Answer:
1 m/s^2
Explanation:
F=ma
2 Newtons=2 Kg * a
a=1 m/s^2
A cryogenic vacuum pump works by condensing vapors onto some absorbent medium. This is an efficient and clean way to pump a system in a research environment. The term cryo means cold, which indicates that these types of vacuum pumps contain a refrigerant cycle to cool the internal parts. The temperature difference between the inside and outside of a typical cryogenic pump is Δ=303 ∘C . Derive an expression to convert this difference into Fahrenheit and express the answer.
Answer: Temperature in Fahrenheit is 577.4
Explanation:
The conversion factor for converting celcius to Fahrenheit is:
[tex]F=\frac{9}{5}\times C+32[/tex]
where F = temperature in Fahrenheit
C = Temperature in Celcius
Given : Temperature difference in Celcius = [tex]303^0C[/tex]
Putting in the values we get:
[tex]F=\frac{9}{5}\times 303+32[/tex]
[tex]F=577.4[/tex]
Thus the answer in Fahrenheit is 577.4
The required conversion of temperature from Celsius to Fahrenheit is 577.4 Degree Fahrenheit.
Given data:
The temperature difference between the inside and outside of a typical cryogenic pump is, [tex]\Delta T=303^{\circ}\rm C[/tex].
The degree measure of the flow of heat from one point to another is known as the temperature at a specific point. The temperature measurements are done in Celsius, Fahrenheit and Kelvin scales. The expression to convert the Celsius scale to Fahrenheit scale is,
[tex]T=(C \times \dfrac{9}{5} + 32)\;\rm ^{\circ}F[/tex]
Solving as,
[tex]T=(303 \times \dfrac{9}{5}+32 )\;\rm ^{\circ}F\\\\T=(303 \times \dfrac{9}{5}+32 )\;\rm ^{\circ}F\\\\T=577.4\;\rm ^{\circ}F[/tex]
Thus, we can conclude that the required conversion of temperature from Celsius to Fahrenheit is 577.4 Degree Fahrenheit.
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3. Materials that lets electricity to pass
Answer:
materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.
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]
in winter soap does not dissolve properly in winter why
Answer:
In winter, the temperature of the water is very low which makes the soap dissolve in a very small amount. ... Since the water molecules move faster, they come in contact with soap more often causing it to dissolve faster
Hope it helps...
Have a great day : )
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
Why was Mars orbiter sent to mars?
Explanation:
to develop the technologies required for designing, planning, management and operations of an interplanetary mission. The secondary objective is to explore Mars' surface features, morphology, mineralogy and Martian atmosphere using indigenous scientific instruments
Answer: Mars Climate Orbiter was lost on arrival September 23, 1999. Engineers concluded that the spacecraft entered the planet's atmosphere too low and probably burned up.
Explanation: The scientific reasons for going to Mars can be summarised by the search for life, understanding the surface and the planet's evolution, and preparing for future human exploration. Understanding whether life existed elsewhere in the Universe beyond Earth is a fundamental question of humankind.
A small dog is trained to jump straight up a distance of 1.1 m. How much kinetic energy does the 7.7 kg dog need to jump this high?
Answer:
83.09 J
Explanation:
The potential energy at the point of the top of the jump is represented by the equation
[tex]mgdeltah[/tex]
when the dog jumps, all the potential energy converts to kinetic energy (1/2mv^2). Plugging in the values:
(7.7)(4.184)(1.1) = 83.0907 J
Hello everyone I have a question for you today. So you know we have discovered black holes and have theorized about white holes so let me ask you without you going to the internet and looking it up what would happen if a white hole and a black hole collided?
Answer:
White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
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
Which time interval has the greatest speed?
Answer:
es la 2
Explanation:
epero que te curva
(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
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
What are the first three harmonics of a note produced on a 32.0 cm long violin string if waves on this string have a speed of 280.0 m/s?
___ Hz
___Hz
___Hz
Answer:
The first harmonics are the fundamental, the first overtone, and the second overtone,
For a string fixed at both ends you have:
N-A-N F/2
N-A-N-A-N F
N-A-N-A-N-A-N 3 F/2 each harmonic increases by F/2
Where N are the anti-nodes and A the antinodes
N-A-N-A-N represents 1 wavelength for string fixed at both ends
So 1 wavelength is 64 cm
280 m/s = F * .64 m and F = 437.5 Hz (second harmonic)
First harmonic = F/2 = 218.75 / sec
Third harmonic = 3 F / 2 = 656.25 /sec
Rhodium is in period 5 of the periodic table. What does this tell you about this element
(It’s number 4 in the picture)
Rhodium (Rh) is in period 5 on the periodic table. What does this tell you about this element ?
option - D
It has 5 energy levels/shells in its electron cloud.
With the information from the periodic table we can find that the correct statement is:
D. it is 5 energy level
The periodic table of the elements is a method of classifying the chemical elements based on their atomic number, it is divided into rows and columns
The rows are called the period and indicate the energy shell that is filled with electrons, all periods start with one electron and end when the shell is filled with 8 electrons.
The columns called groups correspond to elements that have the same number of electrons in their last shell, the chemical properties of the elements of the group are similar.
In this case we have an element the Rhodium, from period 5 and group VIIIB, when reviewing the period table it can be indicated that it has properties similar to cobalt and the inner layer is filled with two electors in its last layer.
When reviewing the different claims
a) False. The periodic table is not about molecules but about atoms
b) False. The metallic characteristics depend on the relationship between the number of electrons and the fermi level that is not indicated in the periodic table.
c) False. The number of protons is equal to the atomic number, in this case the rhodium has Z = 45, therefore it has 45 protons in the nucleus
d) True. It is at level 5 since the energy levels and the periods coincide.
In conclusion, with the information from the periodic table, we can find that the correct statement is:
D. it is 5 energy level
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The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
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.
a car travels at the speed of 117km/h.How far will the car travels in 50 minutes
Answer:
117/60*50
Explanation:
Trust me bc I'm smart
what dose nuclear reactions produce?
Answer:
a new chromebook for you and you will get to know the other one that
An 1120 kg car traveling at 17 m/s is brought to a stop while skidding 40 m. Calculate the work done on the car by friction forces.
Answer:
Approximately [tex](-1.6 \times 10^{5}\; \rm J)[/tex] (assuming that the car was on level ground.)
Explanation:
When an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of that object would be [tex](1/2) \, m \cdot v^{2}[/tex].
Initial kinetic energy of the car:
[tex]\begin{aligned}&\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2}\times 1120\; \rm kg \times (17\; \rm m \cdot s^{-1})^{2} \\ &\approx 1.6\times 10^{5}\; \rm J \end{aligned}[/tex].
After the car comes to a stop, the kinetic energy of this car would be [tex]0\; \rm J[/tex] because the car would not be moving.
Change to the kinetic energy of the car: [tex]\text{Final KE} - \text{Initial KE} = -1.6 \times 10^{5}\; \rm J[/tex].
If the car is traveling on level ground, friction would be the only force that contributed to this energy change. Hence:
[tex]\text{Work of friction} = \text{Energy change} = -1.6\times 10^{5}\; \rm J[/tex].
The value of the work that friction did is negative. The reason is that at any instant before the car comes to a stop, friction would be exactly opposite to the direction of the movement of the car.
The work of a force on an object is the dot product of that force and the displacement of that object. The dot product of two vectors of opposite directions is negative. Hence, in this question, the work that friction did on the car would be negative because the friction vector would be opposite to the movement of the car.
When we add or remove energy from a substance, what kind of changes can we observe? Can they happen at the same time ?
Answer:
When you add energy to a substance?
One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. The temperature at which a substance goes from a solid to a liquid is it melting point.
When you remove energy from a substance?
When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.
Explanation:
The addition of energy increases the kinetic energy of the particles, which reduces the intermolecular forces between the particles. Freezing occurs when a liquid becomes a solid and energy is released.
Atoms lose energy when they change from solid to liquid or gas and quid to gas.
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
A banana peel has lots of friction.
True or False
Answer:
False
Explanation:
I learned it the hard way trust me T^T
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.
?.............................
Please why does a simple pendulum graph fail to pass through the origin ?
Answer:
The friction of the string and its pivotal anchor point cannot be eliminated. The precise measurement of the length of the pendulum is difficult to take by using meter sticks or rulers. The value of the acceleration due to gravity g in the locality is not constant and must be obtained from reliable sources.
You cannot completely remove the friction between the string and its key anchor point. Using meter sticks or rulers makes it challenging to take an accurate measurement of the pendulum's length. Since it is not constant, it is necessary to get accurate information on the local gravity acceleration related to g.
What is a pendulum?The body is suspended from a central object to act as a pendulum that swings back and forth as a result of gravity. Because the period—the amount of time between each full oscillation—is constant, pendulums are employed to control the movement of clocks. The formula for a pendulum's period T is T = 2 Square root of L/g, where L is the pendulum's length and g is the acceleration brought on by gravity.
Galileo, an Italian scientist, made the first observation about the consistency of a pendulum's period by contrasting the motion of a swinging lantern in a cathedral in Pisa with his heart rate.
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A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude is 0.45 m, what is the frequency of the oscillation?
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s[/tex]
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
What is the scientific study of how animals are classified?
Answer:
biology
Explanation:
Animals and plants
What is local business?explain with exampls
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
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.03 mm apart and position your screen 3.19 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 641 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe
Answer:
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 2.978 mm
Explanation:
We have the following data:
λ = wavelength of light = 641 nm = 6.41 x 10⁷ m
L = Distance of Screen from slits = 3.19 m
d = slit separation = 1.03 mm = 1.03 x 10⁻³ m
Δx = distance between consecutive bright fringes = fringe spacing = ?
Using formula:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\Delta x = \frac{(6.41\ x\ 10^{-7}\ m)(3.19\ m)}{1.03\ x\ 10^{-3}\ m}[/tex]
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 1.5Δx
distance between central bright fringe and second dark fringe = (1.5)(1.985 mm)
distance between central bright fringe and second dark fringe = 2.978 mm