A class is learning about states of matter. The students set up the investigation in the diagram.
Which kinds of energy are needed in this investigation to change the state of matter of the owl made of wax?
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answer:
102.5N
Explanation:
Given that a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W
The power = 4.12 × 746 = 3073.52 W
Using the formula
Power = force × velocity
3073.52 = force × 30
Force = 3073.52 / 30
Force = 102.5 N
Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,
the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N
PLEASE HELP The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the cost? In three to five sentences, provide two specific examples of things we have learned from space exploration, and explain how these examples influence your opinion.(4 points)
Answer: I think $20 billion a year it’s worth the cost. The reasoning behind that is because we can conduct research on various things that could help out humanity. Therefore we can conclude that’s spending billions of dollars every year is worth it.
Explanation:
A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW
Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = [tex]\frac{1}{2}[/tex]vμω²A²
p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
A ball is sitting at the top of a ramp. As the ball rolls down the ramp, the potential energy of the ball decreases, what happens to the potential energy as the ball moves
Answer:
the potential energy decreases as it is converted to kinetic energy.
Explanation:
As things move, their potential energy converts to kinetic energy to power them along. When a ball rolls down the top of a ramp, all the potential energy it accumulated at the top of the ramp converts to kinetic energy to help it roll down. In other words, its potential energy decreases as its kinetic energy increases.
You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pushing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks
Answer:
the energy in the oscillations between the blocks is 3.025 J
Explanation:
Given the data in the question;
Force f = 138 N
stiffness of spring k = 605 kg/s²
mass of block = 202 g = 0.202 kg
pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm
i.e
x₁ = 15 cm
x₂ = 5cm
the energy in the oscillations between the blocks will be;
E[tex]_A[/tex] = E[tex]_B[/tex] = [tex]\frac{1}{2}[/tex]k( Δx )²
we substitute
= [tex]\frac{1}{2}[/tex] × k( 15 - 5 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × ( 10 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × 100 × 10⁻⁴
= 3.025 J
Therefore, the energy in the oscillations between the blocks is 3.025 J
In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 seconds. When the car rebounds from the barrier [in the opposite direction], the magnitude of its velocity is 1.5 mi/h. Use the principle of impulse and momentum to determine the magnitude of the average horizontal force (lb) exerted on the car during the impact.
Answer:
F = 2074.13 lb
Explanation:
Given that,
Mass of car, m = 2800 lb = 1270.059 kg
Initial speed, u = 5 mi/h = 2.2352 m/s
Final speed, v = - 1.5 mi/h = -0.67056 m/s (in opposite direction)
Time, t = 0.4 s
We need to find the magnitude of the average horizontal force (lb) exerted on the car during the impact. It can be calculated as :
[tex]F=m\times \dfrac{v-u}{t}\\\\F=1270.059\times \dfrac{-0.67056 -2.2352 }{0.4}\\\\F=9226.21\ N[/tex]
or
F = -2074.13 lb
So, the required force is 2074.13 lb.
Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0
m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?
Answer:
v = 2.57 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. When it is at an angle of 30º
Em₀ = K + U = ½ m v₁² + m g y₁
final point. Lowest position
Em_f = K = ½ m v²
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v₁² + m g y₁ = ½ m v²
Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle
cos 30 = L ’/ L
L ’= L cos 30
y₁ = L -L '
y₁ = L- L cos 30
we substitute
½ m v₁² + m g L (1- cos 30) = ½ m v²
v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]
let's calculate
v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]
v = 2.57 m / s
Can two waves have the
same wavelength but different amplitudes?
Explain
Answer:
I think Yes because they could have different amounts of energy.
Explanation:
TRUE OR FALSE
2 QUESTIONS
NEED HELP ASAP
THX :)
LOTS OF POINTS :>
Answer: Both False
Explanation:
Our Milky Way Galaxy is a spiral galaxy. Some spiral galaxies are what we call "barred spirals" because the central bulge looks elongated
Irregualuar glaxyices are all over the place
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the period of its rotation?
Answer:
[tex]\huge\boxed{T = 0.025\ seconds}[/tex]
Explanation:
Given:
Frequency = f = 40 Hz
Required:
Time period = T = ?
Formula:
[tex]\sf T = 1 / f[/tex]
Solution:
T = 1 / 40
T = 0.025 seconds
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807an inventor makes a clock using a brass rod and a heavy mass as a pendulum.WHAT Happens when the clock get colder?
An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time
What is thermal expansion?The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.
As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.
The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum
T = 2π√(L/g)
As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time
Thus, if a pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder
Learn more about Thermal expansion here
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23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m, (b) the gravitational potential energy at 21.4 m, (c) the total mechanical energy at 21.4 m, (d) the kinetic energy at 0 m, (e) the gravitational potential energy at 0 m, and (f) the total mechanical energy at 0 m.
Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by : [tex]E_k=\dfrac{1}{2}mv^2[/tex]
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by : [tex]E_p=mgh[/tex]
It can be calculated as :
[tex]E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J[/tex]
(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.
The average mean distance of Saturn from the sun is
Answer:
From an average distance of 886 million miles (1.4 billion kilometers), Saturn is 9.5 astronomical units away from the Sun. One astronomical unit (abbreviated as AU), is the distance from the Sun to Earth. From this distance, it takes sunlight 80 minutes to travel from the Sun to Saturn.
we have that from the Question"The average mean distance of Saturn from the sun is" it can be said that Tthe average mean distance of Saturn from the sun is
A distance of 1427 x 10^6 km or 886 696 691 milesFrom the Question we are told
The average mean distance of Saturn from the sun is
Generally
The Sun is the star of the milky way galaxy and its distance from every planet in the milky way determines in one way or another its properties and in-habitability
Saturn being a Planet of the milky way we see that Saturn is a significant distance away from sun
A distance of 1427 x 10^6 km or 886 696 691 miles
Therefore
The average mean distance of Saturn from the sun is
A distance of 1427 x 10^6 km or 886 696 691 miles
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Rhodium is in period 5 of the periodic table. What does this tell you about this element
Answer:
. It is an extraordinarily rare, silvery-white, hard, corrosion-resistant, and chemically inert transition metal. It is a noble metal and a member of the platinum group.
Explanation:
A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp
Answer:
Both.
Explanation:
Given that a block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp ?
Since both of them have the same mass and the same initial velocity, then, they will both have the same kinetic energy.
That is,
K.E = 1/2mv^2
Friction is a force that opposes motion. And since the frictional force is zero,
Both of them will accelerate from Newton's law.
F = ma
We can therefore conclude that both of them will make it further up the ramp.
Two students Tim and Alane travel to South Dakota. Tim stands on Earth’s surface and enjoys some sunshine. At the same time, Alane descends into a gold mine where neutrinos are detected, Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
Explanation:
Neutrinos are otherwise called leptons. They are principal particles. A lepton is a rudimentary half-spin molecule that doesn't go through solid reactions. Neutrinos are not usually charged and exceptionally light weighted so they once in a while interface with other matter. Neutrinos are light weighted. Their mass is around 10⁻⁷ kg. A neutrino possesses a small radius, too little to ever be estimated. A little span and very less mass make them imperceptible. Since neutrinos have next to no mass. they travel at almost the speed of light and thus they arrive at the outside of the Sun in only 2 seconds, dissimilar to photons which take convoluted ways to arrive at the Sun's surface in a huge number of years.
The photon and neutrino, both were made in the Sun's center yet on various occasions. The neutrino is only a couple of minutes old though the photon is around 1,000,000 years of age. At the point when the photon was made in the Sun's center. it needed to venture out to the outside of the Sun. in any case, rather because of its hefty mass and cooperation with other matter, it headed out a crisscross way to the surface. Ordinarily, it was repulsed and it was sent back to the middle where it needed to begin once more. It required a large number of years for a photon to arrive at the outside of the Sun.
Nonetheless, when it arrived at the Sun's surface, it required just 8.8 minutes for the photon to arrive at Earth. The neutrino was anyway made only a couple of minutes prior in the Sun's center. Since it has an entirely irrelevant mass, little size, and no charge, it didn't interface with its environmental factors. So it just required 2 seconds for the neutrino to arrive at the Sun's surface. When it arrived at the Sun's surface, it arrived at the earth in about 8.8 minutes. with the photon. So both, photon and neutrino have various histories as the two of them were made at a hole of around 1,000,000 years.
Diffraction occurs for all types of waves, including sound waves.
a. True
b. False
Answer:
a. True
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
[tex] Speed = wavelength * frequency [/tex]
Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.
Diffraction occurs for all types of waves, including sound waves.
A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?
Answer: North East
Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.
If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1x104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.1 m to 0.5 m amplitude?
Answer: Hello, Mark me as Brainliest! :)
If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .
Your Welcome!
Explanation:
A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.
Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
(1) Which appliance is designed to transfer electrical energy to kinetic energy?
D)
A food mbuer
BB kettle
Clamp
D radio
Answer:
bb kettle
Explanation:
it transfres electricsl to kinetic
Which is an example of a noncontract force?
(A) elastic force
(B) Normal force
(C)Applied force
(D) Electric Force
Answer:
electric force
Explanation:
its a contact and noncontact force
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.
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