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3 years ago

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If the atomic radius of Copper is 0.128 nm, calculate the volume of its unit cell in cubic nanometers. Round off the answer to t

hree significant figures. Do not include the units.

1 answer:

hodyreva [135]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take

kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0

3 years ago

What is the chemical formula for mercury(I) nitrate? Hgmc021-1.jpg(NOmc021-2.jpg) Hg(NOmc021-3.jpg)mc021-4.jpg Hgmc021-5.jpg(NOm

Anit [1.1K]

If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
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3 years ago

Read 2 more answers

A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

Bumek [7]

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s speed of ball in the moving frame (the truck)

U = -20 m/s speed of primed (rest) frame with respect to moving frame

So V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

3 0

3 years ago

3.00 m^3 of water is at 20.0°C.

romanna [79]

Answer:

$\triangle V = 0.02484m^3$

Explanation:

Given

$V_1 = 3.00m^3$ --- initial volume

$T_1 = 20.0^oC$ --- initial temperature

$T_2 = 60.0^oC$ --- final temperature

$\gamma = 207*10^{-6$ --- coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

$\triangle V = \gamma * V_2 * (T_2 - T_1)$

So, we have:

$\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)$

$\triangle V = 207 * 10^{-6} * 3.00 * 40.0$

$\triangle V = 0.02484m^3$

The volume will expand by $0.02484m^3$

7 0

2 years ago

A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is

Anika [276]

Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

$A\pm \Delta A$ and $B\pm \Delta B$

The sum is represented as

$Sum=(A+B)\pm (\Delta A+\Delta B)$

For the the values given to us the sum is calculated as

$Sum=(2.9+3.9)\pm (0.1+0.2)$

$Sum=6.8\pm 0.3$

Now the since the uncertainity inthe sum is $\pm 0.3$

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals $6.8-0.3=6.5$ meters

3 0

3 years ago