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I just want you to find exaples of “Primary, secondary, quartenary, tertianary protien structure and thei functions”
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Protein Structures
The protein structure is divided into four structures;
The primary structure is the simplest among all and is simply the sequence of amino
acids in t...
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I need someone to paraphrase this for me
1. Write a brief explanation of what
happened to the coin before and after you removed the index card in Part 1 in ...
I need someone to paraphrase this for me
1. Write a brief explanation of what
happened to the coin before and after you removed the index card in Part 1 in
terms of forces and Newton’s laws.
Answer:
When the coin is resting on the top of a beaker on an
index card, it is not moving. There is a static friction between the coin and
the index card. When I exerted force on the card and caused the card to move,
the coin fell in to the cup. If this situation was to be explained using
Newton's laws,Newton's first law of motion says that an object will remain at
rest when the net force is zero. The fact that the coin stayed still on the
card was because the net force was zero or balance. This is because the card is
exerting the same magnitude but opposite direction force on the coin as the
coin is exerting on it and so those two forces cancel out each other. The coin
and the card exert force on each other because, for every reaction ( weight of
the coin) there is an opposite but equal reaction (the force the card exerts to
support the coin). But when force was
exerted on the card, the net force was no longer zero and so the object moved.
The coin fell to the glass and the card moved away in the direction the force
was exerted. The coin accelerated
downward because it's weight was no longer supported by the card and it caused
it to accelerate downward to the bottom of the glass.
(5 points)
Score
2. Based on your observations of the
spring scale in Part 2, describe all the forces acting on an elevator as it
moves from rest upward and stops at its floor.
Answer:
When the elevator moves up ward from rest, there is
applied force by the strings that causes it to accelerate. But the applied
force has to be greater than the force of gravity for it to accelerate
upwards. If there are passengers in the
elevator then their weight is applied towards the floor of the elevator and the
elevator force also exerts a normal force on the passenger. When the elevator
is accelerating upward the weight of the passengers appears to be greater than
the real weight. When the elevator is coming to a rest at a floor, then it
begins to decelerate and acceleration is less than the acceleration due to
gravity. The weight of the passenger here appears to be less than the real
weight.
(8 points)
Score
3. Use your data from Part 3 and
Newton’s laws to explain why the force meter measures a force if the cart is
moving at a constant velocity. You should cover the following points.
·
State Newton’s first law.
·
Identify the forces acting on the cart.
·
Describe the relationships between applied force and mass and between applied
force and weight as linear (from data or mathematical derivation by applying
Newton’s first law).
·
Explain why the data are linear in terms of Newton’s first law.
·
Graph F versus mass + cart.
Answer:
Newton's first law of motion says that an object at
rest will remain at rest if the net force acting on it is zero and an object in motion will continue to move in a
constant velocity if the net force acting on it is zero. The cart was moving at
a constant velocity but still I got a value different from zero when I measured
it with the force meter. This is because I was measuring the applied force only
but if I calculated the net force by subtracting or adding other forces acting
on the cart, I would probably get a zero net force. Forces acting on the cart
were the applied force I was measuring, friction in opposite direction of the
applied force, the weight of the cart on the table and the table's force on the
cart that was in opposite direction called normal force. This force is equal to
the weight of the cart.
Mass (g)
Mass + Cart (g)
Mass + Cart (kg)
Weight (N)
Force (N)
100
151
0.15
1.47
0.15
200
251
0.25
2.45
0.25
300
351
0.35
3.43
0.5
400
451
0.45
4.41
0.75
500
551
0.55
5.39
1
Mass of the cart = 51 grams
As more mass was added on the cart, the more applied
force need. This means that applied force had a linear relationship with mass
as both were increasing together. Also the weight and applied force has linear
relationship with the weight of the cart because they were also both increasing
together. They are linear because according to Newton's second law of motion
mass is proportional with force.
PICTURE (8 points)
Score
4. For Part 4, describe the motion
of the ball as it moves from the top of the ramp and comes to rest in the cup.
Explain how the motion changes in terms of the velocity, distance,
acceleration, and force that you measured. Use your data and Newton’s laws to
calculate these values and explain the situation. Include the tables you used
to calculate average initial velocity and average displacement. Also show your
work for your calculations of acceleration and net force.
Here are the kinematic equations:PICTURE 4
Answer:
when the ball was rolling down the ramp, the force of
gravity was acting on it and therefore it was decelerating. Once the ball
reached the end, it collided with the cup and moved a few distance and came to
stop. The ball had a constant acceleration which means the velocity was not
constant. The acceleration of the ball can be calculate using a kinematic
formula vf2 = vi2 + 2aΔd.
This formula can be modified to a = vf2 -vi ^2 /
2Δd.
Final velocity was zero and the first velocity was 1
m/s. Moreover the change in displacement was 0.36 meters
a =( 0m/s)^2 – (1m/s )^2 / 2* 0.36
a= -1.4 m/s^2
Time (s)
Distance (m)
Velocity (m/s)
0.93
1
1.08
1
1
1
1
1
1
1
1
1
1.07
1
0.93
Average velocity (m/s) = 1. 0 m/s
∆d (cm)
∆d (m)
Trial 1
34
0.34
Trial 2
36
0.36
Trail 3
36
0.36
Trial 4
38
0.38
Average
∆d(m) = 0.36 m
The net force of the collision of the ball with cup
can also be calculated using the formula
Fnet= mass x
acceleration
Fnet = ?
a = 1.4
m= 0.67kg
Fnet = 0.67kg * 1.4 m/s^2
Fnet = 0.94 N
EES 2021 Temple University Sedimentary Environments & Structures Discussion
I alrady attched file below...................................................................................
EES 2021 Temple University Sedimentary Environments & Structures Discussion
I alrady attched file below...................................................................................
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Module 8 Activity.pdf download and complete the worksheet. As always, please don't hesitate to follow up with me if you have any questions.
Virtual Lab on Cobalt Chloride and LeChatlier's Principle, chemistry homework help
Click on the link below and complete the virtual lab for Cobalt Chloride and LeChatlier's Principle. After completin ...
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Click on the link below and complete the virtual lab for Cobalt Chloride and LeChatlier's Principle. After completing the lab, you will generate a simple lab report following the outline in the attached document. All data should be included in the report. The lab must be accurate and correct. http://chemcollective.org/vlab/85
Virtual Physics Lab Access via ToolWire, assignment help
Complete the following labs in the Virtual Physics Lab Workbook. Within the workbook, choose the chapter that matches titl ...
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I need someone to paraphrase this for me
1. Write a brief explanation of what
happened to the coin before and after you removed the index card in Part 1 in ...
I need someone to paraphrase this for me
1. Write a brief explanation of what
happened to the coin before and after you removed the index card in Part 1 in
terms of forces and Newton’s laws.
Answer:
When the coin is resting on the top of a beaker on an
index card, it is not moving. There is a static friction between the coin and
the index card. When I exerted force on the card and caused the card to move,
the coin fell in to the cup. If this situation was to be explained using
Newton's laws,Newton's first law of motion says that an object will remain at
rest when the net force is zero. The fact that the coin stayed still on the
card was because the net force was zero or balance. This is because the card is
exerting the same magnitude but opposite direction force on the coin as the
coin is exerting on it and so those two forces cancel out each other. The coin
and the card exert force on each other because, for every reaction ( weight of
the coin) there is an opposite but equal reaction (the force the card exerts to
support the coin). But when force was
exerted on the card, the net force was no longer zero and so the object moved.
The coin fell to the glass and the card moved away in the direction the force
was exerted. The coin accelerated
downward because it's weight was no longer supported by the card and it caused
it to accelerate downward to the bottom of the glass.
(5 points)
Score
2. Based on your observations of the
spring scale in Part 2, describe all the forces acting on an elevator as it
moves from rest upward and stops at its floor.
Answer:
When the elevator moves up ward from rest, there is
applied force by the strings that causes it to accelerate. But the applied
force has to be greater than the force of gravity for it to accelerate
upwards. If there are passengers in the
elevator then their weight is applied towards the floor of the elevator and the
elevator force also exerts a normal force on the passenger. When the elevator
is accelerating upward the weight of the passengers appears to be greater than
the real weight. When the elevator is coming to a rest at a floor, then it
begins to decelerate and acceleration is less than the acceleration due to
gravity. The weight of the passenger here appears to be less than the real
weight.
(8 points)
Score
3. Use your data from Part 3 and
Newton’s laws to explain why the force meter measures a force if the cart is
moving at a constant velocity. You should cover the following points.
·
State Newton’s first law.
·
Identify the forces acting on the cart.
·
Describe the relationships between applied force and mass and between applied
force and weight as linear (from data or mathematical derivation by applying
Newton’s first law).
·
Explain why the data are linear in terms of Newton’s first law.
·
Graph F versus mass + cart.
Answer:
Newton's first law of motion says that an object at
rest will remain at rest if the net force acting on it is zero and an object in motion will continue to move in a
constant velocity if the net force acting on it is zero. The cart was moving at
a constant velocity but still I got a value different from zero when I measured
it with the force meter. This is because I was measuring the applied force only
but if I calculated the net force by subtracting or adding other forces acting
on the cart, I would probably get a zero net force. Forces acting on the cart
were the applied force I was measuring, friction in opposite direction of the
applied force, the weight of the cart on the table and the table's force on the
cart that was in opposite direction called normal force. This force is equal to
the weight of the cart.
Mass (g)
Mass + Cart (g)
Mass + Cart (kg)
Weight (N)
Force (N)
100
151
0.15
1.47
0.15
200
251
0.25
2.45
0.25
300
351
0.35
3.43
0.5
400
451
0.45
4.41
0.75
500
551
0.55
5.39
1
Mass of the cart = 51 grams
As more mass was added on the cart, the more applied
force need. This means that applied force had a linear relationship with mass
as both were increasing together. Also the weight and applied force has linear
relationship with the weight of the cart because they were also both increasing
together. They are linear because according to Newton's second law of motion
mass is proportional with force.
PICTURE (8 points)
Score
4. For Part 4, describe the motion
of the ball as it moves from the top of the ramp and comes to rest in the cup.
Explain how the motion changes in terms of the velocity, distance,
acceleration, and force that you measured. Use your data and Newton’s laws to
calculate these values and explain the situation. Include the tables you used
to calculate average initial velocity and average displacement. Also show your
work for your calculations of acceleration and net force.
Here are the kinematic equations:PICTURE 4
Answer:
when the ball was rolling down the ramp, the force of
gravity was acting on it and therefore it was decelerating. Once the ball
reached the end, it collided with the cup and moved a few distance and came to
stop. The ball had a constant acceleration which means the velocity was not
constant. The acceleration of the ball can be calculate using a kinematic
formula vf2 = vi2 + 2aΔd.
This formula can be modified to a = vf2 -vi ^2 /
2Δd.
Final velocity was zero and the first velocity was 1
m/s. Moreover the change in displacement was 0.36 meters
a =( 0m/s)^2 – (1m/s )^2 / 2* 0.36
a= -1.4 m/s^2
Time (s)
Distance (m)
Velocity (m/s)
0.93
1
1.08
1
1
1
1
1
1
1
1
1
1.07
1
0.93
Average velocity (m/s) = 1. 0 m/s
∆d (cm)
∆d (m)
Trial 1
34
0.34
Trial 2
36
0.36
Trail 3
36
0.36
Trial 4
38
0.38
Average
∆d(m) = 0.36 m
The net force of the collision of the ball with cup
can also be calculated using the formula
Fnet= mass x
acceleration
Fnet = ?
a = 1.4
m= 0.67kg
Fnet = 0.67kg * 1.4 m/s^2
Fnet = 0.94 N
EES 2021 Temple University Sedimentary Environments & Structures Discussion
I alrady attched file below...................................................................................
EES 2021 Temple University Sedimentary Environments & Structures Discussion
I alrady attched file below...................................................................................
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