A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

Answers

Answer 1

Answer: Since volume of each cube is 1 cm^3
Then we can get the volume of 1 mole of cubes, which is $1 * 6.02 * 10^23 cm^3$
The edge $edge = v^1/3$
And the new adge that we are looking for: $new edge = (6.02*10^23)^1/3$ = $= 1.8191 * 46415888.336 = 84435142.472$
So, the final soution for the edge length of cube is 844km.

Do hope it helps!
Regards.

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You're driving along at 25 m/s with your aunt's valuable antiques in the back of your pickup truck when suddenly you see a giant hole in the road 55 m ahead of you. Fortunately, your foot is right beside the brake and your reaction time is zero! Can you stop without the antiques sliding and being damaged? Their coefficients of friction are μs=0.6 and μk=0.3. Hint: You're not trying to stop in the shortest possible distance. What's your best strategy for avoiding damage to the antiques?

Answers

Answer:

Explanation:

initial velocity, u = 25 m/s

distance, s = 55 m

coefficient of static friction = 0.6

coefficient of kinetic friction = 0.3

Let the acceleration is a.

Use third equation of motion

v² = u² + 2as

0 = 25 x 25 - 2 x a x 55

a = 5.68

a = μg

μ = 5.68 / 9.8 = 0.58

so, the coefficient of friction is less then the coefficient of static friction so the antiques are safe.

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

A local ice hockey pond is at 4°C when the air temperature falls, causing the water temperature to drop to 0°C with 10.9 cm of ice forming at the surface at a temperature of 0°C. How much heat was lost if the unfrozen pond is 1 m deep and 50 m on each side?

Answers

Answer: 114.4 GJ

Explanation:

Heat loss Q=U×A×ΔT

Heat loss of size A is determined by the U value of materials and the difference in temperature.

From 10.9cm from the ice

50m= 5000cm

A= 5000×5000

Q== (10.9) (5000) (5000)(4.184)(1×4 + 80)

Q = 95,771,760,000J

Q≈ 95.8 GJ

Linear gradient from the bottom of the pond to the ice:

Q = (89.1)(5000)(5000)(4.184)(1*2)

Q = 18,639,720,000J

Q ≈ 18.6 GJ

Total heat loss:

Q= 95.8GJ + 18.6GJ

Q= 114.4 GJ

What is a light year

Answers

Answer:

A light-year is the distance light travels in one year.

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

$\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)$

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = $(4.53\ cm)/(10)$ = 0.00453 m

Therefore,

$\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)$

λ = 5.734 x 10⁻⁷ m = 573.4 nm

In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.