Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
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Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
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Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? =
Answer:
x = -3 cm
Explanation:
The electrical potential is the sum of the potentials of each charge
V = k ∑ [tex]q_{i} / r_{i}[/tex]
let's apply this to our case where the potential is V = 0 for x = 0
0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))
in our case
q₁ = + 2.0 10⁻⁶ C
q₂ = - 6.0 10⁻⁶ C
q₃ = + 3.0 10⁻⁶ C
x₁ = -1.0 cm = 1.0 10⁻² m
x₂ = +2.0 cm = 2.0 10⁻² m
we substitute in the equation
0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)
3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴
3 10⁻⁶ / x = -1 10⁻⁴
x = - 3 10⁻² m
x = -3 cm
An individual is moving out of his apartment and has several boxes to carry down the stairs. Some of the boxes are light and some of the boxes are heavy. 3. Using the concept of motor units, describe how he generates more or less power to move the boxes.
Answer:
Motor unit is made up of motor neurons. Group of Motor Units work together to coordinate contraction of single muscle. When an individual needs to carry several boxes down stairs then his brain gives signal to neurons who innervates a group of skeletal muscle. When neurons receive signal they stimulate all muscle fibers in that particular motor unit.
Explanation:
Motor unit is made up of motor neurons. Group of Motor Units work together to coordinate contraction of single muscle. When an individual needs to carry several boxes down stairs then his brain gives signal to neurons who innervates a group of skeletal muscle. When neurons receive signal they stimulate all muscle fibers in that particular motor unit.
An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron
Answer:
acting force is the answer
The direction of the magnetic force on the moving electron is upward.
The direction of the magnetic force on the electron can be determined by applying right hand rule.
This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.
Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.
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House voltage is 240 volts. If a light bulb runs a current of 0.5 amps, what's the
resistance of the bulb?
Answer:
480 Ohms
Explanation:
Resistance: This can be defined as a opposition to current flow in an electric circuit. The S.I unit of resistance is ohms(Ω).
From the question,
V = IR................... Equation 1
Make R the subject of the equation
R = V/I............... Equation 2
V = The House voltage, I = Current, R = Resistance of the bulb.
Given: V = 240 Volts, I = 0.5 Amps.
Substitute into equation 2
R = 240/0.5
R = 480 Ohms
Hence the resistance of the bulb is 480 Ohms
One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic field along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]
Therefore, the voltage induced in the coil is 1.25 V.
Equal but opposite charges Q are placed on the square plates of an air-filled parallel-plate capacitor. The plates are then pulled apart to twice their original separation, which is small compared to the dimensions of the plates. Which of the following statements about this capacitor are true?
A) The energy stored in the capacitor has doubled.
B) The energy density in the capacitor has increased.
C) The electric field between the plates has increased.
D) The potential difference across the plates has doubled.
E) The capacitance has doubled.
Answer:
A& D
Explanation:
See attached file
The temperature coefficient of resistivity for copper is 0.0068 (C°)-1. If a copper wire has a resistance of 104 Ω at 20°C, what is its resistance 80°C?
Answer:
R₈₀ = 146.43 Ω
Explanation:
The resistance of a resistor depends upon many factors. One of the main factors of the change in resistance of a resistor is the change in temperature. The formula for the resistance at a temperature other than 20°C is given as follows:
R₈₀ = R₀(1 + αΔT)
where,
R₈₀ = Resistance of wire at 80°C = ?
R₀ = Resistance of wire at 20° C = 104 Ω
α = Temperature coefficient of resistance for copper = 0.0068 °C⁻¹
ΔT = T₂ - T₁ = 80°C - 20°C = 60°C
Therefore,
R₈₀ = (104 Ω)[1 + (0.0068°C⁻¹)(60°C)]
R₈₀ = 146.43 Ω
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
[tex]F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N[/tex]
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
[tex]W = F_{net} *h\\\\W = 15.328 *10^{-3} * h[/tex]
W = K.E
[tex]15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m[/tex]
Therefore, the ball traveled 0.827 m
The height at which the ball goes for the given parameters is; 0.827 m
What is the height of the ball?We are given;
distance between the metal plates; d = 3.1 m
mass of glass; m = 1.1g = 0.0011 kg
charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C
speed of the glass ball; v = 4.8 m/s
voltage of the ceiling; V = +3.0 × 10⁶ V
The repulsive force experienced by the ball is gotten from the formula;
F = qV/d
|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1
|F| = 4.548 × 10⁻³ N
F = -4.548 × 10⁻³ N (negative because it is repulsive force)
The net horizontal force experienced by the ball is;
F_net = F - mg
F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)
F_net = -15.328 × 10⁻³ N
To get the height of the ball, we will use the formula;
F_net * h = ¹/₂mv²
h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)
We took the absolute value of F_net, hence it is not negative
h = 0.827 m
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Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.
Explanation:
It is given that,
The separation between two parallel wires, r = 11 cm = 0.11 m
Current in wire 1, [tex]q_1=54\ A[/tex]
Current in wire 2, [tex]q_2=45\ A[/tex]
Length of wires, l = 4.3 m
We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]
So, the magnetic force on a 4.3 m length of the wire on both of currents is F=0.0189 N.
car travel at the speed of 20 km/hr for 2 hour and 60 km/hr for next 2 hour find average speed
What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t
Answer:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Explanation:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
When you replace helium in a balloon with less-dense hydrogen, does the buoyant force on the balloon change if the balloon remains the same size?
Answer:
No change
Explanation:
First, we hare to understand what we mean by buoyant force.
Archimedes Principle states that
"the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object."
Hot air balloons rise into the air as a result of the density of the air inside the balloon is less dense i.e warmer air than the air outside the balloon i.e cooler air. This is basically how the balloons work. Now applying this to our question
Replacing Helium with less dense Hydrogen will make no difference to the buoyant force because the volume of the balloon did not change. The buoyant force depends on the weight of the displaced air, and not on the force causing the displacement.
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.0050 m) is attached to the end of the hose in units of m/s.
Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
v₂ = 306.12 m/s
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each block is suspended just above the bottom of the aquarium by a thread. Which of the following is true?
A. More information is needed to choose the correct answer.
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
C. The buoyant force on the lead block is greater than the buoyant force on the copper block.
D. The buoyant force is the same on both blocks.
Answer:
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
Explanation:
Given;
mass of lead block, m₁ = 200 g = 0.2 kg
mass of copper block, m₂ = 200 g = 0.2 kg
density of water, ρ = 1 g/cm³
density of lead block, ρ₁ = 11.34 g/cm³
density of copper block, ρ₂ = 8.96 g/cm³
The buoyant force on each block is calculated as;
[tex]F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )[/tex]
The buoyant force of lead block;
[tex]F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173 \ N[/tex]
The buoyant force of copper block
[tex]F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219 \ N[/tex]
Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block
10 pts! :) If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much work did Kyla do on the bag?
Explanation:
work = force x Distance
w = 10 x 1.5 = 15Nm
The amount of work done by Kyla in lifting the bag is 15 J.
What is meant by work done ?Work done on an object is defined as the cross product of the force applied on the object and the vertical displacement of the object.
Here,
Force applied by Kyla to pick up the bag, F = 10 N
Vertical displacement of the bag, s = 1.5 m
The work done by Kyla in lifting the bag,
W = F x s
W = 10 x 1.5
W = 15 J
Hence,
The amount of work done by Kyla in lifting the bag is 15 J.
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A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same.
A)
The new sphere has a density of ρ = ρ0 and a mass of m < m0.
B)
The new sphere has a density of ρ = ρ0 and a radius of r > r0.
C)
The new sphere has a density of ρ < ρ0 and a mass of m = m0.
The options are r, f, and s. Rises, Falls, Stays the same.
Answer:
(a) f
(b) r
(c) s
Explanation:
There are two forces on the sphere: weight and buoyancy.
Sum of forces in the y direction:
∑F = ma
B − mg = 0
B = mg
Buoyancy is equal to the weight of the displaced fluid, or ρVg, where ρ is the density of the fluid and V is the displaced volume.
ρVg = mg
ρV = m
V = m/ρ
(a) The mass decreases, so the displaced volume decreases.
(b) The sphere's density is constant and its radius increases, which means its mass increases, so the displaced volume increases.
(c) The mass stays the same, so the displaced volume is the same.
A 11.7-Ω resistor is made from a coil of copper wire whose total mass is 13.5 g . The resistivity of copper is 1.68×10−8Ω⋅m, the density of copper is 8.9×103kg/m3.
a) What is the diameter of the wire?
b) What is the length of the wire?
Answer:
a) d = 7.62 10⁻⁶ m, b) l = 3.25 10⁴ m
Explanation:
Resistance is expressed by the formula
R = ρ l / A (1)
density is defined by
density = m / V
the volume of a wire is the cross section by the length
V = A l
we substitute
density = m / A l
A = m / density l
we substitute in 1
R = ρ l density l / m
R =ρ density l² / m
l = √ (R m /ρ density)
let's calculate the cable length
l = √(11.7 13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))
l = √(10.56 10⁸)
l = 3.25 10⁴ m
now we can find the cable diameter with the density equation
A = m / density l
A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)
A = 4,557 10⁻¹¹ m²
the area of the circle is
A = π r² = π d² / 4
d = √ (4A /π)
d = √ (4 4,557 10⁻¹¹/π)
d = 7.62 10⁻⁶ m
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
Data;
Resistor = 11.7Ωmass = 13.5gresistivity of copper = 1.68 * 10 ^-8 Ω.mdensity of copper = 8.9*10^3 kg/m^3Resistivity of CopperThe resistivity of copper is calculated by
[tex]R = \frac{\rho L}{A}\\[/tex]
Let's calculated the volume of the wire first;
[tex]\rho = \frac{mass}{volume} \\volume = \frac{mass}{density} \\volume = \frac{13.5*10^-^3}{8.9*10^3} \\v = 1.52*10^-6m^3[/tex]
The diameter of the wire will be
[tex]R = \frac{\rho L}{A}\\ R = \frac{\rho LA}{A^2}\\ R = \frac{\rho V}{A^2} \\A^2 = \frac{\rho V}{R}\\ A^2 = \frac{8.9*10^3 * 1.516*10^-^6}{11.7} \\A^2 = 0.0011\\A = \sqrt{0.0011} \\A = 0.034m^2[/tex]
Taking the area
[tex]a = \pi r^2\\0.034 = 3.14 * r^2\\r^2 = \frac{0.034}{3.14} \\r^2 = 0.01083\\r = \sqrt{0.01083}\\ r = 0.104m\\d = 2r\\d = 2 * 0.104\\d = 0.208m[/tex]
Length of the wire can be calculated as
[tex]V = AL\\L = V/A\\L = \frac{1.52*10^-^6}{0.034}\\ L = 4.47*10^-^5m[/tex]
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
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Two factors that regulate (control) glandular secretion.
Answer:
The factors include age and puberty
Explanation:
Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.
As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.
Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.
A particle might be placed:___________.
1. inside a uniform spherical shell of mass M, but not at the center
2. inside a uniform spherical shell of mass M, at the center
3. outside a uniform spherical shell of mass M, a distance r from the center
4. outside a uniform solid sphere of mass M, a distance 2r from the center
Rank these situations according to the magnitude of the gravitational force on the particle, least to greatest.
A) All tie
B) 1, 2, 3, 4
C) 1 and 2 tie, then 3 and 4 tie
D) 1 and 2 tie, then 3, then 4
E) 1 and 2 tie, then 4, then 3
Answer: D) 1 and 2 tie, then 3, then 4.
Explanation: Gravitational Force is one of the 4 fundamental forces of nature (the others being eletromagnetic, weak force and strong force) and is explained by Newton's Law of Universal Gravitation, in which, every massive particle attracts every other massive particle.
The way to calculate gravitational force is:
[tex]F_{G} = \frac{GMm}{r^{2}}[/tex]
G is Universal Gravitational Constant and it is constant anywhere in the universe.
M and m are mass of each object;
r is distance between the objects;
The formula shows that gravitational force acts at an inverse square law, i.e., if objects are 2 distances apart, Force between them will decrease to one quarter of it was.
So, analysing the question, a particle inside an uniform spherical shell not at the center and at the center will have equal forces.
Outside at a distance 2r, force will be less than at a distance r.
Then the order is 1=2 > 3 > 4
Let’s measure precisely how much electromotive force the generator can produce. The amount of electromotive force is measured in volts. In fact, electromotive force is sometimes referred to as voltage. Use a digital multimeter to measure the voltage of your generator. Connect the free ends of the probe wires to the wires coming out of the generator. What is the voltage reading when the generator is idle (i.e. when the crank is not being turned?)
Answer:
fem = 0
Explanation:
The electromotive force is given by Faraday's law
fem = - d [tex]\phi _{B}[/tex] / dt
where \phi _{B} is the flux of the magnetic field
The low magnetic flux is
\phi _{B} = B. A = B A cos θ
substituting
fem - d (B A cos θ) / dt
therefore for an electromotive force to exist there must be a change with respect to the time of
* The magnetic field
* The area waxed by the loop
* the angle between the magnetic field and the normal loop
* Any combination of the above
In this case, as the generator is stopped (inactive), there is no change in the parameters, therefore the electromotive force is zero (fem = 0).
Consider 4 charges placed at the corners of a square with side 1.25m as shown. What are the magnitude and direction of the electrostatic force on Q1 resulring from other three charges?
(Note: Please put your final answer in 4 decimal places.). pls answer:)
Answer:
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
F = 3.0226 10⁻² N , θ = 21.16º
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r₁₂²
We also use that the force is a vector magnitude, so we must calculate each component of the force , see the adjoint for the direction of the vectors
X axis
Fₓ = -F₁₄ + F₁₃ₓ
Y axis
[tex]F_{y}[/tex] = F₁₂ -F_{13y}
let's look for the expression for each force
where the side of the square is a = 1.25 m
F₁₂ = k Q₁Q₂ / a²
F₁₄ = k Q₁Q₄ / a²
the distance between 1 and 3 is
d = √(a² + a²) = a √2
F₁₃ = k Q₁Q₃ / d²
let's use trigonometry to find the components
cos 45 = F₁₃ₓ / F₁₃
F₁₃ₓ = F₁₃ cos 45
F₁₃ₓ = k Q₁Q₃ / 2a²
sin 45 = F_{13y} / F₁₃
F_{13y} = F₁₃ sin 45
F_{13y} = k Q₁Q₃ / 2a² sin 45
Taking all terms, we substitute in the force for each axis
X axis
Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45
Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2 cos 45)
Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25² (- 4.5 10⁻⁶ + 3.5/2 cos 45 10⁻⁶)
Fₓ = 8.64 10³ (3.2626 10⁻⁶)
Fₓ = 2.8189 10⁻² N
Y axis
F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a² sin 45
F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)
F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25² (2.5 10⁻⁶ - 3.5/2 sin 45 10⁻⁶)
F_{y} = 8.64 10³ (1.26256 10⁻⁶)
F_{y} = 1.0909 10⁻² N
The result can be given in two ways
1) F = Fₓ i ^ + F_{y} j ^
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry
F = √ (Fₓ² + F_{y}²)
F = 10⁻² √ (2,8189² + 1,0909²)
F = 3.0226 10⁻² N
tan θ = F_{y} / Fx
θ = tan⁻¹ (F_{y} / Fₓ)
θ = tan⁻¹ (1.0909 / 2.8189)
θ = 21.16º
If an inductor, a capacitor, and a resistor are connected in series with a sine wave generator, what quantity will be common to all three components
Answer:
CURRENTExplanation:
For series connected elements (an inductor, a capacitor, and a resistor) in a simple AC circuit, the same current will flows through the elements since are are no presence of nodes between the elements. The total current from the source is what will flow through all of them.
For example, let assume the total current flowing in the circuit is 3A, the amount of current that will flow through the inductor, capacitor and resistor will be the same 3A because of the nature of their connection (series). It is the voltage across each of them that differs.
A 4.0 kg mass is attached to a spring whose spring constant is 950 N/m. It oscillates with an amplitude of 0.12 m. What is the maximum velocity of the mass
Answer:
velocity = 2.62m/s
Explanation:
950= (4 x A)/0.12
950 x 0.12 = 4 x A
114 = 4 x A
A = 114/4
A = 28.5m/s²
U²=2asU² = 2 x 28.5 x 0.12U² = 6.84U = √6.84U = 2.62m/sThe maximum velocity of the mass is equal to 1.85,/s when the amplitude of oscillation is 0.12m.
What is the spring force?The spring force will be acting on the spring when the spring is stretched or compressed, which opposes the load force. These springs are divided into many types based on how this load force is applied to them.
F = -kx
where k is the spring constant and x is the displacement of the spring attached with mass.
Given, the mass attached to the spring, m = 4.0 Kg
The value of spring constant, k = 950 N/m
The amplitude of oscillation, A = 0.12m
The maximum velocity can be calculated as:
[tex]\frac{mv_{max}^2}{2} =\frac{kA^2}{2}[/tex]
[tex]v_{max} =\sqrt{\frac{kA^2}{m} }[/tex]
Substitute the values of the m, k, and A in the above equation:
Vmax = [tex]=\frac{950N/m(0.12m)^2}{4Kg}[/tex]
Vmax = √3.42 m/s
Vmax = 1.85m/s
Therefore, the maximum velocity of the mass is equal to 1/85 m/s.
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Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.
You push a shopping cart full of groceries. The shopping cart has a mass of 32
kg.
a. What is the weight of your shopping cart?
b. How much force must you apply to give the cart an acceleration of 1.3 m/s2?
c. If you push with a force of 200 N, what is the acceleration of the cart?
d. You are driving home from the store. If your car has an acceleration of 4.1 m/s2 and its engine is applying a forward force of 7000 N, what is the mass of your car?
Answer:
a) 320N (if you take gravity as 10ms^-2) or 313.6N (if you take gravity as 9.8ms^-2
b) 41.6N
c) 6.25ms^-2
d) 1707.317 kg
Explanation:
a) W=mg
W= 32 x 10 or 9.8
W = 320N or 313.6N
b) F=ma
F= 32 x 1.3
F= 41.6N
c) F=ma
200 = 32 x a
a= 6.25ms^-2
d) F=ma
7000= m x 4.1
m= 1707.317 kg
I am not completely sure about the d) part because I dont whether you will be taking Friction and Normal Reaction too. As per my knowledge, I think no, as no angles nor the gradient of the floor/road is mentioned here.
a) The weight of your shopping cart is 313.6N .
b) The required force is 41.6N
c) The acceleration of the car is 6.25ms^-2
d) The mass of the car is 1707.317 kg.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
a) The weight of the shopping cart: W=mg
W= 32 x 9.8 N
W = 313.6N
b) The required force is to be applied: F=ma
F= 32 x 1.3 N.
F= 41.6N.
c) Let the acceleration of the cart is a.
Then, force: F=ma
200 = 32 x a
a= 6.25ms^-2
Hence, the acceleration of the cart is 6.25ms^-2.
d) Let the mass of the car is m.
Force applied on the car: F=ma
7000= m x 4.1
m= 1707.317 kg
The mass of the car is 1707.317 kg.
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Find an article online or application in your daily life involving rotating objects and physics.
Answer:
the planet Earth is a good example