Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
Drag the titles to the correct boxes to complete the pairs.
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
(i) Final velocity
(ii)The time taken
Answer:
(I)
[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]
Newspapers often talk about an energy crisis-about running out of certain energy sources in the not-so-distant future. About which kind of energy sources are they talking
Answer:
Nonrenewable energy
Explanation:
Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.
Basically, a renewable energy source is sustainable and as such can not be exhausted.
On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.
For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.
In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.
Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
why is the water drawn from the bottom of the dam rather than the top?
Answer:
because minerals can be gotten from the bottom
Explanation:
it's self explanatory
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTA satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.
Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?
Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Answer:
a) A = 1.50 m / s², B = 1.33 m/s³, b) a = 12.1667 m / s²,
c) I = M (1.5 t + 1.333 t²) , d) ΔI = M 2.833 N
Explanation:
In this exercise give the expression for the speed of the rocket
v (t) = A t + B t²
and the initial conditions
a = 1.50 m / s² for t = 0 s
v = 2.00 m / s for t = 1.00 s
a) it is asked to determine the constants.
Let's look for acceleration with its definition
a = [tex]\frac{dv}{dt}[/tex]
a = A + 2B t
we apply the first condition t = 0 s
a = A
A = 1.50 m / s²
we apply the second condition t = 1.00 s
v = 1.5 1 + B 1²
2 = 1.5 + B
B = 2 / 1.5
B = 1.33 m/s³
the equation remains
v = 1.50 t + 1.333 t²
b) the acceleration for t = 4.00 s
a = 1.50 + 1.333 2t
a = 1.50 + 2.666 4
a = 12.1667 m / s²
c) The thrust
I = ∫ F dt = p_f - p₀
Newton's second law
F = M a
F = M (1.5 + 2 1.333 t) dt
we replace and integrate
I = M ∫ (1.5 + 2.666 t) dt
I = 1.5 t + 2.666 t²/2
I = M (1.5 t + 1.333 t²) + cte
in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant
cte = 0
I = M (1.5 t + 1.333 t²)
d) the initial push
For this we must assume some small time interval, for example between
t = 0 s and t = 1 s
ΔI = I_f - I₀
ΔI = M (1.5 1 + 1.333 1²)
ΔI = M 2.833 N
A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
An electric eel can generate a 180-V, 0.1-A shock for stunning its prey. What is the eel's power output
Power output = volts x amps
Power output = 170 volts x 0.1 amps
Power output = 18 watts
Two cylindrical resistors are made from copper. The first one is of length L and of radius r . The 2nd resistor is of length 6L and of radius 2r. The ratio of these two resistances R1/R2 is:
Answer:
[tex]R1/R2=\frac{2}{3}[/tex]
Explanation:
From the question we are told that:
1st's Length [tex]l=L[/tex]
1st's radius [tex]r=r[/tex]
2nd's Length [tex]l=6L[/tex]
2nd's radius [tex]r=2r[/tex]
Generally the equation for Resistance R is mathematically given by
[tex]R=\frac{\rho L}{\pi r^2}[/tex]
Therefore
[tex]R_1=\frac{\rho L}{\pi r^2}[/tex]
And
[tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]
Therefore
[tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]
[tex]R1/R2=\frac{2}{3}[/tex]
0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K
Answer:
c. 250k
Explanation:
The temperature of the sink is approximately 250 K.
To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:
Efficiency = 1 - (Temperature of Sink / Temperature of Source)
Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):
Efficiency = (Q_source - Q_sink) / Q_source
Efficiency = (2003 J - 100 J) / 2003 J
Efficiency = 1903 J / 2003 J
Efficiency = 0.9497
Now, plug the efficiency back into the first equation to solve for T_sink:
0.9497 = 1 - (T_sink / 500 K)
T_sink / 500 K = 1 - 0.9497
T_sink / 500 K = 0.0503
Now, isolate T_sink:
T_sink = 0.0503 * 500 K
T_sink = 25.15 K
Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.
To learn more about sink energy, here
https://brainly.com/question/10483137
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A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
magnitude m/s
direction ° counterclockwise from the +x-axis
Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.
Answer:
Answer
Correct option is
A
5×10
−6
tesla
I=5A
x=0.2m
Magnetic field at a distance 0.2 m away from the wire.
B=
2πx
μ
0
I
=
2π×0.2
4π×10
−7
×5
=10×5×10
−7
=5×10
−6
tesla
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.
a. True
b. False
Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.
Answer: The correct statement is:
--> The voltage across each resistor is the same.
Explanation:
RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:
--> Resistors in parallel and
--> Resistors in series
RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.
