When two heavy blocks were connected by a uniform rope that has a mass of 4.00 kg, acceleration of the system was found to be 2.7 m/s²; the tension at the top of heavy rope was 136.5 N; the tension exerted at the midpoint of the rope was112.5N.
Tension is defined as the act of stretching or straining or the condition of a substance being stretched or strained.
a) Acceleration of the system can be calculated as
a=Fn/m
={200-(16x9.8)}/16
=2.7m/s²
so therefore the acceleration of the system was calculated and found to be=2.7m/s²
b) the tension of the heavy rope at the top can be calculated as
200-50-T1=5(2.7)
T1=136.5N
the tension at the top of heavy rope was 136.5 N
c)the tension at the middle point of the rope was
T2-9g = 9(a)
=9(g+a)
=9(9.8+2.7)
=112.5 N
so here the tension exerted at the midpoint of the rope was 112.5N.
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If the mass of the box is .612 kg, what is the box's acceleration?
Answer:
Net force/.612=acceleraton
Explanation:
Need more information for a specific answer. I derived this general equation using newton's second law: F=MA. The acceleration of the object is calculated by dividing the net force on the object by its mass.
A uniform electric field of magnitude 1.1 x 104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the electric flux through the sheet?
The electric flux through the square sheet is [tex]4.4 * 10^4 Nm^2/C[/tex], when a uniform electric field of magnitude [tex]1.1 * 10^4 N/C[/tex] is perpendicular to a square sheet with sides 2.0 m long.
The electric flux through a closed surface is given by the formula:
[tex]\[ \Phi = \mathbf{E} \cdot \mathbf{A} \][/tex]
where [tex]\(\Phi\)[/tex] is the electric flux, [tex]\(\mathbf{E}\)[/tex] is the electric field, and [tex]\(\mathbf{A}\)[/tex] is the area vector of the surface. In this case, the electric field [tex]\(\mathbf{E}\)[/tex] is perpendicular to the square sheet, and the magnitude of the electric field is given as [tex]1.1 * 10^4 N/C[/tex].
The area of the square sheet is [tex]\(A = (2.0 \, \text{m})^2 = 4.0 \, \text{m}^2[/tex]). Since the electric field is perpendicular to the surface, the angle between the electric field and the area vector is 0 degrees.
Substituting the values into the formula, we have:
[tex]\[ \Phi = (1.1 \times 10^4 \, \text{N/C}) \cdot (4.0 \, \text{m}^2) = 4.4 \times 10^4 \, \text{N} \cdot \text{m}^2/\text{C} \][/tex]
Therefore, the electric flux through the square sheet is [tex]4.4 * 10^4 Nm^2/C[/tex].
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what are the contents of register r1 in decimal after executing the following two instructions: (all numbers are hexadecimal. negative numbers are represented with two's complement.) mov r1, eb //moves data (eb) into register r1 mov r2, 57 //moves data (57) into register r2 sub r1, r2 // r1
The contents of register r1 after executing these instructions is 148 in decimal. To determine the contents of register r1 in decimal after executing the given instructions, let's go through each step.
mov r1, eb: This instruction moves the data "eb" into register r1. Assuming the data is in hexadecimal, "eb" is equivalent to the decimal value 235.
mov r2, 57: This instruction moves the data "57" into register r2. Assuming the data is in hexadecimal, "57" is equivalent to the decimal value 87.
sub r1, r2: This instruction subtracts the value in register r2 (87) from the value in register r1 (235). The result of the subtraction is stored in r1.
Subtracting 87 from 235 gives us 148. Therefore, after executing these instructions, the contents of register r1 in decimal would be 148.
n the given scenario, the contents of register r1 in decimal after executing the instructions "mov r1, eb", "mov r2, 57", and "sub r1, r2" are 148.
The instruction "mov r1, eb" moves the hexadecimal value "eb" into register r1, which is equivalent to the decimal value 235.
The instruction "mov r2, 57" moves the hexadecimal value "57" into register r2, which is equivalent to the decimal value 87.
The instruction "sub r1, r2" subtracts the value in register r2 (87) from the value in register r1 (235). The result, 148, is stored in register r1.
Therefore, the contents of register r1 after executing these instructions is 148 in decimal.
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an arrow is launched 30 degrees above the horizontal with an intial speed of 20 m/s from the top of a watch tower after 3 seconds what is the arrows speed
The arrow's speed after 3 seconds is approximately 43.091 m/s.
The horizontal and vertical components of the velocity can be determined using trigonometry. The initial vertical velocity is given by v₀y = v₀ sin θ, and the horizontal velocity is given by v₀x = v₀ cos θ.
Given, θ = 30°, v₀ = 20 m/s, t = 3 s. The acceleration of gravity is g = 9.81 m/s².Let's now use the kinematic equation vf = v₀ + at. Since the arrow is launched at an angle above the horizontal, we need to calculate the horizontal and vertical components of the velocity.
The initial vertical velocity of the arrow is:
v₀y = v₀ sin θ= (20 m/s) sin 30°= 10 m/s
The horizontal velocity of the arrow is:
v₀x = v₀ cos θ= (20 m/s) cos 30°= 17.3205 m/s
Using the kinematic equation vf = v₀ + at, we can calculate the vertical velocity of the arrow after 3 seconds:
vfy = v₀y + gt
= (10 m/s) + (9.81 m/s²)(3 s)
= 39.43 m/s
The horizontal velocity doesn't change during the flight of the arrow, so vf = v₀x = 17.3205 m/s.
The final velocity of the arrow is given by the vector sum of the horizontal and vertical velocities:
v = √(vf² + v₀x²)
= √(39.43² + 17.3205²)
= 43.091 m/s
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What are three different types of energy?
Answer:
Gravitational energy
Thermal energy
Mechanical energy
what is the distance covered by a free falling body during the first 3 seconds of its motion?
ANSWER MY QUESTION WITH EXPLANATION
PLEASE HELP AND HURRY
Suppose that an object is moving horizontally with a constant velocity. The horizontal acceleration must be __________.
A. constantly decreasing
B. constantly increasing
C. positive
D. equal to zero
Answer:
C. POSITIVE
(Sorry if wrong)
Answer:
C positive
Explanation:
It keeps moving in a positive way over and over
A bullet moving horizontally with a velocity of 40.0 m/s strikes a sandbag and penetrates a distance of 20.0 cm before coming to rest. what is the acceleration of the bullet in m/s?
Answer:
-4000 m/s²
Explanation:
Given:
Δx = 0.200 m
v₀ = 40.0 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (40.0 m/s)² + 2a (0.200 m)
a = -4000 m/s²
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
the acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy
at this point?
O A. 306
B. 294,000 J
O C. 29,400 J
O D. 720 j
Answer:
P.E=29400J
Explanation:
m=150Kg h=20m g=9.8ms⁻²
P.E=mgh
P.E=150×9.8×20
P.E=29400J
The potential energy of the crane that raises a crate with a mass of 150kg is 29,400 Joules. Thus, the correct option is C.
What is Potential energy?Potential energy is the energy which is stored inside the body of an object which is at rest. This energy is transformed into kinetic energy when the body comes into motion. The SI unit of potential energy is Joules.
The potential energy stored in the crate is:
PE = m × g × h
where, PE = Potential energy,
m = mass of the object,
g = acceleration due to gravity,
h = height which the object achieve
PE = 150kg × 9.8m/s² × 20m
PE = 3000 × 9.8
PE = 29,400 Joules
Therefore, the correct option is C. The potential energy of the crate is 29,400 Joules.
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Density of oil is less than the density of water. *
true
False
Answer:
false
Explanation:
because oil is thicker than water
Answer:
answer is false
Explanation:
water is more dense than oil so they can't mix . Oil float above the water
Which event is an example of condensation?
O A. Ice forms on the surface of a puddle.
B. The outside of a glass of ice water becomes moist.
c. Perspiration dries on a person's skin.
D. Fog disappears when the Sun comes out.
Answer:
the correct answer is B.
a rifle bullet with a mass of 11.5 g traveling toward the right at 248 m/s strikes a large bag of sand and penetrates it to a depth of 25.0 cm. determine the magnitude and direction of the friction force (assumed constant) that acts on the bullet.
Magnitude of the friction force acting on the bullet is 6823.68 N and the direction of the friction force is to the left. We can use the work-energy principle to determine the friction force acting on the bullet.
Given data : Mass of the rifle bullet (m) = 11.5 g = 0.0115 kg, Initial velocity of the rifle bullet (u) = 0 m/s, Final velocity of the rifle bullet (v) = 248 m/s, Displacement of the bullet (s) = 25 cm = 0.25 m
We can use the work-energy principle to determine the friction force acting on the bullet.
Work-energy principles states that the work done by the net force on an object is equal to its change in kinetic energy. Wnet = ΔK
Where, Wnet = Net work done by all the forces (applied force and friction force), ΔK = Change in kinetic energy = Kf – KiKf = Final kinetic energy,
Ki = Initial kinetic energy, Initial kinetic energy of the bullet,
Ki = (1/2) mu²
= (1/2) × 0.0115 kg × 0²
= 0 J
Final kinetic energy of the bullet,
Kf = (1/2) mv²
= (1/2) × 0.0115 kg × (248 m/s)²
= 1705.92 J
∴ ΔK = Kf - Ki
= 1705.92 J - 0 J
= 1705.92 J
Now, we know that the displacement of the bullet is 0.25 m. Therefore, the work done by the friction force can be written as follows: Wf = Fs, where, F = Magnitude of friction force acting on the bullet= - F (direction of friction force is opposite to that of the motion)
∴ W net = W f + W applied
∴ - F (0.25 m) = 1705.92 J
∴ F = - (1705.92 J) / 0.25 m
= - 6823.68 J/m
= - 6823.68 N (approx)
The friction force is directed opposite to the direction of motion of the bullet. Therefore, the direction of the friction force is to the left.
Magnitude of friction force acting on the bullet = 6823.68 N
Direction of friction force acting on the bullet = Left
Magnitude of the friction force acting on the bullet is 6823.68 N and the direction of the friction force is to the left.
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The graphs represent the position X, velocity V, and acceleration a as a function of time T for a marble moving in one dimension. Which of the following could describe the motion of the marble? a. Rolling along the floor and then bouncing off a wallb. Rolling down one side of a bowl and then rolling up the other sidec. Rolling up a ramp and then rolling back downd. Falling and then bouncing elastically off a hard floorState the correct answer and then for each wrong choice, draw a sketch that shows what a position vs. time and velocity vs. time graph WOULD look like if that were the right answer. Since B is wrong it has been done for you and an example.
The correct answer is: b. Rolling down one side of a bowl and then rolling up the other side
The position vs. time graph for this scenario would show a U-shaped curve, indicating the marble's motion down one side of a bowl and then up the other side. The velocity vs. time graph would show a parabolic shape, with the marble's velocity increasing as it rolls down the bowl, reaching a maximum at the bottom, and then decreasing as it rolls up the other side.
Now, let's examine the other options and describe the position vs. time and velocity vs. time graphs for each:
a. Rolling along the floor and then bouncing off a wall
In this case, the position vs. time graph would show a linear increase or decrease in position, depending on the direction of motion, until the marble reaches the wall and bounces off. The velocity vs. time graph would show a sudden change in velocity when the marble bounces off the wall.
c. Rolling up a ramp and then rolling back down
The position vs. time graph would show an increasing slope as the marble rolls up the ramp and then a decreasing slope as it rolls back down. The velocity vs. time graph would show a positive velocity while rolling up the ramp, then a decrease to zero at the top, followed by a negative velocity while rolling back down.
d. Falling and then bouncing elastically off a hard floor
The position vs. time graph would show a linear increase in position during the fall until the marble hits the floor, followed by a sudden decrease in position when it bounces back up. The velocity vs. time graph would show a constant negative velocity during the fall and then a sudden change in velocity to a positive value when it bounces off the floor.
By analyzing the motion described by the position vs. time and velocity vs. time graphs, we can conclude that the correct answer is b. Rolling down one side of a bowl and then rolling up the other side.
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How
do you think the outcomes of the Lotka- Volterra models would be
similar or different if the predator fed on several different prey
items? Why?
please explain
If the predator in the Lotka-Volterra models feeds on several different prey items, the outcomes would likely be different compared to a scenario where it feeds on a single prey item.
In the Lotka-Volterra models, the interaction between predators and prey is typically represented by a set of differential equations. When a predator feeds on several different prey items, it introduces additional complexities to the model. Here are a few reasons why the outcomes would likely be different:
Prey Interactions: Different prey items may have different population dynamics and interactions with each other. Some prey may compete for resources, while others may have a mutually beneficial relationship. These interactions can affect the predator's ability to obtain food and impact the population dynamics of both the predator and prey species.
Predator Behavior: A predator feeding on multiple prey items may exhibit different foraging behaviors. It could switch between prey types based on availability or preference. This dynamic foraging behavior can influence the predation pressure on each prey species and lead to different population dynamics.
Resource Partitioning: When a predator consumes multiple prey items, there may be resource partitioning among the prey species. Each prey species may have different resource requirements, leading to variations in their population sizes and dynamics. This partitioning can create niche differentiation and affect the overall stability and dynamics of the predator-prey system.
Trophic Cascades: The presence of multiple prey items can introduce the possibility of trophic cascades, where changes in one prey population can indirectly affect the other prey species and subsequently impact the predator. For example, if the predator's primary prey becomes scarce, it may switch to another prey item, which could lead to population fluctuations and indirect effects throughout the ecosystem.
In summary, when the predator in the Lotka-Volterra models feeds on multiple prey items, the outcomes can be different due to complex prey interactions, predator behavior, resource partitioning, and the potential for trophic cascades. These additional factors introduce more variables and dynamics into the model, leading to different population sizes and interactions between the predator and prey species.
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Which of the following is/are the best example(s) of elastic collisio(s)?
A) A collision between two billiard balls.
B) A collision between two automobiles.
C) A basketball bouncing off the floor.
D) An egg colliding with a brick wall.
Answer: (D) an egg colliding with a brick wall is not the best example of an elastic collision.
An elastic collision is defined as a collision in which both kinetic energy and momentum are conserved. Kinetic energy is transferred from one object to another during a collision in which they both collide. The energy lost in a collision is known as inelastic energy.
The best example of an elastic collision is:
A) a collision between two billiard balls and
C) a basketball bouncing off the floor.
A collision between two billiard balls A collision between two billiard balls is the best example of an elastic collision. When two billiard balls collide, kinetic energy is transferred from one ball to the other. However, because the billiard balls are made of elastic materials, the energy is not lost during the collision. As a result, both the momentum and the kinetic energy of the billiard balls remain unchanged after the collision.A basketball bouncing off the floorWhen a basketball bounces off the floor, it is also an example of an elastic collision. When the basketball hits the floor, kinetic energy is transferred from the basketball to the floor.
However, because the basketball is made of an elastic material, the energy is not lost. As a result, the basketball bounces back up off the floor with the same kinetic energy that it had before it hit the floor.
Therefore, a basketball bouncing off the floor is also an example of an elastic collision. An egg colliding with a brick wall is an example of an inelastic collision. In an inelastic collision, some of the kinetic energy is lost during the collision. As a result, both the momentum and the kinetic energy of the objects are not conserved.
Therefore, D) an egg colliding with a brick wall is not the best example of an elastic collision.
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If the compressibility of the aquifer is 10% m?N and its thickness is 10 m, how much compaction (m) will the aquifer undergo during the head reduction in part (c) ?
If the porosity and the hydraulic conductivity of the aquifer are 0.30 and 1.0 m/s, calculate the transmissivity (m7) and storativity of the aquifer. The compressibility of water is 4.4 x 10-1° m'/N.
a. The aquifer will undergo compaction of 1 meter during the head reduction in part (c).
b. The transmissivity of the aquifer is 0.30 m²/s and the storativity is 3.3 x 10⁻⁵.
a. Given that the compressibility of the aquifer is 10% m/N and its thickness is 10 meters, we can calculate the compaction using the formula:
Compaction = Compressibility x Thickness
Compaction = (10% m/N) x 10 m = 0.10 x 10 m = 1 meter
Therefore, the aquifer will undergo compaction of 1 meter during the head reduction.
b. Transmissivity (T) is calculated by multiplying the hydraulic conductivity (K) of the aquifer by its thickness (h):
Transmissivity (T) = K x h = (1.0 m/s) x 10 m = 10 m²/s
Storativity (S) is calculated as the product of the specific storage (Ss) and the thickness (h):
Storativity (S) = Ss x h = (8.8 x 10⁻⁶) x 10 m = 8.8 x 10⁻⁵
Therefore, the transmissivity of the aquifer is 10 m²/s and the storativity is 8.8 x 10⁻⁵.
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A physicist observes that a cannonball shot horizontally from a cannon travels 1.000 m before falling to the ground. Which of the following statements correctly explains this observation using Newton's laws?
O Objects tend to remain in the state of motion they are in unless acted upon by an unbalanced force. Friction and gravity acted on the cannonball, slowing it down and slamming it into the ground
O Objects at rest tend to remain in their cument state of motion unless acted upon by an unbalanced force, but objects in motion require a continual application of force to stay in motion
The force on the cannonball produced by the cannon was less than the force of friction on the cannonball, causing the cannonball to slow down and fall to the ground.
O The force on the cannonball produced by the cannon exactly equaled the force of friction on the cannonball, balancing the forces and causing it to fall to the ground.
The correct statement that explains the observation using Newton's laws is objects at rest tend to remain in their current state of motion unless acted upon by an unbalanced force, but objects in motion require a continual application of force to stay in motion. Here option A is correct.
According to Newton's first law of motion, an object will continue moving at a constant velocity in a straight line unless acted upon by an external force. In this case, when the cannonball is shot horizontally from the cannon, it initially possesses a forward velocity due to the force applied by the cannon. However, once the cannonball is in motion, the only forces acting on it are gravity and friction.
Gravity acts vertically downward, causing the cannonball to accelerate downward. Friction acts horizontally in the opposite direction to the motion of the cannonball. As the cannonball moves forward, friction opposes its motion and gradually slows it down.
Since there is no force continuously propelling the cannonball forward, and the forces of friction and gravity act on it, the cannonball eventually comes to a stop and falls to the ground. Hence option A is correct.
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when a parachute opens, the air exerts a large drag force on it. the upward force initially greater than the weight of the sky diver and, thus, slows him down. suppose the weight of the sky diver in 967 n and the drag force has a magnitude of 1041 n. the mass of the sky diver is 98.7 kg. take upward to be the positive direction. what is his acceleration, including sign?
The main answer to the question is as follows: Given,Weight of skydiver, W = 967 NMass of skydiver, m = 98.7 kgDrag force, Fd = 1041 NNow, the skydiver is under two forces,Weight of the skydiver, W = 967 NUpward force, FUp = - 1041 N (opposite in direction to the weight of the skydiver)
Net force acting on the skydiver, F = FUp + W= - 1041 N + 967 N= - 74 NAs acceleration, a = F/mSubstituting the given values, we get,a = -74 N/98.7 kg= -0.749 m/s² :When a parachute opens, the air exerts a large drag force on it. The upward force initially greater than the weight of the sky diver and, thus, slows him down. The skydiver is under two forces: the weight of the skydiver and the upward force. The weight of the skydiver is always directed towards the ground, whereas the upward force on the skydiver is always directed away from the ground.
The net force acting on the skydiver is the sum of the weight of the skydiver and the upward force.The acceleration of the skydiver can be determined as a = F/m, where F is the net force acting on the skydiver and m is the mass of the skydiver. The acceleration of the skydiver is negative because the net force is in the opposite direction to the upward force.
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calculate the magnitude of the force that the charge distribution q exerts on q . express your answer in terms some or all of the variables q , q , a , r , and constant k .
The magnitude of the force that the charge distribution q exerts on q is given by (k * |q * q'|) / a^2, where k is the Coulomb's constant, q and q' are the magnitudes of the charges, and a is the distance between them.
To calculate the magnitude of the force that the charge distribution q exerts on q, we can use Coulomb's Law. Coulomb's Law states that the force between two charged objects is given by the equation:
F = (k * |q1 * q2|) / r^2
where:
F is the magnitude of the force
k is the Coulomb's constant (approximately 9 x 10^9 N m^2/C^2)
q1 and q2 are the magnitudes of the charges
r is the distance between the charges
Let's use the variables q, q', a, and r to represent the given charge distribution and distance:
q1 = q
q2 = q'
r = a
Now we can substitute these values into the formula:
F = (k * |q1 * q2|) / r^2
F = (k * |q * q'|) / a^2
Therefore, the magnitude of the force that the charge distribution q exerts on q is given by (k * |q * q'|) / a^2, where k is the Coulomb's constant, q and q' are the magnitudes of the charges, and a is the distance between them.
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write the balanced chemical equation for the reaction between solutions of acetic acid and sodium hydroxide. include phtsical states in the equation
The balanced chemical equation for the reaction between solutions of acetic acid (CH₃COOH) and sodium hydroxide (NaOH) is:
CH₃COOH (aq) + NaOH (aq) → CH₃COONa (aq) + H₂O (l)
In this equation, (aq) represents an aqueous solution, indicating that the substances are dissolved in water, and (l) represents a liquid state.
Acetic acid (CH₃COOH) is a weak acid, while sodium hydroxide (NaOH) is a strong base. When these two solutions are mixed, they undergo a neutralization reaction. The acetic acid donates a hydrogen ion (H+) to the sodium hydroxide, forming water (H₂O) and a sodium acetate (CH3COONa) salt.
The reaction can be understood in terms of the ionization of acetic acid and the dissociation of sodium hydroxide. In water, acetic acid partially ionizes, releasing hydrogen ions (H+) and acetate ions (CHCOO-):
CH₃COOH (aq) → H+ (aq) + CH₃COO- (aq)
Similarly, sodium hydroxide dissociates completely into sodium ions (Na+) and hydroxide ions (OH-):
NaOH (aq) → Na+ (aq) + OH- (aq)
When these ions combine, the hydrogen ion from acetic acid reacts with the hydroxide ion from sodium hydroxide to form water:
H+ (aq) + OH- (aq) → H₂O (l)
The remaining sodium ion from sodium hydroxide combines with the acetate ion from acetic acid to form sodium acetate:
Na+ (aq) + CH₃COO- (aq) → CH₃COONa (aq)
Overall, the reaction between acetic acid and sodium hydroxide results in the formation of sodium acetate and water. The equation represents a balanced chemical equation, meaning that the number of atoms of each element is the same on both sides of the equation.
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If a bus accelerates from 10 m/s to 15 m/s in 10 seconds with a force of 3000N what is its mass?
Answer:
6000kg
Explanation:
Given parameters:
Initial velocity = 10m/s
Final velocity = 15m/s
Time = 10s
Force = 3000N
Unknown:
Mass of the bus = ?
Solution:
Force is the mass multiplied by acceleration.
Force = mass x acceleration
Acceleration = [tex]\frac{final velocity - initial velocity }{time}[/tex]
Force = mass x [tex]\frac{final velocity - initial velocity }{time}[/tex]
Insert the parameters and solve;
3000 = mass x [tex]\frac{15 - 10}{10}[/tex]
3000 = [tex]\frac{mass}{2}[/tex]
Mass = 6000kg
What is the metric unit of work expressed in base units?
Answer:
Energy, work, quantity of heat
Name ; joule
Expression in terms of SI base units ; [tex]m^2\cdot kg\cdot s^-^2[/tex]
Explanation:
At a regional airport with an elevation of 76 meters below mean sea level, the observed station pressure is 1015 mb.
Using the equations presented in the activity, _____ mb must be _____ the station pressure to correctly determine the sea level pressure value in mb.
a. 10.15 mb, subtracted from
b. 10.15 mb, added to
c. 7.6 mb, subtracted from
d. 7.6 mb, added to
The answer is option c: 7.6 mb must be subtracted from the station pressure to correctly determine the sea level pressure value in mb.
To determine the sea level pressure value, we need to adjust the observed station pressure for the difference in elevation between the station and sea level.
In this case, the regional airport is located 76 meters below mean sea level. The standard atmosphere assumes a decrease of approximately 1 mb for every 10 meters of elevation gain. Therefore, we need to subtract the equivalent pressure decrease caused by the elevation difference.
Given that the observed station pressure is 1015 mb, we can calculate the adjustment by dividing the elevation difference (76 meters) by 10 and multiplying it by the pressure decrease per 10 meters (1 mb). So, (76 / 10) * 1 = 7.6 mb. Therefore, we must subtract 7.6 mb from the observed station pressure of 1015 mb to correctly determine the sea level pressure value.
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Is best defined as the process by which a change in one part of a system results in changes to other parts of the system, which in turn affect the first part.
The answer is feedback.
light of frequency 9.95x1014 hz ejects electrons from the surface of silver. if the maximum kinetic energy of the ejected electrons is 0.180x10-19 j what is the work function of silver?
The work function of silver is 1.24 × 10^(-19) J Frequency of light, v = 9.95 × 10¹⁴ Hz Maximum kinetic energy of the ejected electrons, KE = 0.180 × 10⁻¹⁹ JThe work function is defined as the minimum amount of energy required by an electron to be emitted from the surface of a metal.
Therefore, according work function can be calculated using the following formula:W = hν - KEWhere,h = Planck's constant = 6.626 × 10⁻³⁴ J.sν = frequency of light KE = maximum kinetic energy of the ejected electronsPutting the given values in the above formula, we get:W = hν - KE= 6.626 × 10⁻³⁴ J.s × 9.95 × 10¹⁴ Hz - 0.180 × 10⁻¹⁹ J= 6.60 × 10⁻¹⁹ J - 0.180 × 10⁻¹⁹ J= 6.42 × 10⁻¹⁹ JThus, the work function of silver is 6.42 × 10⁻¹⁹ J.
However, the answer is given in joules but the work function is usually expressed in electron volts(eV) which can be calculated by dividing the above answer by the charge on one electron, i.e., 1.6 × 10⁻¹⁹ C. This gives:W = 6.42 × 10⁻¹⁹ J / 1.6 × 10⁻¹⁹ C= 4.01 eVTherefore, the work function of silver is 1.24 × 10^(-19) J.
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Completion Status: A Click Submit to complete this assessment. estion 4 A 12 in diameter rod is subjected to an axial tensile load of 60 kips. Compute. a. The normal stress developed on an inclined plane at an angle of 25 with the cross section of the rod. b. The maximum normal stress developed in the rod.
The normal stress on an inclined plane at 25 degrees can be calculated using σ = F / A * cos²θ, while the maximum normal stress is σ_max = F / A_max.
a. To calculate the normal stress developed on an inclined plane, we use the formula: σ = F / A * cos²θ. Given that the diameter of the rod is 12 inches, the radius (r) is half of the diameter, which is 6 inches or 0.5 feet. The cross-sectional area of the rod (A) can be calculated using the formula for the area of a circle: A = π * r². Substituting the values, we get A = π * (0.5)^2 = π * 0.25 = 0.7854 square feet.
Now, we can calculate the normal stress using the given axial tensile load (F) of 60 kips and the angle (θ) of 25 degrees. Since the load is given in kips (1 kip = 1000 pounds), we convert it to pounds by multiplying by 1000: F = 60 * 1000 = 60000 pounds.
Using the formula σ = F / A * cos²θ, we substitute the values and calculate the normal stress:
σ = 60000 / 0.7854 * cos²25 ≈ 95317.91 psi (pounds per square inch).
b. The maximum normal stress in the rod occurs when the inclined plane is aligned with the maximum cross-sectional area. In this case, the maximum cross-sectional area (A_max) is the same as the cross-sectional area of a circle, which is π * r². Substituting the radius value, we get A_max = π * (0.5)^2 = π * 0.25 = 0.7854 square feet.
To calculate the maximum normal stress (σ_max), we use the formula σ_max = F / A_max. Substituting the given axial tensile load (F) of 60 kips and the maximum cross-sectional area (A_max), we calculate the maximum normal stress:
σ_max = 60000 / 0.7854 ≈ 76448.44 psi (pounds per square inch).
Therefore, the normal stress developed on an inclined plane at an angle of 25 degrees with the cross section of the rod is approximately 95317.91 psi, and the maximum normal stress developed in the rod is approximately 76448.44 psi.
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Why is rust formed on iron
a student is standing on the 5 yard line of a football field. the student throws a frisbee to the 40 yard line of the football field, and the flight of the frisbee is 4 seconds. what is the speed (yards per second) that the frisbee is traveling? round your answer to the nearest tenth (0.0) and do not include units in the answer.
This 8.8 is the speed (yards per second) that the frisbee is traveling we need to use the formula.
The frisbee traveled from 5 yard line to 40 yard line, which means it traveled a distance of 40 - 5 = 35 yards. And the time taken for the frisbee to travel this distance is 4 seconds. Therefore, we can find the speed of the frisbee by dividing the distance by time.
Speed = Distance / Time = 35 yards / 4 seconds = 8.75 yards per second (rounded to the nearest tenth)Therefore, the is 8.8.
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When we jump on a concrete surface,the feet are more seriously hurt than while jumping on sand.Give reasons.
When we jump on a concrete surface ,the feet are more seriously hurt than while jumping on sand because if we fall on concrete surface there will very less time to make momentum zero , force will hit with greater intensity and if we fall on any soft surface ,enough time will be their to make momentum zero and force will decrease
What is Newton's 2nd law of motion?
It states that the time rate of change of momentum of a body is equal to force imposed on it The body whose mass is constant , newton's 2nd law of motion can be written as F = ma
When we jump on a concrete ,the feet are more seriously hurt than while jumping on sand because if we fall on concrete there will very less time to make momentum zero , due to which force doesn't decrease and hit with greater intensity
but if we fall on any soft surface then it will take long time for us to stop and enough time will be their to make momentum zero , in that time span the intensity of force will decrease as rate ( by newtons 2nd law) has increased and feet won't hurt that much
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Restate Newton’s third law of motion.
Answer:
The force exerted by one source object on another target object always creates another force at the target object that pushes back on the source object with the same ... or His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.
