The man pushes/pulls with a force of 200 N. The child and sled
combo have a mass of 30 kg and the coefficient of kinetic friction is
0.15.The angle between the slide and the ground is 30 degrees. For each case:
What is the frictional force opposing his efforts?
What is the acceleration of the child?
The frictional force opposing his efforts can be found using the formula:
frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the child and sled combo, which can be found using:
weight = mass * gravitational acceleration
where gravitational acceleration is approximately 9.8 m/s^2.
So, the weight of the child and sled combo is:
weight = 30 kg * 9.8 m/s^2 = 294 N
Therefore, the normal force is also 294 N.
Now, we can calculate the frictional force:
frictional force = 0.15 * 294 N = 44.1 N
To calculate the acceleration of the child, we need to use Newton's second law:
force = mass * acceleration
We can consider the forces acting on the child and sled combo as follows:
- The force of the man pushing/pulling, which is 200 N and is parallel to the slope.
- The weight of the child and sled combo, which is acting straight down and has a magnitude of 294 N.
- The frictional force, which is opposing the motion and is acting in the opposite direction to the force of the man.
Since the angle between the slope and the ground is 30 degrees, we need to resolve the forces into components parallel and perpendicular to the slope. The weight force can be resolved as follows:
weight parallel = weight * sin(30) = 294 N * sin(30) = 147 N
weight perpendicular = weight * cos(30) = 294 N * cos(30) = 254.6 N
The force of the man is already parallel to the slope, so the parallel component is simply 200 N. The frictional force is opposite to the parallel component, so it has a magnitude of 44.1 N in the opposite direction.
Now, we can calculate the net force parallel to the slope:
net force parallel = force parallel - frictional force
net force parallel = 200 N - 44.1 N = 155.9 N
Using Newton's second law, we can calculate the acceleration:
acceleration = net force parallel / mass
acceleration = 155.9 N / 30 kg = 5.2 m/s^2
Therefore, the acceleration of the child is 5.2 m/s^2.
What is the minimum coefficient of static
friction necessary to allow a penny to rotate
along a 33 𝟏
𝟑
rpm record (diameter= 0.300 m),
when the penny is placed at the outer edge of record?
First, we need to find the speed of the penny as it rotates on the record. The circumference of the record is:
circumference = pi * diameter = 3.14 * 0.300 m = 0.942 m
At 33 1/3 rpm, the penny makes one full revolution in 1/3 of a minute (or 20 seconds). Therefore, the speed of the penny can be found by dividing the circumference by the time:
speed = circumference / time = 0.942 m / 20 s = 0.0471 m/s
Now, we can use the formula for centripetal force:
centripetal force = (mass * speed^2) / radius
Since the penny is not slipping on the record, the centripetal force must be equal to or less than the maximum static frictional force that the record can exert on the penny. Therefore, we can use the formula for static friction:
static friction = coefficient of static friction * normal force
The normal force is equal to the weight of the penny, which can be found using:
weight = mass * gravitational acceleration
where gravitational acceleration is approximately 9.8 m/s^2.
A United States penny has a mass of approximately 2.5 grams, or 0.0025 kg. Therefore, the weight of the penny is:
weight = 0.0025 kg * 9.8 m/s^2 = 0.0245 N
We can now substitute the formulas for centripetal force and static friction, and solve for the coefficient of static friction:
(mass * speed^2) / radius <= coefficient of static friction * normal force
(mass * speed^2) / radius <= coefficient of static friction * mass * gravitational acceleration
coefficient of static friction >= (mass * speed^2) / (radius * mass * gravitational acceleration)
coefficient of static friction >= speed^2 / (radius * gravitational acceleration)
coefficient of static friction >= 0.0471 m/s^2 / (0.150 m * 9.8 m/s^2)
coefficient of static friction >= 0.032
Therefore, the minimum coefficient of static friction necessary to allow the penny to rotate along a 33 1/3 rpm record, when the penny is placed at the outer edge of the record, is 0.032.
What does this words means :imidazole side chain and pka value ?
Imidazole is an organic compound that contains a five-membered ring with two nitrogen atoms. It is commonly found as a side chain in some amino acids, such as histidine.
In this context, imidazole side chain refers to the particular chemical group (or "side chain") that is present in some amino acids and contains the imidazole ring.
pKa (short for "acid dissociation constant") value is a measure of the strength of an acid in solution. For imidazole, the pKa value of the nitrogen atom in the ring is around 7.0, which means that it can act as a weak base and become protonated (pick up a hydrogen ion) under acidic conditions.
The pKa value is important in biochemistry because it determines how the side chains of amino acids will behave in different pH environments. For example, the imidazole side chain of histidine has a pKa value around 6, which means that at physiological pH (around 7.4), it will typically exist in a partially protonated form. This property makes it useful in enzyme catalysis and other biochemical reactions.
protonated
Protonation is a chemical process in which a proton (a hydrogen ion) is added to a molecule or ion. When a molecule or ion gains a proton, it becomes positively charged and is said to be protonated.
In the context of biochemistry, protonation often happens to functional groups in protein side chains, such as amino acids. For example, the imidazole side chain of the amino acid histidine contains a nitrogen atom that can pick up a proton. When the nitrogen is protonated, it gains a positive charge and is able to form new chemical bonds with other atoms or molecules, which can have important effects on protein structure and function.
Difference between virus and bacteria
Virus and bacteria are both microscopic organisms that can cause diseases, but they differ in several key ways:
1. Structure: Bacteria are usually single-celled organisms with a cell wall and a cell membrane that contains DNA and other cellular components. Viruses, on the other hand, do not have a cell structure but are made up of genetic material (either DNA or RNA) surrounded by a protein coat or envelope.
2. Reproduction: Bacteria can reproduce on their own through cell division. They can also transfer genetic information to other bacteria through processes like conjugation, transduction, and transformation. In contrast, viruses are incapable of reproduction on their own - they need to infect a host cell and hijack its machinery in order to replicate.
3. Treatment: Antibiotics can be used to treat bacterial infections, but they are ineffective against viral infections. Viral infections are typically treated symptomatically or with antiviral medications that can help inhibit viral replication.
4. Host range: Bacteria can infect a range of hosts, including humans, animals, and plants, while viruses are usually host-specific and can only infect certain types of cells in certain species.
5. Size: Bacteria are larger than viruses. Most bacteria are between 0.5 and 5 micrometers long, while most viruses are between 20 and 300 nanometers in diameter.
6. Living status: Bacteria are considered living organisms because they can carry out metabolic processes, respond to stimuli and reproduce. In contrast, viruses are considered non-living because they are unable to carry out metabolic processes, move, or reproduce on their own - they need to infect a host cell to do so.