Explanation:
because when diffusion occurs it is as a result of the particles moving around freely in open space
The fact that matter is composed of particles and that these particles are incredibly tiny and have space between them supports the theory that matter is composed of particles.
What is diffusion?The mixing of materials is caused by the free movement of their particles is known as diffusion.
Even with a strong microscope, the substance is composed of microscopic particles that cannot be seen. The matter is anything that is both mass- and space-having. Small particles make up matter.
Because the particle's unrestricted movement in an open environment causes diffusion to happen.
Hence, diffusion supports the idea that matter is made up of particles.
To learn more about diffusion refer to;
https://brainly.com/question/13513898
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A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When released the block m2 falls at 5.88m/s^2. If frictional forces are considered so small they are negligible, the block M1 must have a mass of _______ kg.
Answer:
From the figure,
The free-body diagrams for and are shown in the figures below. The only forces on the blocks are the upward tension and the downward gravitational forces and . Applying Newton’s second law, we obtain:
which can be solved to yield
Substituting the result back, we have
(a) With and , the acceleration becomes
(b) Similarly, the tension in the cord is
Calculate the de Broglie wavelength of a 4.5-MeV α particle. Explain if a relativistic or non-relativistic approach is required
Answer:
4.4x10^-32m
Explanation:
Using
Wavelength = hc/E
= 6.63*10^-34 * 3x10^8/ 4.5 x10^ 6ev
= 4.4*10^-32m
Yes a relativistic approach is used because the particle moves at the speed of light
when we increase the exponent 10 to 10 we must
Answer:
If you say it like this (x^10)^10, then it would be x^100.
A dog runs 300m North and sees the dog catcher and runs 120 m south .Whats the dogs displacement.The same jogger runs 3 miles around the track, starting and finishing in the same time after completing exactly 12 laps. What is the average velocity now
Answer:
1. 180 m
2. 1.5 miles/hour
3. 0 m/s
Explanation:
The complete question is
A dog runs 300 m North and sees the dog catcher and runs 120 m south Whats the dogs displacement.
A jogger runs north for 3.0 miles. If this took 2.0 hours, what is the jogger’s average velocity in miles per hour?
The same jogger runs 3.0 miles on the track, starting and finishing his run after completing exactly 12 laps around the track. What is his average velocity?
1. Displacement is how far the dog moves from the original starting point.
If the dog runs 300 m north, and then 120 m south, then the displacement from the starting point will be
displacement = 300 - 120 = 180 m
2. Displacement of the jogger = 3 miles
Time taken by the jogger = 2 hours
velocity of the jogger = ?
Velocity is = displacement/time
velocity = 3/2 = 1.5 miles/hour
3. velocity = displacement/time
if the jogger completes exactly 12 laps around the track, then the jogger will return to the starting point. This means that the jogger's displacement from the starting point is 0 miles
Jogger's velocity will therefore be = 0 m/s
Convert 300 K into the celsius scale
Convert 220 °C into kelvin scale.
Convert 30 °C into fahrenheit scale
Convert 260 °F into °C.
Answer:
26.85 Celsius.
493.15 Kelvin
86F
26.85
493.15
Tk=Tc+273.15
The atmospheric pressure at normal conditions is equal to 10.33 m under water. Then 1.0mm mercury equals to
mm water.
Answer:13.6
Explanation:
I NEED THIS IN AN HOUR! PLEASE PLEASE PLEASE HELP Tim and Jim measured the length of the same piece of board. Using a certain ruler, Tim obtained a length of 0.96 centimeter. Using a different ruler, Jim obtained a length of 0.9662 centimeter. Which measurement is better? Explain the reason for your answer.
Answer:
Jims is better
Explanation:
His number is more complex which means it is more accurate
What is 1.2 kg converted into mg.
I need to know for step by step please?
=> 1200 000 mg
ExPlaNatIoN :We know that,
1 kg = 1000 g
1 g = 1000 gm
then,
1.2kg = 1.2 × 1000 g = 1200 g
1200 g = 1200 × 1000 mg
=> 1200 000 mgA runner changing speed from 3 meters per second to 5 meters per second is an example of
Answer: acceleration
Explanation:
he is accelarting from 3 meters per second to 5 meters per second
Answer:
acceleration
Explanation:
What is the acceleration of softball if it has a mass of 5 kg and hits the catchers glove with a force of 35 N
Answer:
50m/s2^
Explanation:
We know , by newton's law of motion .
F=ma
a=f/m
a=25/0.5
a=50m/s2^
Therefore , the acceleration of a softball is .
Answer:
7 m/s²
Explanation: Force = Mass x acceleration
Also note that 1 N = 1 kg.m/s²
a=F/ m
a= 35 N/ 5 kg = 35 kg.m/ s² ÷ kg Aside:( so you cross out the kg)
a= 7 m/s²
Which of the following is true of the structures labeled A in the cell membrane?
Choose 1 answer:
(Choice A) A They are membrane proteins.
(Choice B) B They are entirely hydrophobic.
(Choice C) C They are phospholipids.
(Choice D) D They are carbohydrates. Report a problem
Answer:
they are membrane proteins
Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird begins flying straight from the runner to the finish line at vb= 13.6 km/hr (2 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner. What cumulative distance does the bird travel? Even though the bird is a dodo, assume that it occupies only one point in space (a “zero” length bird), travels in a straight line, and that it can turn without loss of speed. Answer in units of km. Q2: After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the beginning (including the distance traveled to the first encounter)? Answer in units of km.
Answer:
Q1: 3.2km
Q2: 4.8K
Explanation:
Q1:
So db is the distance of bird, and dr is the distance of runner
db = 2vr and the distance of bird is going to be 2 times greater than the runner.
formulas: db = 2vr & db = 2dr
db = 2drL + (L - x) = 2x2L - x = 2x2L = 3xx = [tex]\frac{2}{3}[/tex]LInsert it in x = [tex]\frac{2}{3}[/tex]L
[tex]\frac{2}{3}[/tex](2.4km) = 1.6km
Now we use formula db = 2dr
db = 2L - xdb = 2(2.4km) - 1.6kmdb = 3.2kmQ2:
Formulas: Vr = L /Δt & Vb = db/Δt
Vr = L/ Δt ⇒ Δt = [tex]\frac{L}{Vr}[/tex][tex]\frac{2.4km}{6.8km/hr}[/tex][tex]\frac{6}{17}hr[/tex](Km cancel each other)
Vb = db/Δt ⇒ db = VbΔt13.6km/hr[tex](\frac{6}{17}hr )[/tex]4.8km(hr cancel each other)
Hope it helps you :)
A body is projected vertically upwards with a velocity of 10m/s from a vertically upward position at height h another body is dropped down at the same instant
Answer:
0.25 second
Explanation:
Given :
Velocity, v = 10 m/s
The height,
[tex]$ h = \frac{v^2}{2g} $[/tex]
[tex]$ h = \frac{10^2}{2\times 9.8} $[/tex]
= 5 m
Therefore the time at which both the bodies meet is
[tex]$ t = \frac{h}{v+v} $[/tex]
[tex]$ t = \frac{5}{10+10} $[/tex]
[tex]$ t = \frac{5}{20} = 0.25$[/tex]
So, time taken is 0.25 seconds
1. An arrow is shot with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. What is the maximum height it will reach?
a. 23 m
b. 46 m
c. 69 m
d. 92 m
2. A projectile is launched horizontally from a cliff with an initial speed of 40 m/s. The cliff is 125 m high and the projectile travels a horizontal distance of 200 m from the bottom of the cliff. What is the speed of the projectile right before it hits the ground?
a. 48 m/s
b. 52 m/s
c. 56 m/s
d. 60 m/s
e. 64 m/s
Answer:
46m
Explanation:
Given the following :
Initial Velocity (u) = 60
Angle(θ) = 30°
Maximum height it will reach?
The maximum height if a projection is calculated using the formula :
H = u²sin²θ / 2g
Where H = maximum height, g = acceleration due to gravity = 9.8m/s²
H = (60²sin × 0. 5²) / 2(9.8)
H = 900 / 19.6
H = 45.918m
H = 46m
A carpenter apprentice was killed when he was struck in the head by a nail that was fired from a powder actuated tool. The tool operator, while attempting to anchor a plywood form in preparation for pouring a concrete wall, fired the gun, causing the nail to pass through the hollow wall. The nail traveled some twenty-seven feet before striking the victim. The tool operator had never received training in the proper use of the tool, and none of the employees in the area were wearing personal protective equipment. What should be recommended to prevent this accident from happening again?
Answer:
Explanation:
The recommendations that need to be urgently addressed by the workshop and it's employees are
1) Training of all employees (particularly the tool operators) on the safe and proper use of all tools used in the course of the carpentry work.
2) Training of all employees in health, safety and environment (HSE). This training will help them appreciate the importance of always wearing personal protective equipment (PPE) when at the workshop or during the course of a work (because if the apprentice was putting on an engineer's helmet, he could have been safe).
3) Employees who fail to adhere strictly to the use of PPE should be sanctioned or even dismissed from the workplace
please anyone help me to answer this questions.plzz... it's very urgent..
Answer:
1) 3 applications of pressure in daily life are :-
● The area of sharp edge of knife, scissor or handsaws are much less then blunt edge. So, for same total force pressure is more for sharp edges than the blunt one. Hence sharp knife, scissors etc, cuts easily than a blunt one.
●Broad handles in bags and suitcases are provided for the comfort. Broad handles have large area. So, the pressure exerted on hands and shoulders would be small while carrying the bags and the suitcases.
●Trucks carrying heavy loads have more than four tyres. More tyres in case of trucks increase the area of contact with the road. This results in reduced pressure on the tyres.
2) Area of the surface which is on ground = 1.5×1
= 1.5m^2
Mass of the block = 300kg
Force applied by the block = Mass × g = 300×10
= 3000N (where g = acceleration due to gravity )
Pressure = Force applied / Area of the surface
= 3000N / 1.5m^2
= 2000 Pa
3)
a) The above experiment signifies that more the area of the surface of an object , less the pressure an object applies.
b) B exerts the minimum pressure because the area of its surface to ground is greater than others & as it has more area of surface , it exerts less pressure. ( area is inversely proportional to pressure )
c) D exerts the maximum pressure because the area of its surface to ground is lesser than others & as it has less area of surface , it exerts more pressure. ( area is inversely proportional to pressure )
d) It depend upon the way an object is kept on ground. If an object is kept in such a way dat the area of the surface to the ground is more , then pressure will be least exerted .If an object is kept in such a way dat the area of the surface to the ground is less, then pressure will be exerted more .
e) Do it yourself . only i will suggest that make the tip of the cone ( which is to the ground ) more narrower.
Explanation:
HELPPPP PLSSSS A group of students is investigating whether air resistance depends on the size of an object. The students throw two paper pieces each with the same force. One piece of paper is flat and the other is scrunched into a ball. The distance traveled by each paper piece is shown. Experimental Observations Student 1 Student 2 Student 3 Student 4 Distance traveled by flat paper 0.2 m 1.2 m 1.5 m 0.8 m Distance traveled by scrunched paper 1.5 m 2.4 m 3 m 2 m What was the test variable (independent variable) in this experiment? A.Distance traveled by flat paper B.The shape of the paper C.Distance traveled by scrunched paper D.The number of students
Answer:
Explanation: societal law is A statement about how things act in the natural world. the wire and electro magnet need to touch in question 2. a decrease in spee decreases kinetic energy. the test variable was the distance traveled by the scrunched paper.
Answer:
B
Explanation:
The shape is the only thing that changes.
A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 cubic meters and the tension in the cord is 900 N. a) Calculate the buoyant force exerted by the water on the sphere. b) what is the mass of the sphere? c) the cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?
Answer: a) B = 6811N
b) m = 603.2kg
c) 86.8%
Explanation: Buoyant force is a force a fluid exerts on a submerged object.
It can be calculated as:
[tex]B=\rho_{fluid}.V_{obj}.g[/tex]
where:
[tex]\rho_{fluid}[/tex] is density of the fluid the object is in;
[tex]V_{obj}[/tex] is volume of the object;
g is acceleration due to gravity, is constant and equals 9.8m/s²
a) For the hollow plastic sphere, density of water is 1000kg/m³:
[tex]B=10^{3}.0.695.9.8[/tex]
B = 6811N
b) Anchored to the bottom, the forces acting on the sphere are Buoyant, Tension and Force due to gravity:
B = T + [tex]F_{g}[/tex]
B = T + mg
mg = B - T
[tex]m=\frac{B - T}{g}[/tex]
Calculating:
[tex]m=\frac{6811 - 900}{9.8}[/tex]
m = 603.2kg
c) When the shpere comes to rest on the surface of the water, there are only buoyant and gravity acting on it:
B = m.g
[tex]\rho_{w}.V_{sub}.g=m.g[/tex]
[tex]V_{sub}=\frac{m}{\rho_{w}}[/tex]
[tex]V_{sub}=\frac{603.2}{1000}[/tex]
[tex]V_{sub}[/tex] = 0.6032m³
Fraction of the submerged volume is:
[tex]\frac{V_{sub}}{V_{obj}}[/tex] = [tex]\frac{0.6932}{0.695}[/tex] = 0.868
A car slows down uniformly from a speed of 22 m/s to rest in 4.0 seconds. How far did it travel in that time
Answer:
[tex] \boxed{\sf Distance \ travelled = 44 \ m} [/tex]
Given:
Initial speed (u) = 22 m/s
Final speed (v) = 0 m/s (Rest)
Time taken (t) = 4 seconds
To Find:
Distance travelled by car (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:
[tex] \boxed{ \bold{s = (\frac{v + u}{2} )t}}[/tex]
By substituting value of v, u & t in the equation we get:
[tex] \sf \implies s = ( \frac{0 + 22}{2} ) \times 4 \\ \\ \sf \implies s = \frac{22}{2} \times 4 \\ \\ \sf \implies s = 11 \times 4 \\ \\ \sf \implies s = 44 \: m[/tex]
[tex] \therefore[/tex]
Distance travelled by car (s) = 44 m
Energy flow in an ecosystem is best represented by
a. the predator/prey relationship
b. food chains
C. food webs
d. autotrophs
Answer:
It's actually C. food webs
Explanation:
took the test
Answer: c
Explanation:
edge 2020
A scientific LAW has been proven and a THEORY has not. True Or False
Answer:
true
Explanation:
i did the test its right
Answer:
true
mark mine as brainiest, i could use the points
the unit for the measure of distance is
Answer: Well their are many Miles, inches, kilometres, meters, cementers, and so on.
hope it helps you bye
Centimetre, Meter, & Kilometre
Determine the acceleration (in m/s) of an object which
A. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds
B. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m
C. accelerates from 21.5 m/s to 8.9 m/s over a distance of 109 m
Please help!
Explanation:
A=23.5-12.1/7.81
1.46
b)(18.9^2-(33.4)^2)/2(109)
-3.5
c)8.9^2-(21.5)^2/2(109)
-1.76
What is the strength of electric field EpEp 0.60 mmmm from a proton? Express your answer to two significant figures and include the appropriate units.
Answer:
3.99*10^-3N/C
Explanation:
Using
Ep= kq/r²
Where r = 0.6mm = 0.6*10^-3m
K= 8.9*10^9 and q= 1.6*10^-19
So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²
= 3.99*10^-3N/C
3. Which statement best describes the research projects that are funded by private
foundations?
a. Research projects that have little chance of success
b. Research projects for a specific cause
c. Research projects concerning the environment
d. Research projects that benefit consumers
Answer:
b. Research projects for a specific cause
Explanation:
Research funded by private foundations are usually for a specific cause. These type of research are usually properly scrutinized, and must have a high probability of success. Most of these projects may not be to benefit the consumers, but are usually done with a special purpose in mind.
Two point charges are separated by a certain distance. How does the strength of the electric field produced by the first charge, at the position of the second charge, change if the second charge is doubled?
Answer:
The field will remain the same
Explanation:
This is because electric field given as
E1= kq1/r²
And that of second charge
E² = kq2/r²
Is not affected by the size of the second charge q2
How does online gaming affect the health of the student
Answer:
because we more see th online gaming and not eat food and playing that game all the time so it effect the health
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Why is there no ideal time of day when all students should study?
Answer:because each individual works best at a different time of day. D. because scheduling prevents everyone from studying at a set time. emilleexo818 is waiting for your help.
Explanation:
Answer:
The correct and overall question would be: Because each individual works best at a different time of day
Explanation:
(I'm smart and I took the quiz, I hope this helped guys! :)
A ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the distance traveled by the ball? What is the displacement of the ball?
Answer:
8.5m and 3.5mExplanation:
In this problem, we are expected to solve/find the total distance traveled and also the displacement of the ball.
given that the initial height is 1.5m and was thrown to a height of 5m
the total distance covered when the ball falls to the ground
Note (see attached a rough sketch for the pictorial explanation)
1. The distance for A to B= 3.5 m
2. The distance from B to A= 3.5 m
3. The distance fro A to C= 1.5 m
the total distance is 3.5+3.5+1.5= 8.5m
the displacement is simply the difference between the final position and the initial position
displacent= 5-1.5= 3.5m
In a walking investigation, Josephine
walked a total distance of 40 feet. At the
halfway point, she had walked for 25
seconds. She stopped for 5 seconds to tie
her shoe and then continued for 25 more
seconds. Sketch a graph that shows
Josephine's distance from the starting point
over time
Explanation:
She walked 20 feet in the first 25 seconds. Then she stayed at the 20 foot position for 5 seconds. Then she walked another 20 feet in 25 seconds. See attached graph.