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3 years ago

Help me please to do those problems...

Physics

1 answer:

Stells [14]3 years ago

3 0

Keeping in mind that:

- Distance is the length of the total path covered by the person, regardless of the directions of the different parts of motion

- Displacement is just the distance in a straight line between the final point and the initial point

Let's apply these concepts to solve the different parts of the problem:

1. 120 m, 80 m west

The total distance is the sum of the length of the different paths:

distance = 100 + 20 = 120 m

To find the displacement, we need to find the distance between the starting point and the ending point. Assuming the starting point as

x = 0

Michelle moved 100 m westward and 20 m eastward, so the ending point is at

ending point = 100 - 20 = 80 m (westward)

So, the displacement is

displacement = 80 - 0 = 80 m (west)

2. 6400 m, 0

The distance is equal to the length of the track multiplied by the number of laps, so:

$distance = 1600 \cdot 4 = 6400 m$

The track is circular, and the runner completes exactly 4 laps: it means that at the end of the motion, the runner is at her starting point. So ending point and starting point coincide, and so the displacement

displacement = ending point - starting point = 0

3. 40 m; 10 m forward

The total distance is just the sum of the lengths of the different parts of the motion, so:

distance = 20 + 5 + 10 + 5 = 40 m

The find the displacement, we need to assign signs to every direction:

Forward --> positive along the forward-backward direction

Backard --> negative along the forward-backward direction

Right --> positive direction along the right-left direction

Left --> negative direction along the right-left direction

Along the forward-backward direction, the displacement is:

20 m forward and 10 m backward, so 20 - 10 = 10 (forward)

Along the left right direction, the displacement is:

5 m right and 5 m left, so 5 - 5 = 0

So the net displacement is 10 m (forward)

4. 16 km; 4 km south

Again, t total distance is the sum of the lengths of the different parts of the motion, so:

distance = 2 + 4 + 6 + 4 = 16 km

To find the displacement, we assign signs to every direction:

North --> positive along the north-south direction

south --> negative along the north-south direction

east --> positive direction along the east-west direction

west --> negative direction along the east-west direction

Along the north-south direction, the displacement is:

2 km north and 6 km south, so 2 - 6 = -4 km (4 km south)

Along the east-west direction, the displacement is:

4 km east and 4 km west, so 4 - 4 = 0

So the net displacement is 4 km south

5. 37.7 m; 24 m

The distance covered by the skater is the length of half circumference, so given the radius

r = 12 m

The distance is

$distance= \pi r = \pi \cdot (12)=37.7 m$

The displacement is the distance in a straight line between the ending position and the starting position. Since the skater ends her motion halfway around the circle, the distance between the initial and final point is equal to the diameter of the circle (two times the radius). So,

$displacement = 2r = 2 \cdot 12 = 24 m$

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Lead atoms occupy a volume of 3 x 10-29 m3. Each atom contributes two free electrons. Calculate the Fermi velocity of lead.

Juliette [100K]

<u>Answer:</u> The Fermi velocity of lead is 64.4 km/s.

<u>Explanation:</u>

To calculate the Fermi velocity, we use the equation:

$V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}$

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

$m_e$ = mass of electron = $9.1\times 10^{-31}kg$

N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = $\frac{2\times N_A\times V}{M}$

$N_A$ = Avogadro's number = $6.022\times 10^{26}mol^{-1}$ (When the mass is in kilograms)

V = Volume = $3\times 10^{-29}m^3$

M = molecular weight of lead = 207.2 g/mol

Putting values in above equation, we get:

$V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}$

$V_f=0.0644\times 10^6m/s=64.4km/s$ (Conversion factor: 1 km = 1000 m)

Hence, the Fermi velocity of lead is 64.4 km/s

4 0

3 years ago

An air bubble has a volume of 1.70 cm³ when it is released by a submarine 115 m below the surface of a lake. What is the volume

Natalija [7]

Answer:

V = 20.67 cm³

Explanation:

In this case, let's apply the Boyle's law which is:

P1V1 = P2V2

Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.

By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²

We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:

P1 = Po + dgh

Where:

Po: innitial pressure, which we can assume is 1 atm

d = density of the substance, in this case, water (1000 kg/m³)

g = gravity (9.8 m/s²)

h = distance of the bubble from the surface (115 m)

Now replacing this data in the boyle's law we have the following:

P1V1 = P2V2

V2 = P1V1/P2

V2 = (Po + dgh) * V1 / P2

Replacing the data we have:

V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5

V2 = 2,087,600 / 1.01x10^5

V2 = 20.67 cm³

5 0

3 years ago

At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

2 years ago

Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one

Harman [31]

Answer: $5.9(10)^{-8} m$

Explanation:

The equation to calculate the center of mass $C_{M}$ of a particle system is:

$C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}$

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

$C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}$