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2 years ago

7

A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

train move away from the track gang and what was its speed after the 5 min.?

1 answer:

Amanda [17]2 years ago

3 0

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check: V0 = 5 m/s

V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s the speed after 300 sec

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The truck went 1,200 miles
on 55 gallons

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The answer I believe is 6

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In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

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3 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

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3 years ago

A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.

ohaa [14]

Answer:

speed of a particle = 2.31 × $10^{6}$ m/s

energy of proton required = 27.77 KeV

energy of electron required = 15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB ..............1

and force due to electric field E = qE .....................2

so without deflection force due to magnetic field = force due to electric field

so here qvB = qE

and V = $\frac{E}{B}$ ...................3

put here value

V = $\frac{0.51*10^6}{0.22}$

speed of a particle = 2.31 × $10^{6}$ m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × $\frac{V^2}{2}$

put here value

energy of proton required = 1.67 × $10^{-27}$ × $\frac{(2.31*10^6)^2}{2}$

energy of proton required = 4.45 × $10^{-15}$ J

energy of proton required = 4.45 × $10^{-15}$ J ÷ (1.602 × $10^{-19}$

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × $\frac{V^2}{2}$

put here value

energy of electron required = 9.11 × $10^{-31}$ × $\frac{(2.31*10^6)^2}{2}$

energy of electron required = 24.305× $10^{-19}$ J

energy of electron required = 24.305 × $10^{-19}$ J ÷ (1.602 × $10^{-19}$

energy of electron required = 15.171 eV

5 0

3 years ago

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