Answer:
The observed volume of the box is 3.796 cubic meters.
Explanation:
The observed length is determined by the formula for the Length Contraction:
[tex]L = \frac{L_{o}}{\gamma}[/tex]
Where:
[tex]L[/tex] - Proper length, measured in meter.
[tex]\gamma[/tex] - Lorentz factor, dimensionless.
The Lorentz factor is represented by the following equation:
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}} }}[/tex]
If [tex]v = 0.8\cdot c[/tex], then:
[tex]\gamma = \frac{1}{\sqrt{1-\frac{0.64\cdot c^{2}}{c^{2}} }}[/tex]
[tex]\gamma = \frac{1}{\sqrt{1-0.64}}[/tex]
[tex]\gamma = \frac{5}{3}[/tex]
Therefore, the observed length is:
[tex]L = \frac{3}{5}\cdot L_{o}[/tex]
Given that [tex]L_{o} = 2.6\,m[/tex], the observed length is:
[tex]L = \frac{3}{5}\cdot (2.6\,m)[/tex]
[tex]L = 1.56\,m[/tex]
The observed volume of the box is:
[tex]V = L^{3}[/tex]
[tex]V = (1.56\,m)^{3}[/tex]
[tex]V= 3.796\,m^{3}[/tex]
The observed volume of the box is 3.796 cubic meters.
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.
A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2.0-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us). What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
F = 0.078N
Explanation:
In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:
[tex]B=\frac{\mu_oNi}{L}[/tex] (1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 580
i: current in the solenoid = 36A
L: length of the solenoid = 18cm = 0.18m
You replace the values of all parameters in the equation (1):
[tex]B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T[/tex]
Next, you calculate the force exerted on the wire, by using the following formula:
[tex]F=iLBsin\theta[/tex] (2)
i: current in the wire = 27A
L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m
θ: angle between wire and the direction of B
B: magneitc field in the solenoid = 0.145T
The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.
You replace the values of the parameters in the equation (2):
[tex]F=(27A)(0.02m)(0.145T)sin90\°=0.078N[/tex]
The magnitude of the force on the wire is 0.078N
Answer: The magnitude of the force is 0.079N
Explanation: Please see the attachments below
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 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
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.
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
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 :)
Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.
Required:
What is the safe's coefficient of kinetic friction on the bank floor?
Answer:
the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
Explanation:
GIven that:
Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car.
So ,let assume they are sliding the bank safe on an horizontal direction
Clyde → Δ(bank safe) → Bonnie
Also; from the above representation; let not forget that the friction force [tex]F_{friction}[/tex] is acting in the opposite direction ←
where;
[tex]F_{friction}[/tex] = [tex]\mu_k mg[/tex]
The safe slides with a constant speed
If Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.
Thus; since the safe slides with a constant speed if the two conditions are met; then the net force acting on the slide will be equal to zero.
SO;
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - F_{frition}[/tex]
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - \mu_k \ mg[/tex]
Since the net force acting on the slide will be equal to zero.
Then; [tex]F_{net} =0[/tex]
Also; let [tex]F_{Cylde} = F_c[/tex] and [tex]F_{Bonnie} = F_B[/tex]
Then;
[tex]0 = F_c + F_B - \mu_k \ mg[/tex]
[tex]\mu_k \ mg= F_c + F_B[/tex]
[tex]\mu_k = \dfrac{F_c + F_B}{\ mg}[/tex]
where;
[tex]F_c = 377 \ N \\ \\ F_B = 353 \ N \\ \\ mass (m) = 325 \ kg[/tex]
Then;
[tex]\mu_k = \dfrac{377 + 353}{325*9.81}[/tex]
[tex]\mu_k = \dfrac{730}{3188.25}[/tex]
[tex]\mathbf{\mu_k =0.2290}[/tex]
Thus; the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
Which of the following statements are true?
1. Liquid water expands with increasing temperature above 4°C.
2. Liquid water expands with increasing temperature between 0°C and 4°C.
3. Water contracts as it freezes at 0°C.
4. Solid ice is less dense than liquid water.
Answer:
water contracts as it freezes at 0°C
Answer:
weeve
Explanation:
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
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 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
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at 6.00 m/s due south. The second car has a mass of 900 kg and was approaching at 25.0 m/s due west. (a) Calculate the final velocity (magnitude in m/s and direction in degrees counterclockwise from the west) of the cars. (Note that since both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x-axis and y-axis; instead, you must look for other simplifying aspects.) magnitude m/s direction ° counterclockwise from west (b) How much kinetic energy (in J) is lost in the collision? (This energy goes into deformation of the cars.) J
Answer:
a) v = 11.24 m / s , θ = 17.76º b) Kf / K₀ = 0.4380
Explanation:
a) This is an exercise in collisions, therefore the conservation of the moment must be used
Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved
Recall that moment is a vector quantity so it must be kept on each axis
X axis
initial moment. Before the crash
p₀ₓ = m₁ v₁
where v₁ = -25.00 me / s
the negative sign is because it is moving west and m₁ = 900 kg
final moment. After the crash
[tex]p_{x f}[/tex]= (m₁ + m₂) vx
p₀ₓ = p_{x f}
m₁ v₁ = (m₁ + m₂) vₓ
vₓ = m1 / (m₁ + m₂) v₁
let's calculate
vₓ = - 900 / (900 + 1200) 25
vₓ = - 10.7 m / s
Axis y
initial moment
[tex]p_{oy}[/tex]= m₂ v₂
where v₂ = - 6.00 m / s
the sign indicates that it is moving to the South
final moment
p_{fy}= (m₁ + m₂) [tex]v_{y}[/tex]
p_{oy} = p_{fy}
m₂ v₂ = (m₁ + m₂) v_{y}
v_{y} = m₂ / (m₁ + m₂) v₂
we calculate
[tex]v_{y}[/tex] = 1200 / (900+ 1200) 6
[tex]v_{y}[/tex] = - 3,428 m / s
for the velocity module we use the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
v = RA (10.7²2 + 3,428²2)
v = 11.24 m / s
now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)
tan θ = [tex]v_{y}[/tex] / Vₓ
θ = tan-1 v_{y} / vₓ)
θ = tan -1 (3,428 / 10.7)
θ = 17.76º
This angle is from the west to the south, that is, in the third quadrant.
b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship
Kf = 1/2 (m1 + m2) v2
K₀ = ½ m₁ v₁² + ½ m₂ v₂²
Kf = ½ (900 + 1200) 11.24 2
Kf = 1.3265 105 J
K₀ = ½ 900 25² + ½ 1200 6²
K₀ = 2,8125 10⁵ + 2,16 10₅4
K₀ = 3.0285 105J
the wasted energy is
Kf / K₀ = 1.3265 105 / 3.0285 105
Kf / K₀ = 0.4380
this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy
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 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].
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 coil is connected to a galvanometer, which can measure the current flowing through the coil. You are not allowed to connect a battery to this coil. Given a magnet, a battery and a long piece of wire, can you induce a steady current in that coil?
Answer:
Yes we can induce current in the coil by moving the magnet in and out of the coil steadily.
Explanation:
A current can be induced there using the magnetic field and the coil of wire. Moving the bar magnet around the coil can induce a current and this is called electromagnetic induction.
What is electromagnetic induction ?The generation of an electromotive force across an electrical conductor in a fluctuating magnetic field is known as electromagnetic or magnetic induction.
Induction was first observed in 1831 by Michael Faraday, and James Clerk Maxwell mathematically named it Faraday's law of induction. The induced field's direction is described by Lenz's law.
Electrical equipment like electric motors and generators as well as parts like inductors and transformers have all found uses for electromagnetic induction.
Here, moving the bar magnet around the coil generates the electronic movement followed by a generation of electric current.
Find more on electromagnetic induction :
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6. A plane due to fly from Montreal to Edmonton required refueling. Because the fuel gauge on the aircraft was not working, a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane. The plane required 22,300 kg of fuel to make the trip. In order to determine the volume of fuel required during the refueling, the pilot asked for the density of the fuel so he could convert a volume of fuel to a mass of fuel. The mechanic provided a factor of 1.77. Assuming that this factor was in metric units (kg/L), the pilot calculated the volume to be added as 4916 L. This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L). How many liters of fuel should have been added
Answer:
The amount of liters of fuel should have been added is 20093 L
Explanation:
Given that;
a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane
i.e the volume of fuel left in the plane = 7682 L
Required fuel to make a trip = 22,300 kg of fuel
Also from the question; we are being told that in order for the pilot to determine the volume ; he asked for the density of the fuel and the mechanic said 1.77.
This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L).
Now; we can understand that the density of the fuel was 1.77 pound /litre.
SO , let convert 1.77 pound /litre to kg/Litre;
we all know that
1 pound = 0.4536 kg
1.77 pound/litre = x kg
If we cross multiply ; we will have:
1.77 pound/litre × 0.4536 kg = 1 pound × x kg
x kg = (1.77 pound/litre × 0.4536 kg) /1 pound
x = 0.802872 kg/litre
[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]
where ;
mass = 22,300 kg of fuel
volume = unknown ???
density = 0.802872 kg/litre
making volume the subject of the formula from above; we have:
[tex]\mathbf{volume = \dfrac{mass}{Density}}[/tex]
[tex]\mathbf{volume = 22300 \ kg \ of \ fuel *\dfrac{1 \ litre }{0.802872 \ kg \ of \ fuel}}[/tex]
volume = 27775.28672 litre
volume [tex]\approx[/tex] 27775 L
Let not forget that we are being told as well that the volume of fuel left in the plane = 7682 L
Now;
The amount of liters of fuel should have been added is: = 27775 L - 7682 L
The amount of liters of fuel should have been added is 20093 L
Phase Angles: Consider a series circuit in which the capacitive reactance equals the inductive reactance. What is the phase angle between current and voltage for this circuit
Answer:
X = R, therefore the current and the voltage are in phase, that is, the angles are zero
Explanation:
In circuits with capacitors and inductors the phase between current and voltage varies according to frequency , capacitance and induction values, but they go in opposite directions. The first creates a delay in the current and the second an advance.
If it is an RLC type serial circuit, the impedance is
X = √[ R² + (XL- Xc)²
with
XL = wL
Xc = 1 / wc
let's apply this to mention if XL = Xc, the impedance the circuit is resistive
X = R
therefore the current and the voltage are in phase, that is, the angles are zero
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration of 4 m/s2 by reducing the throttle. What is the velocity (in m/s) of the boat when it reaches the buoy
Answer:
7.5 m/s
Explanation:
We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:
[tex]v^2 = u^2 + 2as[/tex]
where v = final velocity
u = initial velocity
a = acceleration
s = distance traveled
From the question:
u = 28 m/s
a = -4 [tex]m/s^2[/tex]
s = 91 m
Therefore:
[tex]v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s[/tex]
The velocity of the boat when it reaches the buoy is 7.5 m/s.
which of the following best describes a stable atom?
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 steel wire with mass 25.3 g and length 1.62 m is strung on a bass so that the distance from the nut to the bridge is 1.10 m. (a) Compute the linear density of the string. kg/m (b) What velocity wave on the string will produce the desired fundamental frequency of the E1 string, 41.2 Hz
Answer:
(a) μ = 0.015kg/m
(b) v = 90.64m/s
Explanation:
(a) The linear density of the string is given by the following relation:
[tex]\mu=\frac{m}{L}[/tex] (1)
m: mass of the string = 25.3g = 25.3*10-3 kg
L: length of the string = 1.62m
[tex]\mu=\frac{25.3*10^{-3}kg}{1.62m}=0.015\frac{kg}{m}[/tex]
The linear density of the string is 0.015kg/m
(b) The velocity of the string for the fundamental frequency is:
[tex]f_1=\frac{v}{2l}[/tex] (2)
f1: fundamental frequency = 41.2 Hz
vs: speed of the wave
l: distance between the fixed extremes of the string = 1.10m
You solve for v in the equation (2) and replace the values of the other parameters:
[tex]v=2lf_1=2(1.10m)(41.2Hz)=90.64\frac{m}{s}[/tex]
The speed of the wave for the fundamental frequency is 90.64m/s
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
Is the friction of the pendulum (catch mechanism, support axis, etc.) a random or systematic error? Will this source of error cause your calculated velocity to be less than or greater than the actual velocity?
Answer:
l these errors believe that the speed of the system is less than that calculated
Explanation:
When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.
We can have random uncertainties, correspondin
g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations
There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.
tith the explanation we see that the errors described are systematic.
In general these errors believe that the speed of the system is less than that calculated
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.
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]
Two wheels initially at rest roll the same distance without slipping down identical planes. Wheel B has twice the radius, but the same mass as wheel A. All the mass is concentrated in their rims so that the rotational inertias are I = mR2. Which has more translational kinetic energy when it gets to the bottom?
Answer:
Their translational kinetic energies are the same
Explanation:
The translational kinetic energy of an object is given by the formula:
[tex]KE = 0.5 mv^2[/tex]
Where m = the mass of the object and
v = the linear speed of the object
From the question, it is stated that wheel A has the same mass as wheel B, that is [tex]m_A = m_B[/tex]
Linear speed is also a function of the distance covered. Since both wheels cover the same distance within the same interval, we can conclude that [tex]v_A = v_B[/tex]
Both wheels A and B have equal speed and mass, this means that their translational kinetic energy is the same.
Note that translational kinetic energy is not a function of the radius
Why can the internal resistance of the DMM can be determined without taking into account the 100 Ω resistor? Could this be done for a resistor of any resistance value? Explain your answer.
Answer:
Explanation:
For resistances in parallel we know that the overall resistance will be smaller in the value than any individual resistance. so, in this case we are concerned about [tex]\frac{1}{R}[/tex] = [tex]\frac{1}{Rs}[/tex] ⁺ [tex]\frac{1}{Rp}[/tex]
where Rs is the series resistor and Rp is the parallel resistor.
so, R = 0.9M, roughly.
So, as the infernal resistance is very high as compared to the resistance conntected in parallel that is 100 Ω is not be used.
The digital multimeter (DMM) can have its internal resistance determined without taking into account the 100 Ω resistor because the internal resistance is very high when compared to that of the parallel connection, and this does not change as the resistance principle remains the same.
What is the reason why the internal resistance of the DMM can be determined without taking into account the 100 Ω resistor?Generally, Singles connection of resistors when connected, it topples the resultant parallel connection of resistors.
Therefore
[tex]\frac{1}{R} = \frac{1}{Rs} * \frac{1}{Rp}[/tex]
Hence, because the internal resistance (R)is very high when compared to that of the parallel (Rp), the 100ohms is not put to use.
In conclusion, This could be done for a resistor of any resistance value as the resistance principle remains the same.
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