Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
a circuit shown below is Wheastone Bridge used to determine the valve of unknown resistor X by comparison with three resistors M,N,P whose resistances can be varied. For each setting, the resistances of each resistor is precisely known. With switches k1and k2 closed, these resistors are varied until the current in the galvanometer G is zero; the bridge is then said to be balanced. (a) if the galvanometer G shows zero deflection when M=850.0, N=15.00 and P=33.48, what is the unknown resistance X?
Answer:
X = 0.6
Explanation:
The resistance of the unknown resistor can be found by using the formula of the Wheatstone bridge:
[tex]\frac{M}{N}=\frac{P}{X}\\\\\frac{850}{15} = \frac{33.48}{X}\\\\X = \frac{(33.48)(15)}{850}[/tex]
X = 0.6
Hence, the unknown value of resistance is found to be 0.6 units.
Increased air pressure on the surface of hot water tends to
A) prevent boiling.
B) promote boiling.
C) neither of these
Two objects attract each other with a gravitational force of magnitude 1.00 3 1028 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each
Answer:
The mass of each object is 2kg and 3 kg.
Explanation:
Given that,
Gravitational force,[tex]F=1\times 10^{-8}\ N[/tex]
The distance between masses, d = 20 cm = 0.2 m
The total mass of the two objects, M + m = 5 kg
M = 5-m
The formula for the gravitational force is :
[tex]F=G\dfrac{Mm}{d^2}\\\\1\times 10^{-8}=6.67\times 10^{-11}\times \dfrac{(5-m)m}{(0.2)^2}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}=\frac{(5-x)x}{(0.2)^{2}}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}\cdot(0.2)^{2}\\\\5.99=(5-x)x\\\\x=2\ kg\ and\ 3 \ kg[/tex]
So, the mass of each object is 2kg and 3 kg.
Where is the sun in relation to earths orbit
What's the speed of a sound wave through water at 25 Celsius?
A. 1,000 m/s
B. 1,500 m/s
C. 1,250 m/s
D. 750 m/s
Answer:
B) 1500m/s
Explanation:
Ans is 1500m/s
If the resistance in a circuit remains constant, what happens to the electric power when the current increases?
The power will increase.
B.
The power will decrease.
Ο Ο Ο Ο
There will be no power.
D
The current does not affect the power.
Answer:
Resistance is inversly proportional to the current.
V=I.R.
P=V.I
An object accelerates from rest, and after traveling 145 m it has a speed of 420 m/s. What was the acceleration of the object?
I am not sure how to calculate acceleration without being given the time directly.
Explanation:
Here,we've been given that,
Initial velocity (u) = 0 m/s (as it starts from rest)Distance (s) = 145 mFinal velocity (v) = 420 m/sWe've to find the acceleration of the object. By using the third equation of motion,
→ v² - u² = 2as
→ (420)² - (0)² = 2 × a × 145
→ 176400 - 0 = 290a
→ 176400 = 290a
→ 176400 ÷ 290 = a
→ 608.275862 m/s² = a
If you know initial speed and final speed, you can find the average speed. Then, knowing distance, you can find the time.
KimYurii posted the first answer to this question.
That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.
My problem is that I can never remember all the different formulas. I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo. Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.
When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:
Starting speed . . . zero
Ending speed . . . 420 m/s
Formula: Average speed . . . (1/2)·(0 + 420) = 210 m/s
Distance covered . . . 145 m
Formula: Time taken = (distance) / (average speed) = (145/210) second
(Now you have the time.)
Formula: Distance = (1/2)·(acceleration)·(time²)
145 m = (1/2)·(acceleration)·(145/210 sec)²
Acceleration = 290 m / (145/210 s)²
Acceleration = 608.28 m/s²
A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
Here,
v = 45m/s
t = 5s
d = v × t
Therefore,
d = 45 × 5
= 225m
If 2cm³ of wood has a mass 0.6g what would be its density
we know density = mass/ volume
as mass = 0.6 g
and volume = 2cm³
so density = (6/20)(g/cm³)
0.3g/cm³ (ans)
Hope it helps
Which of the following is a noncontact force?
O A. Friction between your hands
O B. A man pushing on a wall
O C. Air resistance on a car
D. Gravity between you and the Sun
Answer:
Gravity between you and the sun
A solid piece of clear transparent material has an index of refraction of 1.61. If you place it into a clear transparent solution and it seems to disappear, approximately what is the index of refraction of the solution
Answer:
1.61
Explanation:
According to Oxford dictionary, refractive index is, ''the ratio of the velocity of light in a vacuum to its velocity in a specified medium.''
If the clear transparent solid disappears when dipped into the liquid, it means that the index of refraction of the solid and liquid are equal.
Hence, when a transparent solid is immersed in a liquid having the same refractive index, there is no refraction at the boundary between the two media. As long as there is no refraction between the two media, the solid can not be seen because the solid and liquid will appear to the eye as one material.
I HAVE A PHYSICS TEST, ITS 25 QUESTIONS AND I HAVE ABOUT AN HOUR TO SOLVE IT PLEASE IF YOU'RE GOOD AT PHYSICS CONTACT ME ASAP
Answer:
yes sir
Explanation:
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)
Answer:
Explanation:
Given:
volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]
time taken for the discharge, [tex]t=30~s[/tex]
diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]
length of cylindrical urethra, [tex]l=0.2~m[/tex]
density of urine, [tex]\rho=1000~kg/m^3[/tex]
a)
we have volume flow rate Q:
[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]
where:
[tex]A=[/tex] cross-sectional area of urethra
[tex]v=[/tex] velocity of flow
[tex]A.v=\frac{V}{t}[/tex]
[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]
[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]
[tex]v=1.06~m/s[/tex]
b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:
[tex]P=\rho.g.l[/tex]
[tex]P=1000\times 9.8\times 0.2[/tex]
[tex]P=1960~Pa[/tex]
If the moon started it's orbit around the Earth from a spot in line with a certain star, it will return to that same spot in about _______.
Answer:
1 month
Explanation:
Investigators measure the size of fog droplets using the diffraction of light. A camera records the diffraction pattern on a screen as the droplets pass in front of a laser, and a measurement of the size of the central maximum gives the droplet size. In one test, a 690 nm laser creates a pattern on a screen 30 cm from the droplets. If the central maximum of the pattern is 0.24 cm in diameter, how large is the droplet?
Answer:
the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
Explanation:
Given the data in the question;
Diameter of bright central maxima;
⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )
where D is the distance from the the droplet to the screen ( 30 cm )
d is the diameter of the droplet
λ is the wavelength of light ( 690 nm = 690 × 10⁻⁷ cm )
since the central maximum of the pattern is 0.24 cm in diameter,
we substitute
0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )
solve for d
d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm
d = 0.0050508 cm² / 0.24 cm
d = 0.021045 cm or 2.1 × 10⁻² cm
Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency. _____ joules 4.0 x 10 -19 4.0 x 10 -49 9.9 x 10 -32 1.1 x 10 -48
Answer:
pretty sure its 6.2 x 10^-13
Explanation:
I looked it up I'm not a bigbrain but want to help
The primary coil in a transformer has 250 turns; the secondary coil has 500. Which is correct?
a. This is a step-down transformer.
b. The voltage in the secondary coil will be higher than in the primary.
c. The power in the secondary coil is greater.
d. The power in the primary coil is greater.
Explanation:
option b is the correct one
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
8. If a moving object triples its speed, how much kinetic energy will it have? A. six times as much as before B. three times as much as before C. one third as much as before D. nine times as much as before
D
Explanation:
KE: 0.5mv²
when v is tripled v² is 9 times its original value
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?
Answer:
The answer is "[tex]60.74^{\circ}[/tex]".
Explanation:
Cavity and benzene should be extended in equal quantities.
[tex]\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\[/tex]
[tex]\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\[/tex]
[tex]\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\[/tex]
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant pressure until 60 percent of it, by mass, condenses. Determine (a) the work (W) done during the process. (b) What-if Scenario: What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ
why kg is a fundamental unit?
This above answer helps a lot.
Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
1. Find the period of pendulum A for small oscillations.
2. Find the period of pendulum B for small oscillations.
Answer:
1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].
Explanation:
1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:
[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)
Where:
[tex]l[/tex] - Length of the massless bar.
[tex]g[/tex] - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)
Where:
[tex]m[/tex] - Total mass of the pendulum.
[tex]g[/tex] - Gravitational acceleration.
[tex]l[/tex] - Length of the uniform bar.
[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]
[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)
By applying (3) in (2) we get the following expression:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]
[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]
[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]
1. The period of pendulum A for small oscillations is
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
2. The period of pendulum B for small oscillations.
[tex]T_B=1.137.T_A[/tex]
What is simple harmonic motion?Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where:
l - Length of the massless bar.
g - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2
Where:
m - Total mass of the pendulum.
g - Gravitational acceleration.
l - Length of the uniform bar.
Io- Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]
[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3
By applying (3) in (2) we get the following expression:
[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]
[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]
[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]
[tex]TB=1.137.T_A[/tex]
Thus to know more about Simple harmomnic motion follow
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b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velocity of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.
Answer:
The correct answer is "6.96 rad/s".
Explanation:
The given values are:
Length,
L = 0.6 m
Mass,
m₁ = 0.5 kg
m₂ = 0.2 kg
Initial velocity,
V = 8 m/s
Now,
The final angular velocity will be:
⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]
By substituting the values, we get
⇒ [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]
⇒ [tex]=\frac{9.6}{1.38}[/tex]
⇒ [tex]=6.96 \ rad/s[/tex]
1.Lõi thép máy biến áp được ghép từcác lá thép là để:
(a) Giảm tổn hao công suất do dòng điện xoáy
(b) Giảm tổn hao công suất do từ trễ
(c) Giảm tổn hao công suất do dòng điện chạy qua dây quấn
(d) Giảm tất cảcác loại tổn hao công suất.
Answer:
Option (c)
Explanation:
1.The transformer core is assembled from steel sheets to:
(a) Reduced power loss due to eddy current
(b) Reduced power loss due to hysteresis
(c) Reduced power loss due to current flowing through the winding
(d) Reduce all types of power loss.
A transformer is a device which converts the low voltage into high and vice versa.
There are two types of a transformer.
Step up: It is used to convert low voltage into high.
Step down It is used to convert high voltage into high.
It depends on the number of turns in primary and the secondary coil.
The core of the transformer is laminated and it is in the form of sheets.
By using such type of core, the power loss due to the windings is reduced.
option (c) .
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1180 kHz
Answer:
254 m
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of radio wave, λ = wave length, f = frequency
make λ the subject of the equation
λ = v/f............ Equation 2
From the question,
Given: f = 1180 kHz = 1180000 Hz
Constant: v = 3×10⁸ m/s
Substitite into equation 2
λ = 3×10⁸/1180000
λ = 2.54×10²
λ = 254 m
1. Which one of the following is not an organic compound? Why? CH4 C2H6O CaO
2. Fill in the chart below to identify and describe the functional groups associated with organic chemistry. Name General Structure Properties/Uses Alcohol Aldehyde Ketone Fatty acid Ether
3. Explain why carbon is called “the backbone” molecule of organic chemistry and why organic molecules couldn't easily be based on H or O instead.
Answer:
1. CaO is not an organic compound because it doesn’t contain a carbon molecule.
2.
Name General Structure Properties/Uses
Alcohol R-OH (contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to an oxygen and single-bonded to a hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters
Ether R-O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
3. Carbon is able to make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
Explanation:
pf
CaO is not an organic compound because it doesn’t contain a carbon molecule.
Name General Structure Properties/Uses(which contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-1R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters11
Ether -O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
Carbon can make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
Learn more about organic molecules.
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A television tube can accelerate electrons to 2.00 · 104 ev. Calculate the wavelength of emitted X-rays with the highest energy.
λ = _____ m
9.9 x 10 -30
6.2 x 10 -11
1.6 x 10 10
7.1 x 10 -57
Answer:
6.2 × 10^-11 m
Explanation:
1 eV = 1.602 × 10-19 joule
2.00 × 104 ev. = 2.00 × 10^4 eV × 1.602 × 10^-19 joule/1eV
= 3.2 × 10^-15 J
From;
E= hc/λ
λ = hc/E
λ = 6.6 × 10^-34 × 3 × 10^8/3.2 × 10^-15
λ = 6.2 × 10^-11 m
now suppose that we have attached not just two springs in series, but N springs. Write an equation that expresses the effective spring constant of the combination using the spring constant of the original spring k and the number of springs N
Answer:
[tex]k_{eq} = \frac{k}{N}[/tex]
Explanation:
For this exercise let's use hooke's law
F = - k x
where x is the displacement from the equilibrium position.
x = [tex]- \frac{F}{k}[/tex]
if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs
x_ {total} = ∑ x_i
we substitute
x_ {total} = ∑ -F / ki
F / k_ {eq} = -F [tex]\sum \frac{1}{k_i}[/tex]
[tex]\frac{1}{k_{eq}} = \frac{1}{k_i}[/tex] 1 / k_ {eq} = ∑ 1 / k_i
if all the springs are the same
k_i = k
[tex]\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\[/tex]
[tex]\frac{1}{k_{eq} } = \frac{N}{k}[/tex]
[tex]k_{eq} = \frac{k}{N}[/tex]
