A material in which electrons are not able to move easily is known as an insulator, on the other hand, the material in which electrons are able to move easily is known as the conductor, therefore the correct answer is option B.
What are metalloids?The elements of the periodic tables that behave as metals, as well as the nonmetal in some chemical or physical aspects, are known as metalloids. Some examples of metalloids are Boron, silicon, germanium, arsenic, antimony, tellurium, etc.
As given in the problem we have to find out the term for material for which electrons are not able to move easily,
The term "insulator" refers to a substance in which electrons cannot move around freely, whereas the term "conductor" refers to a substance in which electrons may flow around freely. Therefore, option B is the right response.
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what evidence do you that suggest water waves are transverse wave
Answer:
If you throw a pebble into a pond, ripples
spread out from where it went in. These
ripples are waves travelling through the
water. The waves move with a transverse
motion.
Explanation:
A student made the electromagnet shown and used it to pick up one sewing pin. She
then doubled the number of coils around the nail and used it to pick up two sewing pins.
How could she change the electromagnet again in order to pick up three sewing pins?
A. She could replace the 6V battery with a 3V battery.
B. She could replace the nail in the wire coil with a large steel bolt.
C. She could replace the 6V battery with a 12 V battery.
D. She could replace the nail in the wire coil with a thick roll of paper.
Correct answer is C!
Answer: i think its c but i am not sure tho but i think its she could replace the 6 v battery with a 12 v battery
Explanation: cause a 12 v is stronger than a 6 v battery
A student made the electromagnet shown and used it to pick up one sewing pin. She then doubled the number of coils around the nail and used it to pick up two sewing pins. She could replace the 6V battery with a 12 V battery.
What is electromagnet?Permanent magnets are not the same as electromagnets. Coils of wire with electricity running through them make up electromagnets. An electromagnet's wire coils act like magnets when an electric current flows through them because moving charges produce magnetic fields.
The coils stop functioning as a magnet when the electricity is turned off. A lot of things use electromagnets.
When magnetic forces are only required for brief periods of time, electromagnets are frequently used in electrical equipment.
Therefore, A student made the electromagnet shown and used it to pick up one sewing pin. She then doubled the number of coils around the nail and used it to pick up two sewing pins. She could replace the 6V battery with a 12 V battery.
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How do you determine: how many significant figures should you to round to when doing addition and subtraction?
A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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When a spring is compressed, the energy changes from kinetic to potential. Which best describes what is causing this change?
•work
•power
•gravitational energy
•chemical energy
Answer:
work
Explanation:
Answer:
A
Explanation:
a ball of mass 0.5 kg is at point with initial speed 4 m/s at height 10. what is the total energy
Answer:
The total energy is 53 Joule
Explanation:
Mechanical Energy
The mechanical energy of an object of mass m, speed v, and at a height h is:
[tex]\displaystyle E = m.g.h+\frac{mv^2}{2}[/tex]
The ball has a mass of m=0.5 Kg, a speed v = 4 m/s, and at a height of h=10 m. Thus the total energy is:
[tex]\displaystyle E = 0.5\cdot 9.8\cdot 10+\frac{0.5\cdot 4^2}{2}[/tex]
E = 49 J + 4 J = 53 J
The total energy is 53 Joule
HELP ASAP !!! !!!!!!!
Answer:
they are cooler than the rest if the sun
Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plates are spaced 1.4 mm apart and the voltage across them is changing at a rate of 110 V/s .
Solution:
The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :
[tex]$E=\frac{V}{D}$[/tex]
Differentiating on both the sides with respect to time, we get
[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]
Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :
[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]
[tex]$=\frac{1}{1.4 \times 10^{-3}} \times 110$[/tex]
[tex]$=7.85 \times 10^4$[/tex] V/m-s
A 0.420 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 20.5 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.80 m/s and sticks to it. Determine the amount by which the Play-Doh-block system compresses the spring.
Answer:
leon
Explanation:
leom ofjfjbfbfdnns
An oil refinery uses a Venturi tube to measure the flow rate of gasoline. The density of the gasoline is
ρ = 7.40 ✕ 102 kg/m3,
the inlet and outlet tubes, respectively, have a radius of 3.74 cm and 1.87 cm, and the difference in input and output pressure is
P1 − P2 = 1.20 kPa.
a) find the speed of the gasoline as it leaves the hose
b) find the fluid flow rate in cubic meters per second
Answer:
(a) V₂ = 1.86 m/s
(b) Q = 5.1 x 10⁻⁴ m³/s
Explanation:
(a)
The formula derived for Venturi tube is as follows:
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
where,
P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa
ρ = Density of Gasoline = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
Therefore,
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------------------- equation (1)
Now, we will use continuity equation:
A₁V₁ = A₂V₂
where,
A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
Therefore,
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
using this value in equation (1):
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
(b)
For fluid flow rate we use the following equation:
Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)
Q = 5.1 x 10⁻⁴ m³/s
The formula for finding variables in a Venturi tube is shown below:
The speed of the gasolineP₁ - P₂ = (ρ/2)(V₂² - V₁²)
where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity
P₁ - P₂ = 1.2 KPa = 1200 Pa
ρ = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
We then substitute the variables into this equation.
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------ equation (1)
The continuity equation A₁V₁ = A₂V₂ is then used
where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
We then substitute the value into equation 1
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
The fluid flow rate we use the following equation:This can be calculated using the formula
Flow Rate = Q = A₂V₂
= (2.746 x 10⁻⁴ m²)(1.86 m/s)
= 5.1 x 10⁻⁴ m³/s
When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180 million years ago 300 million years ago
Answer:
b
Explanation:
i took the test
Earth’s landmasses were first recognized as the continents we know today
135 million years ago.
Landmass is defines as a large area of land. It can also be referred to as the
continents we have today. There are seven types of earth landmasses and
they are
Asia, Africa, North America, South America, Antarctica, Europe, and Australia.They were first discovered around 135 million years ago by the early
dwellers of the earth.
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Intense physical activity that requires little oxygen but uses short bursts of energy is called Anaerobic exercise? True or false 
A piston has an external pressure of 15.0 atmatm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.450 liters
Answer: Work done is - 501.56 J
Explanation:
Given that;
External pressure P = 15.0 atm
Volume V1 = 0.120 liters
Volume V2 = 0.450 liters
Work done = ?
we know that; Work = -Pdv
where P is pressure and dv is change in volume
so we substitute our values into the equation
Work = -15.0 × ( 0.450 - 0.120)
= -15 × 0.33
= - 4.95 atm/L
we know that;
1 atm.L = 101.325 J
so
- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷
= - 501.56 J
Therefore Work done is - 501.56 J
An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface
Answer:
What is the normal force on the object (draw a diagram if needed)
784 N
How much force is required to get the object to start to move from rest (max static friction)?
470.4 N
When the object starts moving, what is the force of kinetic friction?
274.4 N
If the moving object has a tension force of 300N to the right pulling it, what is the net force on the object in the horizontal direction? 784 N
What is the acceleration (with direction) of the object based on your answer for part d? Remember that a = net force/m.
Explanation:
We have that for the Question "" it can be said that the normal force and How much force is required to get the object to start to move from rest is
F_N=784NF_{max}480N
From the question we are told
An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface
What is the normal force on the object (draw a diagram if needed) How much force is required to get the object to start to move from rest?
a)
Generally the equation for the Normal Force is mathematically given as
[tex]F_N=mg\\\\F_N=80*9.8\\\\F_N=784N\\\\[/tex]
b)
[tex]F_{max}=0.60*800\\\\\F_{max}480N\\\\[/tex]
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Complete Question
An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface
What is the normal force on the object (draw a diagram if needed) How much force is required to get the object to start to move from rest?
A ball was thrown vertically upward on a planet with a velocity of 2.5 m/sec. If it can rise up to 4m,then calculate gravitational strength of the planet.
pleaseeeeeeeeeeee helppppppppppp
Answer:
0.78m/s^2
Explanation:
using the third equation of motion, the answer is found
v^2=u^2 -2gs
where V= 0 at the maximum point
2gs=u^2
g=u^2
2s
g=2.5^2
2(4)
g=0.78125m/s^2
or g=0.78m/s^2
Claudia stubs her toe on the coffee table with a force of 100. N. a) What is the
acceleration of Claudia's 1.80-kg foot? b) What is the acceleration of the table
if it has a mass of 20.0 kg? (Ignore any frictional effects.) c) Why would
Claudia's toe hurt less if the table had less mass?
Answer:
A.) acceleration= 55.6m/s^2
B.) acceleration of table= 5.0m/s^2
C.) More acceleration
Explanation:
A.) 100N/1.8kg= -55.6
B.) 100N÷20kg= 5
C.) Because since the table would have less mass, it would have had to accelerate more
The acceleration of Claudia's foot is 55.56 m/s². The acceleration of the table is 5 m/s². If the table had less mass then its acceleration will be less and Claudia's toe hurt less.
What is acceleration?Acceleration can be described as the rate of change of velocity with respect to time. Acceleration is a vector quantity with both magnitude and direction. Acceleration is also the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to the second law of Newton, the force is equal to the product of mass and acceleration.
F = ma
or, a = F/m
Given, the force with which Claudia stubs her toe on the coffee table,
F = 100N
The mass of Claudia's foot = 1.80 Kg
The acceleration of the Claudia's foot = 100/1.50 = 55.56m/s²
The mass of the table = 20 Kg
The acceleration of the table = 100/20 = 5 m/s²
The acceleration of any object is inversely proportional to its mass. If the table had less mass then its acceleration will be less and it would less hurt Claudia's toe.
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A shell traveling with speed, v0 , exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0 . What is the velocity of the other fragment? Hint: Velocity has both magnitude and direction.
Answer:
yeah
Explanation:
yeah yeah yeah yeah
Fast and safe heart rate for workouts is called muscular strength? True or false
Answer:
False
Explanation:
Answer:
False
Explanation:
Hope this helped, Have a Wonderful Day/Night!!
The diagram shows the four primary steps in the production of work in a four-stroke heat engine.
What is the correct order in which these steps occur?
A. Y, W, X, z
B. X, Z, W. Y
C. W, X, Y, Z
D. Z, X, Y, W
Answer:
D- Z, X, Y, W
Explanation:
Correct on edge
How much thermal energy (in kcal) is required to change a 43 g ice cube from a solid at - 16.5 oC to steam at 11.5 oC above boiling
Answer:
The total thermal energy required is 8.93 Kcal
Explanation:
Given;
mass of the ice cube, m = 43 g
specific heat capacity of water, Cp = 4.18 J/gc
specific latent heat of fusion of ice, Cf = 334 J/g
First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C
Q₁ = mCp[0 - (-16.5)]
Q₁ = 43 x 4.18(16.5)
Q₁ = 2965.71 J
Second step, determine the latent heat of fusion of ice at 0°C
Q₂ = mCf
Q₂ = 43 x 334
Q₂ = 14362 J
Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.
The final temperature of the water = 11.5 °C + 100° C = 111.5 °C
Q₃ = mCp Δθ
Q₃ = 43 x 4.18 (111.5 - 0)
Q₃ = 20041 J
Total thermal energy required = Q₁ + Q₂ + Q₃
Total thermal energy required = 2965.71 J + 14362 J + 20041 J
Total thermal energy required = 37,368.71 J
Total thermal energy required = 8.93 Kcal
Help me guys please with this question
Answer:
[tex]\mid \vec C\mid=31.9[/tex]
Explanation:
Consider the vectors:
[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]
[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]
Calculate the magnitude of
[tex]\vec C=-2\vec B-\vec A[/tex]
Substitute the values of the vectors:
[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]
Operate and remove parentheses:
[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]
Operating both components separately:
[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]
Now find the magnitude of C:
[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]
[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]
[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]
A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is the frictional force?
If work is shown that would be great
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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.
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In this question ,
surface vertical force = Weight of the object
Thus ;
svf = ( mass ) × ( gravity acceleration )
_________________________________
If gravity acceleration is 10 :
svf = 10 × 10 = 100 N
So ;
frictional force = 100 × 0.20
frictional force = 20 N
##############################
If gravity acceleration is 9.8 :
svf = 10 × 9.8 = 98 N
So ;
frictional force = 98 × 0.20
frictional force = 19.6 N
_________________________________
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Which is the function of a nucleus within a eukaryotic cell?
O A) Transport materials within the cell.
O B) Control what materials move into and out of the cell.
OC) Provide the cell with energy to perform its other functions.
OD) Store genetic information that provides instructions for the cell.
What happens to the mechanical advantage of a machine if the output force is less than the input force? What must happen to output distance? Give an example of a machine that does this?
Romeo is a 68 kg astronaut. Juliet is a beautiful cosmonaut who is standing on the balcony of a 4.58 x 10^5kg space station that is at rest and out of gas. Romeo is floating 25 meters away from the space station’s center of mass, how strong is the force between Romeo and Juliet?
Answer:
F = 3.32 x 10⁻⁶ N
Explanation:
The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:
F = Gm₁m₂/r²
where,
F = Force between Romeo and Juliet = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of Romeo = 68 kg
m₂ = mass of space station = 4.58 x 10⁵ kg
r = distance = 25 m
Therefore,
F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²
F = 3.32 x 10⁻⁶ N
If the velocity of Homer the astronaut (mass =200 kg) is 5 m/s and he runs into and grabs his stationary pal Larry (mass = 150 kg), what is the new velocity of the astronauts after the collision?
We are given:
Homer the Astronaut:
Mass of Homer the astronaut(m1) = 200 kg
initial velocity of Homer the astronaut(u1) = 5 m/s
Larry the Pal:
Mass of Larry the Pal (m2) = 150 kg
initial velocity of Larry the Pal (u2) = 0 m/s
Since they will move together after the collision, they will have the same velocity:
v1 = v2 = V
Solving for the Final velocity:
from the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
since v1 = v2 = V:
m1u1 + m2u2 = V(m1 + m2)
replacing the variables with the given values
200 * 5 + 150 * 0 = V(200 + 150)
1000 = 350V
V = 1000 / 350
V = 2.86 m/s
A child cart is given an initial velocity of 8 m/s and undergoes a constant acceleration of 4 m/s2. Calculate the distance moved.
please hellppppp 30points
Answer: This is Yazan From 10-C
Explanation
At what speed does a 1500 kg compact car have the same kinetic energy as a 18000 kgkg truck going 21 km/hr km/hr
Answer:
speed = 72.75km/hr
Explanation:
Let us first determine the Kinetic energy (KE) of the 18000kg truck
[tex]KE = \frac{1}{2} mv^2\\where:\\m = mass = 18000\\v = velocity = 21km/hr\\\therefore KE = \frac{1}{2} \times 18000 \times (21)^2\\= 9000 \times 441\\KE = 3,969,000\ Joules\\= 3,969\ KJ[/tex]
Next we will substitute this value of kinetic energy into the KE equation for the 1500kg compact car
[tex]KE = \frac{1}{2} mv^2\\3969000 = \frac{1}{2} \times 1500 \times v^2\\3969000 = 750 \times v^2\\v^2 = \frac{3969000}{750} = 5292\\v = \sqrt{5292} \\v = 72.75km/hr[/tex]
The speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
The kinetic energy of an object is known to be the energy at work and in motion. It is usually expressed as;
[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]
To determine the speed at which the car will have the same kinetic energy as the truck, we can say that:
[tex]\mathbf{\implies \dfrac{1}{2}m_c v_c^2= \dfrac{1}{2}m_t v_t^2}[/tex]
[tex]\mathbf{m_c}[/tex] = mass of the car[tex]\mathbf{v_c}[/tex] = speed of the car[tex]\mathbf{m_t}[/tex] = mass of the truck[tex]\mathbf{v_t}[/tex] = speed of the truck∴
[tex]\mathbf{\implies \dfrac{1}{2}\times 1500\times v_c^2= \dfrac{1}{2}\times 18000 \times 21^2}[/tex]
[tex]\mathbf{v_c^2= \dfrac{3969000}{ 750}}[/tex]
[tex]\mathbf{v_c^2=5292}[/tex]
[tex]\mathbf{v_c=\sqrt{5292}}[/tex]
[tex]\mathbf{v_c\simeq 73 km/h}[/tex]
Therefore, we can conclude that the speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
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answer pls urgent pls
Answer:
Brittleness
Explanation:
Lustrous means shiny
sonorous means capable of producing a deep or ringing sound
and iron isn't brittle or weak
Answer:
[tex]\huge\boxed{\sf brittleness}[/tex]
Explanation:
Iron is a metal which have luster (shine) so it is lustrous (shiny).
Iron is sonorous. When it is hit with a hard object, it produces ringing sounds.
Iron is not brittle. It cannot be easily broken. So, Iron does not show brittleness (It cannot be easily broken)
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AnonymousHelper1807How does the abundance of hydrogen and helium support the Big Bang Theory?
It is the proportion predicted to be present in the early universe.
The hydrogen and helium abundance helps us to model the expansion rate of the early universe.
In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.
In big Bang model, the universe is mostly light or protons.
This abundance of hydrogen and helium is consistent with this big bang model. The process of forming this hydrogen and helium is often called big bang nucleosynthesis.The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.
Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.
Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.
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