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3 years ago

Carry out the following conversions using the factor-label method. Show all your work.

1 answer:

7 0

1) 3600 sec in an hour (3600 x 24 x 365) = 31,536,000 sec
2) 100 cm in 1 m (.14 x 100) = 14 cm
3) 1000 g in 1 kg (20/1000) = .02 kg
4) 10^-9 m = 1 nm (637 x 10^-9) = 6.37 x 10 ^-7m
5) 1000 m in 1 km (1000 x 3.2) = 3200 m
6) 10^-6 m in 1 micrometers (24 x 10^-6) = 2.4 x 10^-5

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A train traveling at 27.5 accelerates to 42.4 m s over 75.0 s What is the displacement of the train in this time period A train

jarptica [38.1K]

Answer:

The displacement of the train in this time period is 2,616.86 m.

Explanation:

A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.

In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).

An independent equation of useful time in this type of movement is:

$vf^{2} =vi^{2} +2*a*d$ Expression A

where:

vf = final velocity
vi = initial velocity
a = acceleration
d = distance

The equation of velocity as a function of time in this type of movement is:

vf=vi + a*t

So the velocity can be calculated as: $a=\frac{vf-vi}{t}$

In this case:

vf=42.4 m/s
vi=27.5 m/s
t=75 s

Replacing in the definition of acceleration: $a=\frac{42.4 m/s-27.5 m/s}{75 s}$

a=0.199 m/s²

Now, replacing in expression A:

$(42.4 m/s)^{2} =(27.5 m/s)^{2} +2*(0.199 m/s^{2}) *d$

Solving:

$d= \frac{(42.4 m/s)^{2} - (27.5 m/s)^{2} }{2*(0.199 m/s^{2}) }$

d= 2,616.86 m

The displacement of the train in this time period is 2,616.86 m.

4 0

3 years ago

Assume that before separation, the weight of the milk bottle's contents (mixed milk and cream) is wmix. How does the combined we

finlep [7]

Answer:

Wsep = Wmix

Explanation:

When Cream and milk up in a bottle they might appear homogeneous but after the bottle must have settled down for a while i.e kept in a position without shaking the bottle. the contents ( cream and milk ) will be separated. this is because Milk and cream do not mix up just like some other liquids that don't mix-up. Wmix represents the weight of the bottle before separation while Wsep represents the weight after separation. but since both liquids are in the same bottle the weight after separation would remain the same

Wsep = Wmix

3 0

3 years ago

Solid are _______, usually in a regular repeating pattern

Temka [501]

Answer:

particles or arranged will be the answer depending on the previous line.

4 0

3 years ago

Read 2 more answers

When a 5 kilogram mass is

salantis [7]

Answer:If a 5-kilogram mass is raised vertically 2 meters from the surface of the Earth, its gain in potential energy is approximately 1) 0 J 2) 10 J 3) 20 J 4) 100 J 43.

Explanation:

7 0

3 years ago

An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus

frutty [35]

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

Given:

$u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

$v$ = final velocity of the electron = $3u$

$x$ = initial position of the electron from the proton = very distant = $\infty$

Assume:

$m$ = mass of an electron = $9.1\times10^{-31}\ kg$

$e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

$p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

$k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

$y$ = final position of the electron from the proton

$\Delta K$ = change in kinetic energy of the electron

$W$ = work done by the electrostatic force

$F$ = electrostatic force

$r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.

8 0

4 years ago