To calculate the disintegration energies for the given decay processes, we need to determine the mass differences between the initial and final nuclei. The disintegration energy (Q-value) can be calculated using the equation:
Q = (initial mass) - (final mass) - (mass of emitted particle)
Let's calculate the disintegration energies for the three decay processes:
(a) ${^{235}U} → {^{232}Th} + {^{3}He}$:
Q = (235.0429 u) - (232.0381 u) - (3.0160 u) = 0.9888 u
(b) ${^{235}U} → {^{231}Th} + {^{4}He}$:
Q = (235.0429 u) - (231.0363 u) - (4.0026 u) = -0.0032 u
(c) ${^{235}U} → {^{230}Th} + {^{5}He}$:
Q = (235.0429 u) - (230.0331 u) - (5.0122 u) = -0.0036 u
The disintegration energies for processes (a), (b), and (c) are 0.9888 u, -0.0032 u, and -0.0036 u, respectively.
Discussion:
The disintegration energy represents the energy released or absorbed during the decay process. A positive Q-value indicates that energy is released, while a negative Q-value indicates that energy is absorbed. In the case of alpha decay, the emitted alpha particles (helium nuclei) have a relatively high binding energy and are very stable compared to other combinations of nucleons.
From the calculated disintegration energies, we observe that process (a) releases energy (0.9888 u), confirming that the emission of a helium nucleus (alpha particle) contributes to a more stable configuration. On the other hand, processes (b) and (c) have negative Q-values (-0.0032 u and -0.0036 u), indicating that energy must be supplied for these decay processes to occur.
In summary, the preference for alpha particle emission in large radionuclides is due to the alpha particle's stable and tightly bound structure, which results in the release of energy during the decay process. This stability is reflected in the positive Q-value for the decay process (a), where an alpha particle is emitted.
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Full Question -
Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: $$ (a) {^{235}U} → {^{232}Th} + {^{3}He}, \ \ \ (b) {^{235}U} → {^{231}Th} + {^{4}He} $$ $$ (c) {^{235}U} → {^{230}Th} + {^{5}He} $$ The needed atomic masses are $$ ^{232}Th\ \ \ 232.0381 \u \ \ \ \ ^{3}He\ \ \ 3.0160\ u $$ $$ ^{231}Th\ \ \ 231.0363 \u \ \ \ \ ^{4}He\ \ \ 4.0026\ u $$ $$ ^{230}Th\ \ \ 230.0331 \u \ \ \ \ ^{5}He\ \ \ 5.0122\ u $$ $$ ^{235}U\ \ \ 235.0429 \u $$
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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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.
Why is rust formed on iron
Glucose is first broken down in the cytoplasm and then converted to cellular energy in the mitochondria during cellular respiration.
True
False
Answer:
True
Explanation:
Mitochondria is the power-house of almost every cell due to the fact that the cytoplasm stores energy, then sends it to the mitochondria. (at the needed times, of course.)
What are three different types of energy?
Answer:
Gravitational energy
Thermal energy
Mechanical energy
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 2.00 kg body is attached to a psring of negliglible mass and oscillates with a period 1.00s. the force constant of the spring is
Given that the mass of the body attached to the spring is 2.00 kg and the time period of oscillation is 1.00 s.
We can substitute these values in the above equation to obtain:
[tex]1.00 s = 2π√(2.00 kg/k)[/tex]
Squaring both sides, we get: [tex]1.00 s^2 = 4π^2(2.00 kg/k)[/tex]
Simplifying, we get: [tex]k = (4π^2)(2.00 kg)/(1.00 s^2)k = 4π^2 × 2.00 kN/m[/tex]
Therefore, the force constant of the spring is [tex]4π^2 × 2.00 kN/m.[/tex]
This value can be further simplified to 125.66 kN/m.
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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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In 5 seconds a car moving in a straight line increases its speed from 50 km/go to 65 km/h, while a trust goes from rest to 15 km/h in a straight line. Explain how the car and truck have the same acceleration
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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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.
Tension is maintained in a string as in Figure P16.29. The observed wave speed is v = 24.0 m/s when the suspended mass is m = 3.00 kg. (a) What is the mass per unit length of the string? (b) What is the wave speed when the suspended mass is m = 2.00 kg?
The wave speed when the suspended mass is m = 2.00 kg is 31.7 m/s. To solve this problem, we'll use the wave equation:
v = √(T/μ)
where:
v is the wave speed,
T is the tension in the string, and
μ is the mass per unit length of the string.
(a) To find the mass per unit length of the string, we'll rearrange the equation:
μ = T / [tex]v^2[/tex]
Given:
v = 24.0 m/s
m = 3.00 kg
We need to find the tension T. The tension in the string is equal to the weight of the suspended mass, which is given by:
T = m * g
where g is the acceleration due to gravity (approximately 9.8[tex]m/s^2[/tex]).
Substituting the values, we have:
T = (3.00 kg) * (9.8[tex]m/s^2[/tex])
T ≈ 29.4 N
Now, we can calculate the mass per unit length:
μ = (29.4 N) / (24.0[tex]m/s^2[/tex])
μ ≈ 0.06125 kg/m
Therefore, the mass per unit length of the string is approximately 0.06125 kg/m.
(b) To find the wave speed when the suspended mass is m = 2.00 kg, we can use the same equation:
v = √(T/μ)
We already know the tension T from part (a), which is 29.4 N. We can substitute this value and the new mass into the equation:
m = 2.00 kg
v = √((2.00 kg * 9.8 [tex]m/s^2[/tex]) / (0.06125 kg/m))
v ≈ 31.7 m/s
Therefore, the wave speed when the suspended mass is m = 2.00 kg is
31.7 m/s.
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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 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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a toy rocket is shot straight up into the air with an initial speed of 45.0 m/s. how long does it take the rocket to reach its highest point? how high does the rocket rise above the ground?
Answer:
the rocket rises approximately 104.11 meters above the ground.
Explanation:
v = u + at
0 = 45.0 m/s + (-9.8 m/s²) * t
t = -45.0 m/s / -9.8 m/s²
t ≈ 4.59 seconds
To find the height the rocket rises above the ground, we can use the kinematic equation:
s = ut + (1/2) * a * t²
h=ut+(1/2) * g * [tex]t^{2}[/tex]
h = 45.0 m/s * 4.59 s + (1/2) * (-9.8 m/s²) * (4.59 s)²
h ≈ 104.11 meters
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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water is moving at a velocity of 1.30 m/s through a hose with an internal diameter of 4.30 cm. what is the flow rate in l/sec?
The flow rate of water is 0.0056 L/sec (rounded to 3 significant figures). To determine the flow rate in L/sec, we can utilize the concept of continuity equation.
The continuity equation is described as: A₁V₁ = A₂V₂
where A₁ and V₁ are the cross-sectional area and velocity of fluid respectively at the first point, while A₂ and V₂ are the cross-sectional area and velocity of fluid respectively at the second point.
Since we have only one point, we can solve for flow rate using the formula for volume flow rate as:
Q = Av, where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the velocity of fluid.
Substituting the known values in the formula:
We are given that velocity of water, v = 1.30 m/s
and internal diameter of hose, d = 4.30 cm.
To obtain the cross-sectional area, A in m², we need to convert the diameter from centimeter (cm) to meter (m) using the formula:
1 cm = 0.01 m
Substituting the values of d and v:
Thus, the flow rate of water is 0.0056 L/sec (rounded to 3 significant figures).
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two pipes can fill a tank in 10 h and 15 h respectively. however, leakage at the bottom of the tank delays the filling of the tank by 3 h when both the pipes are open simultaneously. how much time would the leak take to empty the full cistern?
The leak will take 120/11 hours to empty the full cistern. When both pipes are open, then the rate of flow of water = (R₁ + R₂) = C/10 hours
Let the capacity of the tank be C. Let the rate of the first pipe be R₁ and the rate of the second pipe be R₂. When both pipes are open, then the rate of flow of water = (R₁ + R₂) = C/10 hours (i)When the first pipe is open for 3 hours, then it fills = 3R₁ volume of water.
Let x be the volume of the tank leaked in 3 hours. Now, the effective rate of the first pipe = R₁ - x/3.
The effective rate of flow of water when both the pipes are open = (R₁ - x/3 + R₂) liters per hour. This effective rate fills the tank in 10 + 3 = 13 hours.
Using the formula of the flow of water, we get: C = (R₁ - x/3 + R₂) * 13 C/10
= (R₁ - x/3 + R₂)13/10
= R1 + R₂ - x/3x/3
= R₁ + R₂ - 13/10C
Now, we know that when the first pipe is open for 3 hours, then it fills 3R₁ volume of water.
Therefore, 3R₁ = C - x... (ii)
From equations (i) and (ii), we get:
C/10 = R₁ + R₂C/15
= R₁ + R₂ - 13/10C + x/3
Adding both equations, we get:
C/10 + C/15
= 2(R₁ + R₂) - 13/10C + x/3C
= 1/6 [(20R₁ + 20R₂) - 13C + 2x]
But we know that 3R₁ = C - x
Therefore, x = C - 3R₁
Therefore, C = 1/6 [(20R₁ + 20R₂) - 13C + 2(C - 3R₁)]
Solving the above equation, we get: C = 120R₁/11
Therefore, the leak will take 120/11 hours to empty the full cistern.
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What is the strength of the electric field Ep1 1.3mm from a proton?
Express your answer to two significant figures and include the appropriate units.
The strength of the electric field Ep1 at a distance of 1.3mm from a proton is 1.05 × 10^14 N/C.
The electric field is defined as the force that one proton exerts on another proton per unit charge. Coulombs or N/C are the units of electric field.
A proton is an atomic particle with a positive charge.
The strength of the electric field is determined by the distance between the charges and the charge itself.
The distance between the electric field and the proton is 1.3 mm.
Proton charges are positive, and the electric field direction is from the positive charges to the negative charges. The electric field intensity is determined by the Coulomb's law.
The formula is as follows;
E= kq/r²Where k = 9 × 10^9 N⋅m²/C² is the Coulomb constant.
q is the electric charge, and r is the distance between the two charges.
Ep1 = k * q/r1^2 = (9 × 10^9 N⋅m²/C²) * (1.6 × 10^-19 C)/(1.3 × 10^-3 m)²= 1.05 × 10^14 N/C
So, the strength of the electric field Ep1 at a distance of 1.3mm from a proton is 1.05 × 10^14 N/C.
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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
Am i the only one that realizes that DONALD j TRUMP is like 45 years older than his wife?? comment down below if u agree with me :)
Answer: I agree and it’s really gross
Explanation:
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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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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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
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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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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.What two horizontal forces are present as a car is pushed down the driveway?
push, velocity
power, friction
power, weight
push, friction
The two horizontal forces that are present as a car is pushed down the driveway are push and friction.
What is force?
A force is a push or pull upon an object that alters the object's state of motion. This occurs as a result of a body interacting with its surroundings. Forces can cause an object to accelerate, decelerate, change direction, or change shape.When an object is in motion, there are two main factors that affect its kinetic energy: the push or force applied to it and the friction it experiences.
What is friction?
Friction is a force that opposes the motion of an object when it comes into contact with another surface.Friction acts in the opposite direction to the object's motion, and it can reduce the object's kinetic energy. When an object moves through a surface with friction, some of its energy is converted into heat and other forms of energy, leading to a decrease in kinetic energy. The amount of friction depends on factors such as the roughness of the surfaces in contact and the normal force between them.What is velocity?
The velocity of an object refers to the rate at which it moves in a particular direction. It can be calculated by determining the distance travelled by an object over time, as well as the direction in which it is moving.Learn more about velocity:
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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
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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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
