A freight train has a mass of 3.8 x 10^7 kg. The locomotive can exert a force of 6.5 x 10^5 N to pull the train.

Required:
a. How long does it take the train to go from rest to 50 km/hr?
b. If there is air resistance of 2.0 x 10^5 N, what is the acceleration of the train, and how long does it take the train to go from rest to 50 km/hr?

Answers

Answer 1

Answer:

(a) The time taken for the plane to reach 50 km/hr is 811.97 s

(b) (i) the acceleration of the plane is 0.012 m/s²

    (ii) The time taken for the plane to reach 50 km/hr is 1172.85 s

Explanation:

Given;

mass of the freight train, m = 3.8 x 10⁷ kg

force applied to pull the train, F = 6.5 x 10⁵ N

a. How long does it take the train to go from rest to 50 km/hr?

Given;

initial velocity, u = 0

final velocity, v = 50 km/hr = 13.889 m/s

Apply Newton's second law of motion to determine the time traveled by the train at the given speed;

[tex]F = ma = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{6.5*10^5}\\\\t = 811.97 \ s[/tex]

(b)

(i) the acceleration of the plane

Given;

air resistance, = 2.0 x 10⁵ N

Apply Newton's second law of motion;

∑F = ma

6.5 x 10⁵ N - 2.0 x 10⁵ N = ma

4.5 x 10⁵ N = ma

[tex]a = \frac{4.5 * 10^5 }{m}\\\\a = \frac{4.5 * 10^5 }{3.8*10^7}\\\\a = 0.012 \ m/s^2[/tex]

(ii) time taken for the plane to reach 50 km/hr is given by;

[tex]F = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{4.5*10^5}\\\\t = 1172.85 \ s[/tex]


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It is the proportion predicted to be present in the early universe.

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Answers

Answer:

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Explanation:

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Answers

Answer:

Answer is: c. It must lose two electrons and become an ion.

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Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

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Answers

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

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F = Force between Romeo and Juliet = ?

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m₂ = mass of space station = 4.58 x 10⁵ kg

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Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

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Answer:

False

Explanation:

Answer:

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If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
how much time did the lawnmower run?

Answers

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

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Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

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Note: we don't use the mass at all

Our school needs to offer healthier options in the lunchroom. Elever High School has recently updated its cafeteria menu to include whole wheat pasta and breads, a fresh salad bar, and other healthy menu items. Students there claim that they have more energy and focus throughout their school day. Let's encourage healthier menus in our lunchroom!

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Answer:

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Explanation:

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Answers

Answer:

Explanation:

The total work done by the wave is expressed as;

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Workdone = mgh + 1/2mv²

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g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

The non-conservation work done will be "3833.83 Joules".

Given:

Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kg

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

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Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

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The total distance is

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[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

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Two identical conducting spheres each have a charge of 2 C. They have a radius of 0.1 m and are separated by 0.5 meters. If you were to increase the radius of these spheres to 0.2 meters, the electrostatic force between them would

Answers

Answer:

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