Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
the graph is tied to the reading
here is the reading for the question
students in a physics class observing the race note that each runner maintains the same acceleration from the start ofthe 8.0 second interval until they each reach the finish line where each of the runners slows to a stop. Two students inthe class make the following claims about the runners’ accelerations from the start of the 8.0 second interval until bothrunners are at rest.
Student 1:“Runner A has a positive acceleration until reaching the finish line and then a negative acceleration aftercrossing the finish line in order to “undo” all of the positive velocity she gained. Since Runner B neverhad any positive acceleration, she doesn’t need any negative acceleration to slow to a stop; heracceleration will be zero the whole time.”
Student 2:“But negative values mean away south in this case – how could runner A be turning around? I think bothrunners must have positive accelerations the whole time since they are always moving north.”
For part (c), do not simple repeat the students’ arguments in your answers.
here are the questions:
i.Which aspects of student 1’s reasoning, if any, are correct? Explain your answer.
ii.Which aspects of student 1’s reasoning, if any, are incorrect? Explain your answer.
iii.Which aspects of student 2’s reasoning, if any, are correct? Explain your answer.
iv.Which aspects of student 2’s reasoning, if any, are incorrect? Explain your answer
THANK YOU THANK YOU THANK YOU
It requires a 70.4 N force (parallel to the inclined plane) to pull a 5.86 kg box up a 58.1° inclined plane with a rope at a constant speed. (a) What is the coefficient of kinetic friction between the inclined plane and the box?
(b) If the rope were to break, what acceleration would the box experience as it slid down the ramp?
Answer:
0.667; 4.965
Explanation:
Look at the picture I attached for the force analysis.
a) The coefficient=Friction/Normal Force. Because it's at constant speed, the force of friction + mgsin58.1° (because it's on an inclined plane and has split forces) is equal to the applied force (70.4N). Normal force is not equal to weight force though, because the box is on an inclined plane; it's equal to mgcos58.1°.
b) If it were to break, then the box no longer has an applied force, and the direction of friction has changed to up the inclined plane. F=m/a, so acceleration = mgsin58.1°- Friction/mass
Is mechanical energy the result of both kinetic and potential energy?
Answer:
Yes
Explanation:
Because putting them together would make a type of energy making the answer yes
pls help it's due today
Answer:
Breh seriously. Ugh fine.
1.B
2.D
3.C
4.C
5.D,A and B
6.A,C and D
Write a summary paragraph discussing this experiment and the results. Use the following questions and topics to help
guide the content of your paragraph.
1. According to your data, was your hypothesis for each experiment correct? (Be sure to refer to your data and graphs
when answering this question.)
2. Summarize the conclusions that you can draw from this experiment. Use the questions above to gulde your ideas.
3. Summarize any difficulties or problems you had in performing the experiment that might have affected the results.
Describe how you might change the procedure to avoid these problems.
4. Give at least one more example from real life where the principles demonstrated in this lab are evident.
Answer:
Sample answer: part 1:
What was your hypothesis?
This is what you wanted to prove or study with the experiment. Reading the question I can see that the experiment seems to study the relation between the viscosity of honey and the temperature, so the hypothesis may be something like:
"The viscosity of honey decreases as temperature increases".
According to your data, do you think your hypothesis was correct?
Here you need to see your experimental data, probably, you found that yes, as you increase the temperature the viscosity decreases.
The last thing we need to do here is to summarize difficulties and problems that you had in the experiment, and propose how you could solve them in order to have a better experiment.
Obviously, this depends on how you performed the experiment, but for example, if you saw that the temperature of the honey increased too fast, you could say:
"one problem was that the temperature of the honey increased too fast"
And one way to solve that would be submerging the honey in a water bath so the temperature increases a bit slower.
part 2:
Viscosity depends strongly on temperature, but this relation depends also on the given material.
Generally, we would see that the viscosity decays exponentially as the temperature increases.
So for the first question:
What effect did the temperature have on the viscosity of the honey?
You should have seen that, as the temperature increased, the viscosity decreased.
Now, we need to give some practical examples where knowledge of viscosity is important.
The first one is for cleaning: We know that if we increase the temperature, the viscosity decreases. Thus if we have some object with some viscous thing on it, increasing the temperature will cause it to be easier to clean.
Another example is to store/transport viscous things, for example, honey, it is actually a lot easier to pour honey in bottles if you first heat it a little, so it becomes less viscous.
So there you have two examples where knowing about viscosity (and how it relates to temperature) may be important.
Explanation:
Why do magnets move objects in different ways?
Answer:
Magnets have two poles, called north and south. The like poles are attracted to unlike poles, but like poles repel each other. For example, the north pole of one magnet is attracted to the south pole of another. ... To make objects move with a magnet attach a piece of metal, or another magnet, to it.
19. A beach ball is rolling in a straight line toward you at a speed of 9 m/sec. Its momentum is 3 kg'm/sec.
What is the mass of the beach ball? moss - 0.33
aroto from 10mle to speed of 14 m/s The mass of the bicycle
The mass of the beach ball is 0.33 kg
From the question,
We are to determine the mass of the beach ball.
Using the formula
ρ = mv
Where ρ is the momentum
m is the mass
and v is the velocity
From the given information,
ρ = 3 kg m/sec
v = 9 m/sec
Putting the parameters into the formula, we get
3 = m × 9
∴ m = 3 ÷ 9
m = 0.33 kg
Hence, the mass of the beach ball is 0.33 kg
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What did Ernest Rutherford expect to happen when he aimed a beam of particles at a thin gold foil
Answer:
he expected a fire to happen I think
Answer:
When Ernest Rutherford aimed a beam of particles at a thin gold foil he expected that the particles would be deflected slightly after passing through the foil.
Explanation:
hope it helped
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center, as shown above. The coin doesn’t slip and the time it takes for the coin to make a complete revolution is 1.5 s.
A)The figure below shows the disk and coin as viewed from above. Draw and label vectors on the figure below to show instantaneous acceleration and linear velocity vectors for the pin when it is at the position shown below.
B)Determine the linear speed of the coin
C)The rate of rotation of the disk is gradually increased. The coefficient of stats if friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
D)If the experiment in part c were repeated with a second, identical coin glued to the top of the first coin, how would this affect the answer to part c? Explain.
A) Figure attached below
B) The linear speed of the coin = 0.59 m/s
C) Linear speed as coin begins to slip = 0.83 m/s
D) The tangential speed will remain the same as seen in part C
Given data :
mass of coin = 0.0050 kg
Distance of coin from the center of disk = 0.14 m
Time to make a complete revolution = 1.5 s
A) Diagram showing the vectors on the figure is attached below
B) Determine the Linear speed of the coinLinear speed of coin = 2 * π * ( 0.14 ) / 1.5
= 0.59 m/s
C) Determine the linear speed of the coin when it just begins to slipgiven that: friction between coin and disk = 0.50
Friction becomes maximum when coin begins to slip
Maximum frictional force (Fmax) = uV
where V = mg
∴ Fmax = u*mg ---- ( 1 )centripetal force = [tex]\frac{mv^{2} }{r}[/tex] ---- ( 2 )Equating equations ( 1 ) and ( 2 ) to determine the linear speed ( v )
v² = u*r*g
∴ v = √(u*r*g ) = √( 0.5 * 0.14 * 9.8 )
= 0.83 m/s
D) If the experiment is repeated with a second coin glued to the top of the first coin the tangential speed will remain the same
Hence we can conclude that The linear speed of the coin = 0.59 m/s Linear speed as coin begins to slip = 0.83 m/s , The tangential speed will remain the same as seen in part C
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Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a factor of five and the mass of object 2 is multiplied by a factor of two, what will the new gravitational force be between the objects?
A. F2
B. 1/2F
C. 0.1F
D. 2F/5
E. 5F
J. 10F
Please help will mark brainliest
If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is
F' = G (5m₁) (2m₂) / d²
F' = 10 Gm₁m₂ / d²
but this is really just F' = 10F. So J is the correct choice.
Hi there!
Using Newton's Law of Universal Gravitation:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{d^2}}[/tex]
Fg = Force of gravity (N)
G = Gravitational Constant
m₁ = mass of object 1 (kg)
m₂ = mass of object 2 (kg)
d = distance between the objects (m)
There is a direct relationship between the masses of the objects and the resulting force of gravity, so we can plug new values in:
[tex]F_g = G\frac{5m_12m_2}{d^2}} = 10G\frac{m_1m_2}{d^2}} = 10F_g[/tex]
Thus, the correct answer is J. 10Fg.
The function of the liver in the metabolism of amino acids is to
Answer:
Produce energy, or make carbohydrates or fats.
Explanation:
Which particles within the atom are electricity charged?
a. Electrons only.
b. Neutrons only.
c. Protons only.
d. Electrons and protons.
e. Protons and neurons.
Answer:
I think b sorry if wrong
Explanation:
it b sorry if wrong
Answer:
Given that these particles make up atoms, they are often referred to as subatomic particles. There are three subatomic particles: protons, neutrons and electrons. Two of the subatomic particles have electrical charges: protons have a positive charge while electrons have a negative charge:
Compare and contrast synthesis and decomposition reactions.
Answer:
Synthesis reactions are chemical reactions where two elements combine to make a product. Decomposition reactions are chemical reactions where a reactant produces another product, usually two, or even more. This is when the bonds in the compound are broken apart to make new compounds.
Explanation:
sue and betti both ski straight down a hill, both starting from rest. sue weighs more than betti. neglecting friction and wind resistance, which skier will be moving the fastest at the bottom of the hill?
Answer:
They are both moving at the same speed since they will have the same acceleration
F = M a describes acceleration
M a = M g sin theta where theta is the slope
One can easily see that
a = g sin theta and the weight does not enter into the equation
1 kg ball has eight joules of kinetic energy. what is its speed?
[tex]\\ \sf\Rrightarrow KE=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \sf\Rrightarrow 8=\dfrac{1}{2}v^2[/tex]
[tex]\\ \sf\Rrightarrow v^2=16[/tex]
[tex]\\ \sf\Rrightarrow v=√16[/tex]
[tex]\\ \sf\Rrightarrow v=4m/s[/tex]
Answer:
the answer is 4m/s
What process primes a molecule to change in a way that increases its activity, produces motion, or does work
Answer:
cellular respiration.
explanation: cellular respiration, the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.
Now you need to put handles on your cookware! You have five different options, including metals and nonmetals such as wood. Based on specific heat, which material would be the best choice? the material with the smallest specific heat the material with the greatest specific heat the material whose specific heat is farthest from the specific heat of the metal pan the material whose specific heat is closest to the specific heat of the metal pan.
Answer: B the material with the greatest specific heat
Explanation:
Just did the assignment
Answer:
B. The material with the greatest specific heat
Explanation:
An object is dropped from a 42 m tall building. How long does it take to reach the ground?
Answer:
2.1 seconds
Explanation:
let D represent the height in meters
let g represent gravity in meters per second squared
let t represent time in seconds
D= -42 meters (negative because it falls down)
g= -9.8m/s^2
t= [tex]\frac{-42 meters}{9.8 m/s^2}[/tex]
this gives t= 4.2857 s^2
the meter units cancel out in the division
now we must simplify the units to singular seconds so...
[tex]\sqrt{4.3 s^{2} }[/tex] which simplifies to [tex]\sqrt{4.3}[/tex] x [tex]\sqrt{s^{2} }[/tex]
this equals 2.1 seconds
An object is moving with a constant velocity of 311 m/s. How long will it take it to travel 9000 m?
[tex]\text{Given that,}\\\\\text{Velocity, v= 311 m s}^{-1}\\\\\text{Displacement, s = 9000 m}\\\\\text{Since velocity is constant,}\\\\s=vt\\\\\implies t =\dfrac{s}v = \dfrac{9000}{311} = 28.94 ~ \text{sec}.[/tex]
[tex] \huge \rm༆ question ༄[/tex]
Calculus proof of second equation of motion ~
Newton's second equation of motion :-
S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]
This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.
So let's see what pieces of information (bundles of equations) do we have with us, initially.
We have, a very primary equation with us,
dS/dt = v… (I)
(Considering motion in a straight line only)
And we also have the equation
dv/dt = a…(II)
Simply replacing the v in eqn (II) by eqn (I), we find
d2S/dt^2 = a…(III)
This is what we need to solve. It's easy.
You know,
d2S/dt^2 = d/dt(dS/dt) = a
⟹ dS/dt = ∫adt = at+c1
Since, dS/dt is the velocity of the particle,
Therefore, at t = 0, dS/dt|t = 0 = u
⟹ u = a∗0 + c1 = c1
⟹ c1 = u
Therefore, dS/dt = u + at
Thus, S = ∫(udt + atdt)
⟹ S = ut + 1/2at^2 +c^2
If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0
This makes S = S0 +ut + 1/2at^2
Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),
We get
S = ut + 1/2at^2
Hope it helps :)
3. Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth's orbit remained
the same as it is now, then compared with gravitational force between Earth and the sun, the gravitational force
between Earth and the star would be -
A. half as much
B. the same
C. two times as much
D. four times as much
Answer:
F1 = G M m / R^2 current force on earth
F2 = G M * 2 m / R^2 force if mass were doubled
F2 / F1 = 2 the force would be twice as large
The gravitational force between two objects not only depends on the mass but also the distance between them. If the distance to the star is same as to sun, then by doubling mass force also doubles. Thus option C is correct.
What is universal law of gravitation?Gravitational force is the attractive force by which an object attracts other objects into its centre of mass. We are all standing in the ground because of the gravitational pull by earth.
The gravitational force between two objects of mass M1 and M2 at a distance r is written as follows:
g = G M1 M2/ r ² . where, G is gravitational constant.
As per this relation, gravitational force is directly proportional to the mass and inversely proportional to the square of the distance.
If the star has a distance from earth equal to the distance between sun and earth, but its mass is double the mass of sun, then the gravitational force will be two times compared with the sun.
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Physicists call any change in energy an impulse true or false?
Why would poor clusters of galaxies be more likely to have irregular shapes then rich
clusters
These Milky Way companion galaxies are easily visible from dark locations in the Southern Hemisphere. Prime examples of erratic galaxies are the Large and Small Magellanic clouds (left and right, respectively).
What clusters of galaxies likely to have irregular shapes?In comparison to a rich cluster, the poor cluster typically has a slightly more erratic shape. A number of smaller galaxies orbit each major spiral. The Small and Large Magellanic clouds are the two most well-known examples of atypical galaxies.
When two galaxies collide, irregular galaxies frequently result. This unusual Cartwheel Galaxy was created when a tiny galaxy slid through the centre of a massive spiral galaxy.
Therefore, Rich clusters are other clusters that include hundreds to thousands of galaxies. A weak cluster can't cling to its members strongly because of its low bulk.
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which term is defined by the interaction between two charged particles?
Answer: Electromagnetism
The term which is defined by the interaction between two charged particles is known as electromagnetism.
What is an electromagnet?The study of charge and the fields and forces it generates is known as electromagnetism. Electromagnetism has two components: electricity and magnetism.
The special theory of relativity, developed by Albert Einstein in 1905, proved beyond a shadow of a doubt that both are components of the same reality. But in reality, magnetic and electric forces behave very differently and are modeled by various equations. Electric charges can produce forces whether they are stationary or moving. On the other hand, magnetic forces are only created by moving charges and only affect charges that are moving.
Thus, the term is known as electromagnetism.
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The weight of a person in an elevator at rest = 500 N. Acceleration due to gravity is 9.8 m/s2. When lift accelerated, the tension force is 750 N. What is the acceleration of lift.
Answer:If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.
The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.
T – w = m a
750 – 500 = 50 a
250 = 50 a
a = 250 / 50
a = 5.0 m s–2
Explanation:
The Sun has a mass of 1.99 x 10^30kg [^30 is an exponent] and is 1.5 x 10^11m [^11 is an exponent] from the earth. The planet Earth is 5.98 x 10^24kg [^24 is an exponent]. What is the gravitational attraction between the sun and the earth? G=6.67×10^-11 (-11 is an exponent)
Answer:
F = G M m / R^2 = 6.67E-11 * 1.99E30 * 5.98E24 / (1.5E11)^2
F = 6.67 * 1.99 * 5.98 / 2.25 E21 Newtons
F = 3.53E22 N
A car weighing 8000N is traveling at 45 m/s on a perfectly flat, frictionless road. If the driver slams on the brakes, how far will thw car slide before it comes to a stop?
Without friction, the car cannot stop...
how much brighter is Vega than the Sun? (astronomy)
Answer: 60 times brighter
Drag each tile to the correct location.
Match each process to where it occurs in the carbon cycle.
Ingestion
fossilization
decomposition
combustion
light
energy
oxygen
carbon
dicadde
- sugar
minerak
water
All rights reserved
The order of the location of each process in the carbon cycle are as follows;
Ingestion → Decomposition → Fossilization → CombustionReasons:
Ingestion; During the ingestion process, the products of the direct Sunlight and photosynthesis, which
are energy rich chemical substances are ingested by herbivores.
Decomposition; The plants and animals die and undergo decomposition.
Fossilization; A long period the decomposed plants and animals manifest
into hard petrified materials, with their parts transformed to fossils.
Combustion; The remains of plants and animals, subjected to heat and
pressure due to compression, as when as bacteria decomposition that
removes several other elements, leaving a carbon and hydrogen rich
sludge that can be processed into fuel used for combustion. The
combustion process produces carbon dioxide which restarts the cycle.
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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/a to the left (-x direction).what is the magnitude of the racquetball's change in momentum?
The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.
Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).
mass m = 42g = 42/1000 = 0.042kg
initial velocity before collision u = 7 m/s
It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,
velocity after collision v = 7 m/s
To calculate the magnitude of the racquetball's change in momentum, we will use the formula below
Change in momentum = Mv - Mu
Since momentum is a vector quantity, we will consider the direction.
Change in momentum = 0.042 x 7 - ( 0.042 x - 7)
Change in momentum = 0.294 + 0.294
Change in momentum = 0.588 kgm/s
Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.
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