Answer:
Work = 1167.54 J
Explanation:
The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:
Gain in K.E = Loss in P.E + Work
(0.5)(m)(Vf² - Vi²) - mgh = Work
where,
m = mass of boy = 60 kg
Vf = Final Speed = 8.5 m/s
Vi = Initial Speed = 1.6 m/s
g = 9.8 m/s²
h = height drop = 1.57 m
Therefore,
(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work
Work = 2090.7 J - 923.16 J
Work = 1167.54 J
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°. N · m2/C
Answer:
Φ = 361872 N.m^2 / C
Explanation:
Given:-
- The area of the two plates, [tex]A_p = 180 cm^2[/tex]
- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]
- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]
- The radius for the flux region, [tex]r = 3.3 cm[/tex]
- The angle between normal to region and perpendicular to plates, θ = 4°
Find:-
Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.
Solution:-
- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:
[tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]
- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):
σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]
σ = 0.00094 C / m^2
- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.
[tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]
- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:
[tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]
- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:
Φ = E_net * Ar * cos ( θ )
Φ = (106214689.26553) * (0.00342) * cos ( 5 )
Φ = 361872 N.m^2 / C
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket motor is ignited and the rocket burns fuel at a rate of 95 kg/s. The exit speed of the exhaust gas relative to the rocket is 2900 m/s. Neglecting drag and friction forces, determine the acceleration and the velocity of the car at time t = 10 s. Plot the acceleration and velocity from time t0 to t = 10 s.
Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.
Explanation:
The expression used will be as follows.
[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]
[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]
= [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]
[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]
[tex]v_{o} = 0[/tex]
As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]
u = -2900 m/s
[tex]M_{f} = M_{o} - m \times t_{f}[/tex]
= [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]
= [tex](3500 - 95t_{f})s[/tex]
[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]
Also, we know that
a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]
= [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]
= [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]
At t = 10 sec,
[tex]v_{f}[/tex] = 918.34 m/s
and, a = 108 [tex]m/s^{2}[/tex]
The Pauli exclusion principle states that Question 1 options: the wavelength of a photon of light times its frequency is equal to the speed of light. no two electrons in the same atom can have the same set of four quantum numbers. both the position of an electron and its momentum cannot be known simultaneously very accurately. the wavelength and mass of a subatomic particle are related by . an electron can have either particle character or wave character.
Answer:
no two electrons in the same atom can have the same set of four quantum numbers
Explanation:
Pauli 's Theory of Exclusion specifies that for all four of its quantum numbers, neither two electrons in the same atom can have similar value.
In a different way, we can say that no more than two electrons can take up the identical orbital, and two electrons must have adversely spin in the identical orbital
Therefore the second option is correct
A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 4.5 μg is floating a distance d = 25 cm above charge The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.
|Q2| = m g d2/( k Q1 )
Calculate the magnitude of Q2 in units of nanocoulombs.
Answer:
( About ) 6.8nC
Explanation:
We are given the equation |Q2| = mgd^2 / kQ1. Let us substitute known values into this equation, but first list the given,
Charge Q2 = +45nC = (45 × 10⁻⁹) C
mass of charge Q2 = 4.5 μg, force of gravity = 4.5 μg × 9.8 m/s² = ( 4.41 × 10^-5 ) N,
Distance between charges = 25 cm = 0.25 m,
k = Coulomb's constant = 9 × 10^9
_______________________________________________________
And of course, we have to solve for the magnitude of Q2, represented by the charge magnitude of the charge on Q2 -
(4.41 × 10^-5) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] / 0.25²
_______________________________________________________
Solution = ( About ) 6.8nC
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm .What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Complete Question:
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Answer:
[tex]\triangle E = 12.79 J[/tex]
Explanation:
Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]
Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]
Spring constant for the two groups, k = 31 N/mm = 3100 N/m
Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]
Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]
Difference in maximum stored energy between the sprinters and the non-athlethes:
[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations in 33.0 s. Also, a given maximum travels 424 cm along the rope in 15.0 s. What is the wavelength
Answer:
0.218
Explanation:
Given that
Total vibrations completed by the wave is 43 vibrations
Time taken to complete the vibrations is 33 seconds
Length of the wave is 424 cm = 4.24 m
to solve this problem, we first find the frequency.
Frequency, F = 43 / 33 hz
Frequency, F = 1.3 hz
Also, we find the wave velocity. Which is gotten using the relation,
Wave velocity = 4.24 / 15
Wave velocity = 0.283 m/s
Now, to get our answer, we use the formula.
Frequency * Wavelength = Wave Velocity
Wavelength = Wave Velocity / Frequency
Wavelength = 0.283 / 1.3
Wavelength = 0.218
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a force of 79 N, the second a force of 92 N, kinetic friction is 5.5 N, and the mass of the third child plus sled is 24 kg.
1. Using a coordinate system where the second child is pushing in the positive direction, calculate the acceleration in m/s2.
2. What is the system of interest if the accelaration of the child in the wagon is to be calculated?
3. Draw a free body diagram including all bodies acting on the system
4. What would be the acceleration if friction were 150 N?
Answer:
Please, read the anser below
Explanation:
1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:
[tex]F_2-F_1-F_f=Ma[/tex] (1)
Where is has been used that the motion is in the direction of the applied force by the second child
F2: force of the second child = 92N
F1: force of the first child = 79N
Ff: friction force = 5.5N
M: mass of the third child = 24kg
a: acceleration of the third child = ?
You solve the equation (1) for a, and you replace the values of the other parameters:
[tex]a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}[/tex]
The acceleration is 0.48m/s^2
2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.
3. An image with the diagram forces is attached below.
4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.
Then, the acceleration is zero
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
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the
Answer:
F₂ = 925.92 N
Explanation:
In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,
σ₁ = σ₂
F₁/A₁ = F₂/A₂
F₂ = F₁ A₂/A₁
where,
F₂ = Initial force that must be applied to narrow arm = ?
F₁ = Load on Wider Arm to be raised = 12000 N
A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²
A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²
Therefore,
F₂ = (12000 N)(25π cm²)/(324π cm²)
F₂ = 925.92 N
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
Water molecules are made of slightly positively charged hydrogen atoms and slightly negatively charged oxygen atoms. Which force keeps water molecules stuck to one another? strong nuclear gravitational weak nuclear electromagnetic
Answer:
The answer is electromagnetic
Answer:
electromagnetic
Explanation:
edge 2021
A charged Adam or particle is called a
Answer:
A charged atom or particle is called an ion :)
The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repulsion must be overcome. This can be done using the kinetic energy of high temperature gas ions or by accelerating the nuclei toward one another.
Required:
a. Calculate the potential energy of two singly charged nuclei separated by 1.00×10^−12m
b. At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
a
[tex]PE = 2.3 *10^{-16} \ J[/tex]
b
[tex]T = 1.1 *10^{7} \ K[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 1.00 *10^{-12} \ m[/tex]
Generally the electric potential energy can be mathematically represented as
[tex]PE = \frac{k * q_1 q_2 }{d}[/tex]
Given that in a nuclei the only charged particle is the proton who charge is
[tex]p = 1.60 *10^{-19} \ C[/tex]
Hence
[tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]
And k is the coulomb constant with values [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]
So we have that
[tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]
[tex]PE = 2.3 *10^{-16} \ J[/tex]
The relationship between the electrical potential energy and the temperature is mathematically represented as
[tex]PE = \frac{3}{2} kT[/tex]
Here k is the Boltzmann's constant with value [tex]k = 1.38*10^{-23} JK^{-1}[/tex]
making T the subject
[tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]
substituting values
[tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]
[tex]T = 1.1 *10^{7} \ K[/tex]
What do behaviorism and cognitive psychology have in common?
O Both rely on the scientific method.
Both attempt to explain human behavior.
Both note the differences between human and animal behavior
Behaviorism focuses on actions only.
Answer:
Both attempt to explain human behavior
Explanation:
Psychology is generally regarded as the science of human behavior. Behaviourism is the psychological theory which holds that behaviour can be fully understood in terms of conditioning, without actually considering thoughts or feelings. The theory holds that psychological disorders can be aptly handled by simply altering the behavioural patterns of the individual. It involves the study of stimulus and responses.
Cognitive psychology attempts to decipher what is going on in people's minds. That is, it looks at the mind as a processor of information. Hence we can define cognitive psychology as the study of the internal mental processes. This according to behaviorists, cannot be studied in measurable terms as in behaviourism (stimulus response approach) even though mental processes are known to influence human behavior significantly.
Hence, both behaviourism and cognitive psychology attempt to study human behavior from different perspectives.
g The Trans-Alaskan pipeline is 1,300 km long, reaching from Prudhoe Bay to the port of Valdez, and is subject to temperatures ranging from -71°C to +35°C. How much does the steel pipeline expand due to the difference in temperature?
Answer:
ΔL = 1.653 km
Explanation:
The linear expansion of any object due to change in temperature is given by the following formula:
ΔL = αLΔT
where,
ΔL = Change in length or expansion of steel pipe line = ?
α = coefficient of linear expansion of steel = 12 x 10⁻⁶ /°C
L = Original Length of the steel pipe = 1300 km
ΔT = Change in temperature = 35°C - (- 71°C) = 35°C + 71°C = 106°C
Therefore,
ΔL = (12 x 10⁻⁶ /°C)(1300 km)(106°C)
ΔL = 1.653 km
Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long (in s) did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher's mound
Answer:
t = 0.414s
Explanation:
In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:
[tex]t=\frac{d}{v}[/tex] (1)
d: distance to the plate = 18.4m
v: speed of the ball = 160.0km/h
You first convert the units of the sped of the ball to appropriate units (m/s)
[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]
Then, you replace the values of the speed v and distance s in the equation (1):
[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]
THe ball takes 0.414s to reach the home plate
If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What is the ratio of the magnitude of C to that of D?
a) 1.3
b) 1.6
c) 1.8
d) 2.2
e) 3.2
Answer:
(e) 3.2
Explanation:
We are given that vector C and D.
Let R be the magnitude of C+D.
According to question
R=3D
We have to find the ratio of the magnitude of C to that of D.
By using right triangle property
[tex]C^2=R^2+D^2[/tex]
[tex]C^2=(3D)^2+D^2[/tex]
[tex]C^2=9D^2+D^2[/tex]
[tex]C^2=10D^2[/tex]
[tex]C=\sqrt{10D^2}=3.2D[/tex]
[tex]\frac{C}{D}=3.2[/tex]
Hence, the ratio of the magnitude of C to that of D=3.2
(e) 3.2
As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by
U = 12k x 2
where k is the force constant of the spring and x is the distance from the equilibrium position. The kinetic energy of the system is, as always,
K = 12mv2
where m is the mass of the block and v is the speed of the block.
A) Find the total energy of the object at any point in its motion.
B) Find the amplitude of the motion.
C) Find the maximum speed attained by the object during its motion.
Answer:
a) [tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex], b) Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex], c) The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].
Explanation:
a) The total energy of the object is equal to the sum of potential and kinetic energies. That is:
[tex]E = K + U[/tex]
Where:
[tex]K[/tex] - Kinetic energy, dimensionless.
[tex]U[/tex] - Potential energy, dimensionless.
After replacing each term, the total energy of the object at any point in its motion is:
[tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex]
b) The amplitude of the motion occurs when total energy is equal to potential energy, that is, when objects reaches maximum or minimum position with respect to position of equilibrium. That is:
[tex]E = U[/tex]
[tex]E = \frac{1}{2} \cdot k \cdot A^{2}[/tex]
Amplitude is finally cleared:
[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]
Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex].
c) The maximum speed of the motion when total energy is equal to kinetic energy. That is to say:
[tex]E = K[/tex]
[tex]E = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]
Maximum speed is now cleared:
[tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex]
The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].
A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3 for the density of air, determine the height of the building
Answer:
The height of the building is 158.140 meters.
Explanation:
A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:
[tex]\Delta P \propto \rho \cdot \Delta h[/tex]
[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]
Where:
[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.
[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.
[tex]\Delta h[/tex] - Height difference, measured in meters.
Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:
[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]
[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]
Where:
[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.
[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.
[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.
[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.
If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:
[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]
[tex]\Delta h_{air} = 158.140\,m[/tex]
The height of the building is 158.140 meters.
158.13m
Explanation:
Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e
P = ρhg
Let;
Pressure measured at the roof top = ([tex]P_{R}[/tex])
Pressure measured at the ground level = ([tex]P_{G}[/tex])
Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P ---------------(i)
(a) P = ρhg -----------(***)
But;
ρ = density of air = 1.29kg/m³
h = height of column of air = height of building
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (***)
P = 1.29 x h x 10
P = 12.9h Pa
(b) [tex]P_{G}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (*)
[tex]P_{G}[/tex] = 13600 x 0.76 x 10
[tex]P_{G}[/tex] = 103360 Pa
(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g --------------(**)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (**)
[tex]P_{R}[/tex] = 13600 x 0.745 x 10
[tex]P_{R}[/tex] = 101320 Pa
(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P
103360 = 101320 + 12.9h
Solve for h;
12.9h = 103360 - 101320
12.9h = 2040
h = [tex]\frac{2040}{12.9}[/tex]
h = 158.13m
Therefore, the height of the building is 158.13m
A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe
Answer:
[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Explanation:
Given that
Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]
slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]
slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]
We know that for double slit the maxima condition is that
[tex]\operatorname{dsin} \theta=m \lambda[/tex]
[tex]\sin \theta=\frac{m \lambda}{d}[/tex]
[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]
For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]
[tex]\tan \theta=\frac{y_{m}}{D}[/tex]
[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]
Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]
Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]
Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Suppose you are chatting with your friend, who lives on the moon. He tells you he has just won a Newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a Newton of gold. Who is richer
Answer:
The friend on moon is richer.
Explanation:
The value of acceleration due to gravity changes from planet to planet. So the weight of 1 Newton of gold carries different mass on different places. So we need to calculate the mass of gold that both persons have.
FRIEND ON MOON:
W₁ = m₁g₁
where,
W₁ = Weight of Gold won by friend on moon = 1 N
m₁ = mass of gold won by friend on moon = ?
g₁ = acceleration due to gravity on moon = 1.625 m/s²
Therefore,
1 N = m₁(1.625 m/s²)
m₁ = 0.62 kg
ON EARTH:
W₂ = m₂g₂
where,
W₂ = Weight of Gold won by me on Earth = 1 N
m₂ = mass of gold won by me on Earth = ?
g₂ = acceleration due to gravity on Earth = 9.8 m/s²
Therefore,
1 N = m₁(9.8 m/s²)
m₁ = 0.1 kg
Since, the friend on moon has greater mass of gold than me.
Therefore, the friend on moon is richer.
according to newtons second law of motion, what is equal to the acceleration of an object
Answer: According to Newtons second Law of motion ;
F = ma (Force equals mass multiplied by acceleration.)
The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force
Explanation:
In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight can the other piston slowly lift if its cross sectional area is 25 m2
Answer:
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
Explanation:
By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:
Hydraulic Lift - Before change
[tex]P = \frac{F}{A}[/tex]
Hydraulic Lift - After change
[tex]P + \Delta P = \frac{F + \Delta F}{A}[/tex]
Where:
[tex]P[/tex] - Hydrostatic pressure, measured in pascals.
[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.
[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.
[tex]F[/tex] - Hydrostatic force, measured in newtons.
[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.
The additional weight is obtained after some algebraic handling and the replacing of all inputs:
[tex]\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}[/tex]
[tex]\Delta P = \frac{\Delta F}{A}[/tex]
[tex]\Delta F = A\cdot \Delta P[/tex]
Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:
[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]
[tex]\Delta F = 2500\,N[/tex]
The additional mass needed for the additional weight is:
[tex]\Delta m = \frac{\Delta F}{g}[/tex]
Where:
[tex]\Delta F[/tex] - Additional weight, measured in newtons.
[tex]\Delta m[/tex] - Additional mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:
[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta m = 254.92\,kg[/tex]
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
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 56.0 g ball of copper has a net charge of 2.10 μC. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)
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].
A 2.0-kg object moving at 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20 J
Explanation:
From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.
m'u'+mu = V(m+m')........... Equation 1
Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.
make V the subject of the equation above
V = (m'u'+mu)/(m+m')............. Equation 2
Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).
Substitute into equation 2
V = [(2×5)+(8×0)]/(2+8)
V = 10/10
V = 1 m/s.
Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision
Total kinetic energy before collision = 1/2(2)(5²) = 25 J
Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J
Lost in kinetic energy = 25-5 = 20 J
The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.
Since there is a lost in kinetic energy, the collision is inelastic collision.
m'u'+mu = V(m+m')
[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]
Where
m' = mass of the moving object = 2 kg
m = mass of the stick = 8 kg,
u' = initial velocity of the moving object = 5 m/s
V = common velocity after collision= ?
u = 0 m/s (at rest).
put the values in the formula,
[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]
kinetic energy before collision
[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]
kinetic energy after collision
[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]
Lost in kinetic energy = 25-5 = 20 J
Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.
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The magnet has an unchanging magnetic field: very strong near the magnet, and weak far from the magnet. How did the magnetic field through the coil change as the magnet fell toward it? How did the magnetic flux through the coil change as the magnet fell toward it?
Answer:
The magnetic field through the coil at first increases steadily up to its maximum value, and then decreases gradually to its minimum value.
Explanation:
At first, the magnet fall towards the coils; inducing a gradually increasing magnetic field through the coil as it falls into the coil. At the instance when half the magnet coincides with the coil, the magnetic field magnitude on the coil is at its maximum value. When the magnet falls pass the coil towards the floor, the magnetic field then starts to decrease gradually from a strong magnitude to a weak magnitude.
This action creates a changing magnetic flux around the coil. The result is that an induced current is induced in the coil, and the induced current in the coil will flow in such a way as to oppose the action of the falling magnet. This is based on lenz law that states that the induced current acts in such a way as to oppose the motion or the action that produces it.
The electric potential of a charge distribution is given by the equation V(x) = 3x2y2 + yz3 - 2z3x, where x, y, z are measured in meters and V is measured in volts. Calculate the magnitude of the electric field vector at the position (x,y,z) = (1.0, 1.0, 1.0)
Answer:
The magnitude of the electric field is [tex]|E| = 8.602 \ V/m[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 3x^2y^2 + yz^3 - 2z^3x[/tex]
Generally electric filed is mathematically represented as
[tex]E = - [\frac{dV }{dx} i + \frac{dV}{dy} j + \frac{dV}{dz} \ k][/tex]
So
[tex]E =- ( [6xy^2 - 2z^3] i + [6x^2y+ z^3]j + [3yz^2 -6xz^2])[/tex]
at (x,y,z) = (1.0, 1.0, 1.0)
[tex]E = [6(1)(1)^2 - 2(1)^3] i + [6(1)^2(1)+ (1)^3]j + [6(1)(1)^2 -6(1)(1)^2][/tex]
[tex]E =- ([4] i + [7]j + [-3])[/tex]
[tex]E =-4i -7j + 3 k[/tex]
The magnitude of the electric field is
[tex]|E| = \sqrt{(-4)^2 + (-7)^2 + (3^2)}[/tex]
[tex]|E| = 8.602 \ V/m[/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]
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
5.0 seconds
b) 0.50 seconds
c) 0.05 seconds
Explanation:
F = ma, and a = Δv / Δt.
F = m Δv / Δt
Given: m = 60 kg and Δv = -30 m/s.
a) Δt = 5.0 s
F = (60 kg) (-30 m/s) / (5.0 s)
F = -360 N
b) Δt = 0.50 s
F = (60 kg) (-30 m/s) / (0.50 s)
F = -3600 N
c) Δt = 0.05 s
F = (60 kg) (-30 m/s) / (0.05 s)
F = -36000 N
360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
To find the force, we need to know about the mathematical formulation of force.
What is force?According to Newton's second law of motion, force is defined as mass times acceleration. Its SI unit is Newton (N).What is the mathematical formulation of force?Mathematically, it is written as
F= m×a= m×(∆V/∆t)
What is the force needed to stop 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds?Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s
When ∆t=5 seconds, F= 60×(30/5)=360N
When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N
When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N
Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
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