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4 years ago

Show that 1.0 m/s = 3.6 km/h . Hint: Show the explicit steps involved in converting 1.0 m/s = 3.6 km/h.

1 answer:

5 0

Just multiply the "1.0 m/s" by ' 1 ' a few times. (Remember that a fraction with the same quantity on top and bottom is equal to ' 1 ' .)

(1.0 m/sec) · (1 km/1000 m) · (60 sec/min) · (60 min/hr) =

(1.0 · 60 · 60 / 1,000) (m · km · sec · min / sec · m · min · hr) =

(3,600 / 1,000) (km / hr) =

3.6 km/hr .

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What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?

Dvinal [7]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth: (0.0001 kg) x (9.81 m/s²) = 0.000981 Newton

On the Moon: (0.0001 kg) x (1.62 m/s²) = 0.000162 N

The presence or absence of atmosphere makes no difference. In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup. Weight is just the force of gravity between the feather and the Earth. It's not affected by what's around the feather, or what's happening to it.

6 0

3 years ago

A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

Ad libitum [116K]

Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × $10^{-19}$ C

and charge on iron will be q3 = 2 × 1.602 × $10^{-19}$ C

so charge on electron is q1 = - 1.602 × $10^{-19}$ C

and we know F = $k\frac{q*q}{r^{2} }$

so now by equilibrium

Fu = Fi

$k\frac{q*q}{r^{2} }$ = $k\frac{q*q}{r^{2} }$

put here k = $9*10^{9}$ and find r

$9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }$ = $9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }$

$\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}$

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

6 0

3 years ago

A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it mai

Sloan [31]

Answer: Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

U1 = 2321 KJ/kg

at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa , T = 500°C

h1 = 3434.7 KJ/Kg

calculate final mass ( m2 )

M2 = V2 / v'2

= 0.2 / 0.2481 = 0.806 kg

therefore the mass added = m2 - m1

= 0.806 - 0.6 = 0.206 kg

3 0

3 years ago

Now that you have learned about measuring electricity, complete and submit the following projects Complete the following: 1. Exa

allochka39001 [22]

Answer:

in the attached documents are the answers i made a 100 on it

Explanation:

Download pdf

3 0

3 years ago

Liquids and gases are highly compressible. a)True b)False

densk [106]

True.
Gas -> Liquid
At the GAS stage, the particles are wildly spread.
Liquid -> Solid
At a LIQUID stage, the particles are slightly more compressed.
and yeah you get the idea.

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3 years ago