estimate the weight of his package so he doesn't exceed the weight limit. He has a cast iron skillet that weighs 5.67 lbs, a dictionary that weighs 8.34 lbs, and a set of dishes that weighs 37.88 lbs. What will the estimated weight of Donovan's package be if he rounds each item to the nearest pound before totaling the weight?

Answers

52 Lb

Explanation

Step 1

round each itme to the nearest pound.

in this case, Rounding a price to the nearest pound is the same as rounding a decimal to the closest whole number

so

for example, If the price is $3.80 you can round up to $4 because the number in the tenths position is 8. The closest whole number to 3.8 is 4

hence

[tex]\begin{gathered} cast\text{ iron=5.67 Lbs}\rightarrow6\text{ LB} \\ dictionary=8.34\text{ Lb}\rightarrow8\text{ Lb} \\ a\text{ set of disehes=}37.88\text{ Lb}\rightarrow38\text{ Lb} \end{gathered}[/tex]

Step 2

now, add the cost of the items to find the weight of the package

[tex]\begin{gathered} Weight_{package}=Weight_{cast}+Weight_{dictironay}+Weight_{dishes}\text{ } \\ \text{replace} \\ Weight_{package}=6\text{ lb+8 lb +38 lb=52 lb} \end{gathered}[/tex]

therefore, the answer is

52 Lb

I hope this helps you

A dog must be at least 17 pounds to enter the dog park.Which description best represents the weight the dog needs to be?Any value less than or equal to 17Any value equal to 17Any value greater than 17Any value greater than or equal to 17

Answers

ANSWER:

4th option: Any value greater than or equal to 17

STEP-BY-STEP EXPLANATION:

According to the statement, the first must be at least 17 pounds to enter the dog park, which means that the minimum weight is 17 pounds.

Which means the best way to represent it is any value greater than or equal to 17.

. A person pushes on a hockey puck with their stick at an angle so the vertical force is 22 N[down] and the horizontal force is 45 N [forward]. Assume the ice is frictionless.a) What is the actual force the hockey player transmits to the puck?b) What is the work done by the person pushing the hockey stick if they push the puck for 3.0s as it moves with a constant velocity of 22 m/s [forward]?c) What is the significance of the fact that both the horizontal force and motion are bothforwards?

Answers

50.08 Newton is the actual force the hockey player transmits to the puck and  3305.93 Joule is the work done by the person pushing the hockey stick if they push for 3.0s as it moves with a constant velocity of 22 m/s

(a)

Given, horizontal force Fx = 45 N, vertical force Fy = 22 N

Thus, the total force acting on the  is F = [(Fx)^2 + (Fy)2]^1/2

Therefore, F = [(45N)^2 + (22 N)2]^1/2 = 50.08 N

(b)

Constant velocity v = 22 m/s and time interval t = 3.0 s

The horizontal distance travelled by the is x = v t = (22 m/s) (3.0 s) = 66 m

The work done by the person pushing the hockey stick is W = F. x = (50.08 N) (66 m) = 3305.93 J

(c) The magnitude of the horizontal force is greater than (almost double) the vertical force, so the motion of  is in the same direction as the horizontal force. The horizontal force dominates here.

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Homework: Action-reaction forces area. equal in magnitude and point in the same directionb. equal in magnitude and point in opposite directionsc. unequal in magnitude but point in the same directiond. unequal in magnitude and point in opposite directions

Answers

We will have that they are:

Equal in magnitude and point in opposite directions. {Option B]

To the correct number of significant figures: 2999 - 0.1 =

Answers

ANSWER:

2999, 4 significant figures

STEP-BY-STEP EXPLANATION:

We have the following:

[tex]\begin{gathered} 2999-0.1=2998.9 \\ 2998.9\cong2999\rightarrow Decimals\colon0 \end{gathered}[/tex]

Therefore, we can determine that in total there are 4 significant figures

A 100-N ball suspended by a rope A is pulled to one side horizontally by another rope B and supported so that the rope A makes an angle of 30º with the vertical wall (see figure). Find the tensions of the ropes A and B. You solve it for me by fi uwu

Answers

Tension in A rope = 58.115 N

Tension in B rope =  116.23 N

What is tension?

Tension is defined as the force transmitted through a rope, cord, or wire when pulled by forces acting from opposite sides. Tension is transmitted along the length of the wire, drawing energy evenly into the bodies at both ends. T = mg + ma

Where;

T = tension, (N)

m = mass (kg)

g = gravitational force, 9.8 m/s²

A = acceleration (m/s²)

As, F = mg

100 = m x 9.8

m = [tex]\frac{100}{9.8}[/tex]

m = 10.20 kg

Let the tension in A rope be T₁ and in the B rope be T₂ which is making angle of 30⁰

The vertical component of tension T₂ will balance the weight

= T₂ cos 30 = 10.20 x 9.8

T₂ = [tex]\frac{99.96}{cos 30}[/tex]

Since cos 30 = [tex]\frac{\sqrt{3} }{2}[/tex]

T₂ = [tex]\frac{99.96}{0.86}[/tex]

T₂ = 116.23 N

The horizontal component of T₂ will balance T₁

T₂ sin 30 = T₁

116.23 sin 30 = T₁

T₁ = 116.23 × [tex]\frac{1}{2}[/tex]

T₁ = 58.115 N

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In 1897,J.J Thompson would gain recognition for discovering the electron and would later win a noble prize for his ________ model of the atom?

Answers

Answer:

Plum pudding

Explanation:

Note these points:

• J.J Thompson performed the Cathose ray experiment

,

• From this experiment, he discovered that all atoms contain tiny negatively charged subatomic particles called electrons

,

• J.J Thompson proposed the plum pudding model of the atom

,

• This model contained negatively-charged electrons embedded within a positively charged soup

,

• It was on April 30, 1897 that J.J Thompson announced the presence of the electrons in the atom, a result of his cathode ray experiment

From the points highlighted above, we can conclude that J.J Thompson gained recognition fpr discovering the electron and also won a noble prize for his plum pudding model of the atom

a student throws a rock straight down words from a bridge into a river below if initial speed of the rock is 10.0 Ms and it takes 2.1 s to reach the river how high is the bridge

Answers

Since this is a free fall and all the motion is going down, let's make down the positive direction (that way we won't use minus signs). A free fall is an uniform accelerated motion, this means that we can use the formula:

[tex]y-y_0=v_0t+\frac{1}{2}at^2[/tex]

In this case we know the initial velocity is 10 m/s, the time is 2.1 and the acceleration of gravity is 9.8 m/s^2. Plugging this values we have that:

[tex]\begin{gathered} y-y_0=10(2.1)+\frac{1}{2}(9.8)(2.1)^2 \\ y-y_0=42.609 \end{gathered}[/tex]

This means that the rock fell 42.609 meters. Therefore, the bridge is 42.609 m tall.

1) The net external force on a golf cart is 390 N north. If the cart has a total mass of 270 kg, what arethe magnitude and directions of its acceleration?

Answers

Given data

*The net external force on a golf cart is F = 390 N

*The cart has a total mass is m = 270 kg

The formula for the magnitude of the acceleration of the cart is given as

[tex]a=\frac{F}{m}[/tex]

Substitute the known values in the above expression as

[tex]\begin{gathered} a=\frac{390}{270} \\ =1.44m/s^2 \end{gathered}[/tex]

Hence, the magnitude of the acceleration of the cart is 1.44 m/s^2. The direction of the acceleration is towards the north direction.

A uniform 500 N/C electric field points in the positive y-direction and acts on an electron initially at rest. After the electron has moved4.00 cm in the field, what is the energy of electron in eV?

Answers

Given data:

* The electric field in the y-direction is,

[tex]E=500\text{ N/C}[/tex]

* The distance traveled by the electron is,

[tex]\begin{gathered} d=4\text{ cm} \\ d=0.04\text{ m} \end{gathered}[/tex]

Solution:

The work done in terms of electric field is,

[tex]W=\text{Edq}[/tex]

where q is the charge on an electron,

Substituting the known values,

[tex]\begin{gathered} W=500\times0.04\times1.6\times10^{-19}\text{ J} \\ W=\frac{500\times0.04\times1.6\times10^{-19}}{1.6\times10^{-19}}\text{ eV} \\ W=500\times0.04\text{ eV} \\ W=20\text{ eV} \end{gathered}[/tex]

This work done is stored in the charge in form of energy.

Thus, the energy of the electron in eV is 20 eV.

What is the period of the wave?
2.5 seconds
0.4 cycles/second (hertz)
3.9 cm
1.6 cm/s

Answers

Answer:

2.5 seconds

Explanation:

Perios is time divided by number of oscilliations. From the diagram the number of oscilliations is two. This can be seen by counting successive crests of the wave from the 0s mark. The 2 oscilliatoions ends in 5s. 5 divided by 2 is 2.5s.

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?Hz(b) What is its wavelength in glass that has an index of refraction of 1.63?nm(c) What is its speed in the glass?Mm/sNeed Help?Read ItWatch ItSubmit Answer

Answers

ANSWER:

(a) 4.74*10^14 Hz

(b) 388.22 nm

(c) 184 Mm/s

STEP-BY-STEP EXPLANATION:

We have the following information:

[tex]\begin{gathered} \lambda=632.8\text{ nm} \\ n=1.63 \end{gathered}[/tex]

(a)

To calculate the frequency we use the following formula:

[tex]\begin{gathered} f=\frac{c}{\lambda} \\ c=3\cdot10^8\text{ m/s} \\ \lambda=632.8\text{ nm }=632.8\cdot10^{-9}\text{ m} \\ \text{replacing:} \\ f=\frac{3\cdot10^8}{632.8\cdot10^{-9}} \\ f=4.74\cdot10^{14}\text{ Hz} \end{gathered}[/tex]

(b)

In this case, we apply the following:

[tex]\begin{gathered} \lambda_1=\frac{\lambda}{n} \\ \text{ replacing} \\ \lambda_1=\frac{632.8}{1.63} \\ \lambda_1=388.22\text{ nm} \end{gathered}[/tex]

(c)

To calculate the speed it would be:

[tex]\begin{gathered} v=\frac{c}{n} \\ \text{ replacing} \\ v=\frac{3\cdot10^8}{1.63} \\ v=1.84\cdot10^8 \\ v=184\text{ Mm/s} \end{gathered}[/tex]

If a convex lens (converging) has a focal length of 12 cm, where would you place an object in order to produce an image that has the same height, but inverted?Less than 12 cm6 cm12 cm24 cm

Answers

Given that the focal length is F = 12 cm.

We need to find the position of the object such that the image obtained should be inverted, the same height as that of the object.

In a convex lens, for the image to be inverted and the same height as the object, it should be placed at 2F.

So, the object position will be

[tex]\begin{gathered} 2F=2\times12 \\ =24\text{ cm} \end{gathered}[/tex]

Thus, the object position should be 24 cm.

Imagine an asteroid at rest in space that cracks into two pieces. The first piece moves at 1.4 times the speed of the second piece. Calculate the ratio of the first piece’s mass to the second piece’s mass

Answers

Since the total linear momentum of the system was 0 before the asteroid cracking into two pieces, then the magntude of the linear momentum of each piece must be the same after the cracking.

Then:

[tex]m_1v_1=m_2v_2[/tex]

If the speed of the first piece is 1.4 times the speed of the second, then:

[tex]\begin{gathered} v_1=1.4v_2 \\ \Rightarrow m_1\times1.4v_2=m_2v_2 \\ \Rightarrow1.4m_1=m_2 \\ \Rightarrow m_1=\frac{m_2}{1.4} \\ \Rightarrow\frac{m_1}{m_2}=\frac{1}{1.4} \end{gathered}[/tex]

Then, the ratio of the first piece's mass to the second piece's mass is 1:1.4

what is the greatest mass of groceries that can be lifted safely with this bag given that the bag is raised with an acceleration of 1.80 m/s^2

Answers

We will have the following:

According to the image:

First, we remember that:

[tex]F=m\cdot a[/tex]

Now, we will determine he maximum mass will be:

[tex]52.0N=m\cdot(1.8m/s^2)\Rightarrow m=\frac{^{}52.0N}{1.80m/s^2}[/tex][tex]\Rightarrow m=\frac{260}{9}kg\Rightarrow m\approx28.9kg[/tex]

So, the maximum mass will be approximately 28.9 kg.

A boat with initial position (5i -2j+ 4k) metres relative to port, acceleratesuniformly from initial velocity (4i+ j-3k) m s^-1, for 8 seconds until reaching finalvelocity (12i-7j+13k) m s^-1.a)Find the position of the object after 8 seconds.b) Find the acceleration of the object.

Answers

Given,

The initial position of the boat, d_i=(5i-2j+4k) m

The initial velocity of the boat, u=(4i+j-3k) m/s

The time period, t=8 s

The final velocity of the boat, v=(12i-7j+13k) m/s

To find:

a) The position of the object after 8 s

b) The acceleration of the object.

Explanation:

a)

From the equation of motion, the total distance traveled by the object is given by,

[tex]d=\frac{1}{2}(v+u)t[/tex]

On substituting the known values,

[tex]\begin{gathered} d=\frac{1}{2}[(12\hat{i}-7\hat{j}+13\hat{k)}+(4\hat{i}+\hat{j}-3\hat{k})]8 \\ =4\times(16\hat{i}-6\hat{j}+10\hat{k}) \\ =(64\hat{i}-24\hat{j}+40\hat{k})\text{ m} \end{gathered}[/tex]

Thus the final position of the boat is,

[tex]\begin{gathered} d_f=d_i+d \\ =\left(5\hat{i}-2\hat{j}+4\hat{k}\right)+(64\hat{i}-24\hat{j}+40\hat{k}) \\ =(69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{gathered}[/tex]

b)

The acceleration of an object is given by,

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

On substituting the known values,

[tex]\begin{gathered} a=\frac{(12\hat{i}-7\hat{j}+13\hat{k)}-(4\hat{i}+\hat{j}-3\hat{k})}{8} \\ =\frac{8\hat{i}-8\hat{j}+16\hat{k}}{8} \\ =(\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{gathered}[/tex]

Final answer:

a) Thus the position of the boat after 8 s is

[tex]\begin{equation*} (69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{equation*}[/tex]

The acceleration of the boat is

[tex]\begin{equation*} (\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{equation*}[/tex]

A woman holds a makeup mirror with a radius of curvature of 120 cm a distance of 20 cm from her face. What is the magnification of the observed image?

Answers

Given:

The radius of curvature is R = -120 cm

The object distance is u = -20 cm

Required: Magnification of the observed image.

Explanation:

The focal length can be calculated using the concave mirror as

[tex]\begin{gathered} f=\frac{R}{2} \\ =\frac{-120}{2} \\ =-60\text{ cm} \end{gathered}[/tex]

The image distance can be calculated using the mirror formula as

[tex]\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =\frac{1}{-60}-\frac{1}{-20} \\ v=30\text{ cm} \end{gathered}[/tex]

The

A series circuit has more than one path for the electric current to follow true or false

Answers

Answer:

False, because a series circuit is one loop.

A newly invented ride called Crazy Box in an amusement park has a strong magnet. The magnet accelerates the boxcar and its riders from zero to 35 m/s in 5 seconds. Suppose the mass of the boxcar and riders is 6,000 kg. What is the acceleration of the boxcar and its riders? What is the average net force exerted on the boxcar and riders by the magnets?

Answers

Answer:

Below

Explanation:

   Acceleration = change in velocity / change in time = 35/5 = 7 m/s^2

F = m * a

  = 6000 kg  *   7  m/s^2 = 42 000 N

The average net force exerted on the boxcar and its riders by the magnets is 42,000 N.

What is the average net force?

The average net force exerted on the boxcar and riders by the magnets is calculated as;

F(net) = ma

where;

m is massa is acceleration

The acceleration is calculated as;

a = Δv / Δt

a = 35 m/s / 5 s

a = 7 m/s²

The acceleration of the boxcar and its riders is 7 m/s².

The average net force exerted on the boxcar and riders by the magnets is;

F = 6,000 kg x 7 m/s²

F = 42,000 N

Thus, the average net force exerted on them is determined as 42,000 N.

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The force of attraction between m1 and m2 is 76 N. What will this force become if m2 is tripled?

Answers

The new force is obtained as 228 N since the product of the masses is directly proportional to the force.

What is the new force?

From the Newton law of universal gravitation, we can see that the gravitational force that exists between two masses is directly proportional to the product of the masses and inversely proportional to the distance between the masses.

Where the distance between the masses, remain the same, we can see that the new force would be three times the initial force since the the force and the mass are directly proportional.

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Please help me!A boy on a skateboard travels the first 600m of a trip at an average speed of 2 m/s. He then travels the next 800m in 200s and spends the last 100s at a speed of 5 m/s. Find the average speed of the bicyclist for this trip.

Answers

average speed = total distance / total time

speed x time = distance

part 1

D1= 600 m, s1=2m/s

Part 2

d2=800m, t2=200s

Part 3

s3=5m/s ,t3=100s

time1 = Distance1/speed1 = 600m/2m/s = 300s

Distance3 = speed3 x time 3 = 5 m/s x 100s = 500m

Total distance = d1+d2+d3 = 600 + 800 + 500 = 1900m

Total time = t1+t2+t3 =300 + 200 + 100 = 600 s

Avg speed = 1900m/600s = 3.17 m/s

A park ranger driving on a back country road suddenly sees a deer in his headlights 20m ahead. The ranger, who is driving at 11.4 m/s, immediately applies the brakes andslows down with an acceleration of 3.80 m/s2. How much distance is required for theranger's vehicle to come to rest? Only enter the number, not the units,

Answers

Given data

*The speed of the ranger who is driving at 11.4 m/s

*The given acceleration is a = -3.80 m/s^2

The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as

[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]

*Here v = 0 m/s is the initial speed of the ranger's vehicle

Substitute the values in the above expression as

[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]

The distance is required for the ranger's vehicle to come to rest is calculated as

[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]

The stopping time is calculated as

[tex]v=v_0+at[/tex]

Substitute the values in the above expression as

[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]

A farm tractor tows a 3300 kg trailer up a 14 degree incline with a steady speed of 2.8 m/s. What force does the tractor exert on the trailer?

Answers

We are asked to determine the force that the tractors is exerting on a trailer up an incline. A free-body diagram of the situation is the following:

Where:

[tex]\begin{gathered} F=\text{ force of the tractor} \\ m=\text{ mass of the trailer} \\ g=\text{ acceleratio of gravity} \end{gathered}[/tex]

Now, we add the forces in the direction of the incline:

[tex]\Sigma F_x=F-mg_x[/tex]

To determine the x-component of "mg" we use the following right triangle:

Now, we use the function sine to determine the value of "mgx":

[tex]\sin14=\frac{mg_x}{mg}[/tex]

Now, we multiply both sides by "mg":

[tex]mg\sin14=mg_x[/tex]

Now, we substitute the values of "m" and "g":

[tex](3300kg)(9.8\frac{m}{s^2})\sin14=mg_x[/tex]

Solving the operations:

[tex]7823.75N=mg_x[/tex]

Now, we substitute the value in the sum of forces:

[tex]\Sigma F_x=F-7823.75N[/tex]

Since the object is moving at a steady speed this means that the sum of forces is zero:

[tex]F-7823.75N=0[/tex]

Now, we add 7823.75N to both sides:

[tex]F=7823.75N[/tex]

Therefore, the tractor exerts a force of 7823.75N

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