Answer:
The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].
Explanation:
An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:
[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]
[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]
The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:
[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]
Where:
[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.
[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.
[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.
If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:
[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]
[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]
As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:
[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]
[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]
The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:
[tex]x = \frac{n_{R}}{n_{E}}[/tex]
[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]
[tex]x = 8.5222 \times 10^{-11}[/tex]
The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].
The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
The distance is [tex]d = 1.5 *10^{15} \ km[/tex]
Explanation:
From the question we are told that
The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]
Generally a grid unit is [tex]\frac{1}{10}[/tex] of an arcsec
This implies that 0.2 grid unit is [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as
[tex]d = \frac{1}{k}[/tex]
substituting values
[tex]d = \frac{1}{0.02}[/tex]
[tex]d = 50 \ parsec[/tex]
Note [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]
So [tex]d = 50 * 3.08 *10^{13}[/tex]
[tex]d = 1.5 *10^{15} \ km[/tex]
A soccer ball is released from rest at the top of a grassy incline. After 2.2 seconds, the ball travels 22 meters. One second later, the ball reaches the bottom of the incline. (Assume that the acceleration was constant.) How long was the incline
Answer:
x = 46.54m
Explanation:
In order to find the length of the incline you use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the soccer ball = 0 m/s
t: time
a: acceleration
You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):
[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex] (2)
Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:
[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex] (3)
The length of the incline is 46.54 m
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potential is 150 V. What is the radius of the sphere
Answer:
The radius of the sphere is 4.05 m
Explanation:
Given;
potential at surface, [tex]V_s[/tex] = 450 V
potential at radial distance, [tex]V_r[/tex] = 150
radial distance, l = 8.1 m
Apply Coulomb's law of electrostatic force;
[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]
[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]
Therefore, the radius of the sphere is 4.05 m
A long horizontal hose of diameter 3.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 14 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
A) What is the velocity of the water in the hose ?
B) What is the pressure differential between the water in the hose and water in the nozzle ?
C) How long will it take to fill a tub of volume 120 liters with the hose ?
Answer:
a) v₁ = 3.92 m / s , b) ΔP = = 9.0 10⁴ Pa, c) t = 0.0297 s
Explanation:
This is a fluid mechanics exercise
a) let's use the continuity equation
let's use index 1 for the hose and index 2 for the nozzle
A₁ v₁ = A₂v₂
in area of a circle is
A = π r² = π d² / 4
we substitute in the continuity equation
π d₁² / 4 v₁ = π d₂² / 4 v₂
d₁² v₁ = d₂² v₂
the speed of the water in the hose is v1
v₁ = v₂ d₂² / d₁²
v₁ = 14 (1.8 / 3.4)²
v₁ = 3.92 m / s
b) they ask us for the pressure difference, for this we use Bernoulli's equation
P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2
as the hose is horizontal y₁ = y₂
P₁ - P₂ = ½ ρ (v₂² - v₁²)
ΔP = ½ 1000 (14² - 3.92²)
ΔP = 90316.8 Pa = 9.0 10⁴ Pa
c) how long does a tub take to flat
the continuity equation is equal to the system flow
Q = A₁v₁
Q = V t
where V is the volume, let's equalize the equations
V t = A₁ v₁
t = A₁ v₁ / V
A₁ = π d₁² / 4
let's reduce it to SI units
V = 120 l (1 m³ / 1000 l) = 0.120 m³
d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m
let's substitute and calculate
t = π d₁²/4 v1 / V
t = π (3.4 10⁻²)²/4 3.92 / 0.120
t = 0.0297 s
Dr. Jones performed an experiment to monitor the effects of sunlight exposure on stem density in aquatic plants. In the study, Dr. Jones measured the mass and volume of stems grown in 5 levels of sun exposure. The data is represented below.
Sun exposure Stem mass (g) Stem volume (mL)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742
a. Convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (mº).
b. Calculate the density of the samples using the equation d = m/v. d = density m = mass (kg) v = volume (m)
c. Convert the density values to scientific notation.
Given that,
Sun exposure = 30%, 45%, 60%, 75%, 90%
Stem mass (g) = 275, 415, 563, 815, 954
Stem volume (ml) = 1100, 1215, 1425, 1610, 1742
(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters
Using conversion of mass
[tex]1\ g=0.001\ kg[/tex]
Conservation of volume
[tex]1\ Lt=0.001\ m^3[/tex]
[tex]1\ mL=1\times10^{-6}\ m^3[/tex]
So, mass in kg
Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954
Volume in m³,
Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742
(b). We need to calculate the density of the samples
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
Where, m = mass
V = volume
If the m = 0.275 kg and V = 0.0011 m³
Put the value into the formula
[tex]\rho=\dfrac{0.275}{0.0011}[/tex]
[tex]\rho=250\ kg/m^3[/tex]
If the m = 0.415 kg and V = 0.001215 m³
Put the value into the formula
[tex]\rho=\dfrac{0.415}{0.001215}[/tex]
[tex]\rho=341.56\ kg/m^3[/tex]
[tex]\rho=342\ kg/m^3[/tex]
If the m = 0.563 kg and V = 0.001425 m³
Put the value into the formula
[tex]\rho=\dfrac{0.563}{0.001425}[/tex]
[tex]\rho=395.08\ kg/m^3[/tex]
If the m = 0.815 kg and V = 0.001610 m³
Put the value into the formula
[tex]\rho=\dfrac{0.815}{0.001610}[/tex]
[tex]\rho=506.21\ kg/m^3[/tex]
If the m = 0.954 kg and V = 0.001742 m³
Put the value into the formula
[tex]\rho=\dfrac{0.954}{0.001742}[/tex]
[tex]\rho=547.6\ kg/m^3[/tex]
[tex]\rho=548\ kg/m^3[/tex]
(c). We need to convert the density values to scientific notation
In scientific notation
The densities are
[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]
Hence, This is required solution.
A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend to the sixth floor, which is a known distance h above the starting point. The elevator undergoes an unknown constant acceleration of magnitude a for a given time interval T. Then the elevator moves at a constant velocity for a time interval 4T. Finally the elevator brakes with an acceleration of magnitude a, (the same magnitude as the initial acceleration), for a time interval T until stopping at the sixth floor.
Answer:
The found acceleration in terms of h and t is:
[tex]a=\frac{h}{5(t_1)^2}[/tex]
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
Stage 1Constant acceleration, starts from rest.
Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]
Velocity = [tex]v_1=at_1[/tex]
Stage 2Constant velocity where
Velocity = [tex]v_o=v_1=at_1[/tex]
Distance =
[tex]y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\[/tex]Stage 3Constant deceleration where
Velocity = [tex]v_0=v_1=at_1[/tex]
Distance =
[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]
Total HeightTotal height = y₁ + y₂ + y₃
Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]
AccelerationFind acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
[tex]a=\frac{h}{5(t_1)^2}[/tex]
A spherical balloon is made from a material whose mass is 4.30 kg. The thickness of the material is negligible compared to the 1.54-m radius of the balloon. The balloon is filled with helium (He) at a temperature of 289 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m3. Find the absolute pressure of the helium gas.
Answer:
P = 5.97 × 10^(5) Pa
Explanation:
We are given;
Mass of balloon;m_b = 4.3 kg
Radius;r = 1.54 m
Temperature;T = 289 K
Density;ρ = 1.19 kg/m³
We know that, density = mass/volume
So, mass = Volume x Density
We also know that Force = mg
Thus;
F = mg = Vρg
Where m = mass of balloon(m_b) + mass of helium (m_he)
So,
(m_b + m_he)g = Vρg
g will cancel out to give;
(m_b + m_he) = Vρ - - - eq1
Since a sphere shaped balloon, Volume(V) = (4/3)πr³
V = (4/3)π(1.54)³
V = 15.3 m³
Plugging relevant values into equation 1,we have;
(3 + m_he) = 15.3 × 1.19
m_he = 18.207 - 3
m_he = 15.207 kg = 15207 g
Molecular weight of helium gas is 4 g/mol
Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles
From ideal gas equation, we know that;
P = nRT/V
Where,
P is absolute pressure
n is number of moles
R is the gas constant and has a value lf 8.314 J/mol.k
T is temperature
V is volume
Plugging in the relevant values, we have;
P = (3802 × 8.314 × 289)/15.3
P = 597074.53 Pa
P = 5.97 × 10^(5) Pa