Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
what is the application of a spherometer in the medical field?
Answer:
To correct the defects of vision by measuring the radius of curvature and thus the power of the lenses.
Explanation:
A spherometer is an instrument used to measure the curvature of objects such as lenses and curved mirrors.
Generally it consists of a fine screw which is moving in a nut carried on the center of a 3 small legged table or frame. The feet forms the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres.
If the screw has two turns of the thread to the milli meter the head is generally divided into 50 equal parts, so that differences of 0.01 millimeter may be measured without using a vernier scale.
The spherometer is used to measure the radius of curvature of the lenses so that the opthalmologist find the focal length of the lens and then give the power to the lens to correct the defects of vision.
L Pretest Unit 7
Question 13 of 20
Andrew is trying to identify an unknown element. The element is shiny, but it
shatters when hammered and cannot be hammered into different shapes.
Where on the periodic table is this element most likely found?
A. On the left side
B. In one of the series below the main body of the table
C. On the right side
D. Along the metalloid stairstep line
SURMIT
Answer:
C
Explanation:
I think it would be there, it sounds like silicone and thats on the right side
What is 11 divided by 73370 I need to know how you found the answer
a car is moving at 56 miles per hour the kinetic energy of that car is how much energy does that same car have when it moves at
Complete Question
A car is moving at 56 miles per hour. The kinetic energy of that car is 5 × 105 J. How much energy does the same car have when it moves at 90 miles per hour? Answer in units of J
Answer:
[tex]K.E_2=640000J[/tex]
Explanation:
From the question we are told that:
Initial Velocity [tex]v_1=56 mil/h=>25.0342m/s[/tex]
[tex]K.E_1=5*10^5J[/tex]
Final Velocity [tex]V_2=90mil/h=>40.2336m/s[/tex]
Generally the equation for Kinetic Energy is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
[tex]K.E_1=0.5mv_1^2[/tex]
[tex]5*10^5J=m(25.0342)^2[/tex]
[tex]m=797.82kg[/tex]
Therefore The Final K.E_2 is
[tex]K.E_2=0.5mv_2^2[/tex]
[tex]K.E_2=0.5*(797.82kg)(40.2336m/s)^2[/tex]
[tex]K.E_2=640000J[/tex]
What is the source of geothermal power
Answer: Geothermal energy is produced by the heat of Earth's molten interior. This energy is harnessed to generate electricity when water is injected deep underground and returns as steam (or hot water, which is later converted to steam) to drive a turbine on an electric power generator.
Explanation: Hope this helps!
What happens in the gray zone between solid and liquid?-,-
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
Answer: 1 cal is 4.186 J, 1 kcal = 4186 J A : 1014 m , B 200 m
Explanation: A) Work done by climber is change in potential energy.
W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.
Solve h = 160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m
B Energy is only 20 % : Then h = 0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.
Actually, muscles also produce heat from most of the energy provided by food.
what is the difference between compulsory and voluntary deductions
A Child stands on the bus Remains Still When The bus is at rest. When the bus moves forward AndeaThe bus is at rest. When the bus moves forward And then slows down, the children the Contnues moving forward at the original speed. This is an example of
Answer:
inertia
Explanation:
What does it mean when work is positive?
O Velocity is greater than kinetic energy
O Kinetic energy is greater than velocity
The environment did work on an object.
O An object did work on the environment.
d. An object did work on the environment.
Explanation:Work is defined in many contexts. Some of these are;
i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.
ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.
In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).
Answer:
its c
Explanation:
What is the Voltage of a circuit that has a resistance of 50 and a current of 2 A?
A 25 V
B. 107
C. 2.5 V
D. 0.4V
Answer:
100 V
Explanation:
V = I * R
V = 2 * 50
V = 100 V
{ I think there is a mistake in the options. }
according to the law of conservation of vhange , what must always be true in a nuclear reaction?
Answer:
The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same
Explanation:
Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.
state any five branches of physics
Answer:
ThermodynamicsQuantum mechanicsNuclear physicsMechanicsAstrophysicsAnswer:
1hjkeiidjjishhhhbsvvvabzbzbshshshsjsjsjsjsjs
Helppp how a deer gets it’s energy
Answer:
option C is the correct one
Explanation:
I hope it helps u
A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.
Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s
A body starts from rest and moves with a uniform acceleration of 2 m/s2 in a
straight line. What is the velocity after 5s, How far has it travelled in this time and When will it be 100m from its starting point?
Answer:
Final velocity = 10 m/s
Time taken to travel 100 meter = 8.16 second (Approx.)
Explanation:
Given:
Acceleration = 2 m/s²
Initial velocity = 0 m/s
Find:
Velocity after 5 seconds
Time taken to travel 100 meter
Computation:
Using first equation of motion
v = u + at
v = 0 + (2)(5)
Final velocity = 10 m/s
Using Second equation of motion
s = ut + (1/2)(a)(t²)
100 = (0)(t) + (1/2)(3)(t²)
100 = (1/2)(2)(t²)
100 = (t²)
t = 10
Time taken to travel 100 meter = 8.16 second (Approx.)
Which pair of magnets has the strongest attraction between them?
Answer in your PE notebook
I have learned that
I have realized that
I will apply
Answer:
physical science is important
hety
civil engineering
How can a wire become magnetic?
add a resistor
point it north
heat it up
run a current through it
Answer:
Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.
Explanation:
Where are the nasal cavities found?
O in the forehead
O behind the chin
O between the eyes
O behind the nostrils
One end of a horizontal spring with the spring constant 1900 N/m is attached to the wall, the other end is attached to a block of mass 1.15 kg. Initially, the spring is compressed 4.5 cm. When released, the spring pushes the block away and is no longer in contact with the block. The block slides along a horizontal frictionless plane.
a/ Compute the maximum speed of the block.
b/ The block goes off the edge of the plane and falls down from the plane to reach the floor with speed of
7 m/s. How high is the plane with respect to the floor?
(a) When the spring is compressed 4.5 cm = 0.045 m, it exerts a restoring force on the block of magnitude
F = (1900 N/m) (0.045 m) = 85.5 N
so that at the moment the block is released, this force accelerates the block with magnitude a such that
85.5 N = (1.15 kg) a ==> a = (85.5 N) / (1.15 kg) ≈ 74.3 m/s²
The block reaches its maximum speed at the spring's equilbrium point, and this speed v is such that
v ² = 2 (74.3 m/s²) (0.045 m) ==> v = √(2 (74.3 m/s²) (0.045 m)) ≈ 2.59 m/s
(b) There is no friction between the block and plane, so the block maintains this speed as it slides over the edge. At that point, it's essentially in free fall, so if y is the height of the plane, then
(7 m/s)² - (2.59 m/s)² = 2gy ==> y = ((7 m/s)² - (2.59 m/s)²) / (2g) ≈ 2.16 m
You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure that it takes a time of 2.91 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the ball weigh on the surface of Planet X?
Answer:
0.3817 N
Explanation:
Remark
One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.
Givens
m = 101 g
vi = 0 That's what at rest means.
t = 2.91 s
d = 16 m
F= ?
Formulas
d = vi*t + 1/2*a * t^2
Force = m * a
Solution
16 = 0 + 1/2 a * 2.91^2
16 = 4.234 a Divide by 4.234
16/4.234 = a
a = 3.779
F = m * a
a = 3.779
m = 101 g = 1 kg / 1000 grams
m = 0.101 kg
F = 0.101 * 3.779
F = 0.3817N
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Answer:
[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]
Explanation:
Given that,
The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]
A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :
[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]
Put all the values,
[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]
So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].
For a different power plant the power input (not output) is 3570 MW. The coal has an energy content of 28,000 kJ per kg. How much coal must be put into the plant each day
Answer:
31475404.8 kg/day
Explanation:
From the information given:
The power plant capacity W = 3570 MW
Energy content of the coal = 28000 kJ/kg
let assume that the thermal efficiency = 35%
Recall that:
1 kw = 3600 kJ/hr provided that the energy conversion is 100% efficient
But assuming the thermal efficiency = 35%.
Then:
Heat input = 3600/0.35 = 10286 kJ/kw.hr
Now, for producing 1 kw.hr, the quantity of the required coal = 10286/28000
= 0.36736 kg
For 3570 MW, the amount of coal that must be input is:
= 0.36736 kg × 3570000
= 1311475.2 kg/hr
= 1311475.2 × 24 kg/day
= 31475404.8 kg/day
Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of 5.00 m/s, four have speeds of 8.00 m/s, three have speeds of 9.00 m/s, and two have speeds of 15.0 m/s. Find:
a. the average speed
b. the rms speed
c. the most probable speed of these particles.
Answer:
a) [tex]V=7.5m/s[/tex]
b) [tex]rms=8.4m/s[/tex]
c) Generally the most probable speed is 8m/s as it the most posses by particles being the average
Explanation:
From the question we are told that:
Sample size N=15
[tex]Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\[/tex]
Generally the equation for Average speed is mathematically given by
[tex]V_{avg}=\frac{\sum(nv)}{N}[/tex]
Therefore
[tex]V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}[/tex]
[tex]V=7.5m/s[/tex]
b)
Generally the equation for RMS speed of the particle is mathematically given by
[tex]rms=\sqrt{\frac{\sum(nv^2)}{N}}[/tex]
[tex]rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}[/tex]
[tex]rms=\sqrt{69.73}[/tex]
[tex]rms=8.4m/s[/tex]
c
Generally the most probable speed is 8m/s as it the most posses by particles being the average
A compact car has a maximum acceleration of 3.0 m/s2 when it carries only the driver and has a total mass of 1300 kg. What is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Answer:
[tex]a_2=3.88m/s^2[/tex]
Explanation:
From the Question we are told that:
Initial Mass [tex]m_1=1300kg[/tex]
Final mass [tex]m_2=1300+400=>1700kg[/tex]
[tex]a_1=3.0m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=m_1a_1[/tex]
[tex]F=1300*5[/tex]
[tex]F=6500N[/tex]
Generally the equation for Final acceleration is mathematically given by
[tex]F'=m_2*a_2[/tex]
[tex]a_2=\frac{6500}{1700}[/tex]
[tex]a_2=3.88m/s^2[/tex]
A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.
Answer:
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
Explanation:
By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:
[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass of the ball, in kilograms.
[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.
[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.
[tex]Imp[/tex] - Impulse, in meters per second.
If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:
[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]
[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
When a ball rolls down a hill, what energy conversion occurs?
A. Kinetic energy to potential energy
ОО
B. Chemical energy to kinetic energy
c. Potential energy to kinetic energy
D. Potential energy to chemical energy
Answer:
C
Explanation:
When a ball rolls down a hill, Potential Energy Conversion takes place to Kinetic Energy.
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
To know more about magnification follow
https://brainly.com/question/1599771
What is binding energy?
A.' The attractive forces between the protons in the nucleus and the
electrons
B. The energy required to force two nuclei to undergo nuclear fusion
C. The amount of energy stored in the strong nuclear forces of the
nucleus
D. The amount of energy required to overcome an activation energy
barrier
Please help me out.
Answer:
the answer is B i hope it helps :)
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
B. The energy required to force two nuclei to undergo nuclear fusion. ✅
They are usually expressed in terms of [tex]\sf\purple{kJ/mole}[/tex] of nuclei or [tex]\sf\pink{MeV's/nucleon}[/tex].[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]