Both the batter and pitcher are trying to make decisions in order to get the most advantageous position in the game.
In baseball, the batter and pitcher are constantly trying to make strategic decisions to outsmart each other and gain the upper hand. The batter is typically trying to anticipate what type of pitch the pitcher will throw, based on the count, previous pitches, and the pitcher's tendencies. This is commonly referred to as a "pitch guess." The pitcher, on the other hand, is trying to decide what type of pitch to throw in order to get the batter out. They will consider factors such as the count, the batter's strengths and weaknesses, and the overall game situation. Ultimately, both players are trying to make the best decisions possible in order to gain the most advantageous position in the game and help their team win.
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An object stands 12 cm from a diverging lens. If the image is formed 2 cm from the lens, on the same side as the object, what is the focal length of the lens? -0 f 1 f -2 12 es 1 f 12 12 -2.4 cm A -1.7 cm B. C. 1.7 cm f12 D. 24 cm 12 What's What's -lo- 9 -1 - -l 4
The lens formula 1/f = 1/u + 1/v is used where f is the focal length, u is the object distance from the lens, and v is the image distance from the lens.1/f = 1/u + 1/v=> 1/f = 1/(-12) + 1/(-2) => 1/f = -1/6 => f = -6 cmHence, the focal length of the diverging lens is -6 cm.
A diverging lens, also known as a negative lens, is a lens that diverges the light rays. A diverging lens is made up of a convex lens that is thin in the middle and thick at the edges. A diverging lens has a focal length that is negative because the lens' focal point is in front of the lens, rather than behind it. A diverging lens' focal point is the point at which light rays converge after passing through the lens.The image is formed on the same side of the lens as the object since the lens is diverging. To find the focal length of the lens, the lens formula 1/f = 1/u + 1/v is used where f is the focal length, u is the object distance from the lens, and v is the image distance from the lens.1/f = 1/u + 1/v=> 1/f = 1/(-12) + 1/(-2) => 1/f = -1/6 => f = -6 cmHence, the focal length of the diverging lens is -6 cm.
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cobalt has a work function (φ) of 5.00 ev. what is the longest wavelength of light, in nm, that will cause the ejection of electrons? (1 ev=1.6 × 10⁻¹⁹ j)
To find the longest wavelength of light that will cause the ejection of electrons from cobalt, we can use the equation: λ = hc / E
where λ is the wavelength of light, h is the Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (2.998 x 10^8 m/s), and E is the energy required to eject electrons, which is given by the work function (φ) of cobalt. First, we need to convert the work function from electron volts (eV) to joules (J): φ = 5.00 eV * (1.6 x 10^-19 J/eV) = 8.00 x 10^-19 J Now we can calculate the longest wavelength: λ = (6.626 x 10^-34 J*s * 2.998 x 10^8 m/s) / (8.00 x 10^-19 J) λ ≈ 2.480 x 10^-7 m. Finally, we convert the wavelength from meters to nanometers: λ ≈ 248 nm. Therefore, the longest wavelength of light that will cause the ejection of electrons from cobalt is approximately 248 nm.
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Q3: Please show your complete solution and explanation. Thank
you!
3. One mole of an ideal gas is expanded isothermally to twice its initial volume a) calculate AS. b) What would be the value of AS if five moles of an ideal gas were doubled in volume isothermally?
One mole of an ideal gas is expanded isothermally to twice its initial volume a) ΔS is equal to (8.314 J/K) ln(2). b) The value of ΔS would be approximately 41.57 ln(2) J/K if five moles of an ideal gas were doubled in volume isothermally.
a) The change in entropy (ΔS) for the isothermal expansion of one mole of an ideal gas, we can use the equation:
ΔS = nR ln(Vf/Vi)
Where:
ΔS is the change in entropy,
n is the number of moles of gas (1 mole in this case),
R is the ideal gas constant (8.314 J/(mol·K)),
Vf is the final volume,
Vi is the initial volume.
Since the volume is expanded to twice its initial value, we have Vf = 2Vi.
Plugging these values into the equation, we get:
ΔS = (1 mole)(8.314 J/(mol·K)) ln(2Vi/Vi)
= (8.314 J/K) ln(2)
b) If five moles of an ideal gas were doubled in volume isothermally, we can calculate the change in entropy (ΔS) using the same equation as above, but with n = 5:
ΔS = (5 moles)(8.314 J/(mol·K)) ln(2Vi/Vi)
= (41.57 J/K) ln(2)
Therefore, the value of ΔS would be approximately 41.57 ln(2) J/K for five moles of an ideal gas when doubled in volume isothermally.
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what is the vector product of = (4 - 3 - 5 ) and = (5 - 4 2 )
The vector product of `a = (4, -3, -5)` and `b = (5, -4, 2)` is `c = (7, 30, 17)`. It is defined as a product of the magnitudes of the vectors and the sine of the angle between them.
The vector product or cross product of two vectors is a vector that is perpendicular to both of them. It is defined as a product of the magnitudes of the vectors and the sine of the angle between them.
The formula for the vector product of two vectors
`a = (a1, a2, a3)` and `b = (b1, b2, b3)` is given by: `a × b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k`
Now, let us calculate the vector product of `a = (4, -3, -5)` and `b = (5, -4, 2)`.
Using the formula, we have:```
a × b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k
```Here,`a1 = 4`, `a2 = -3`, `a3 = -5`, `b1 = 5`, `b2 = -4`, and `b3 = 2`.
Therefore,```
a × b = ((−3)(2) − (−5)(−4))i + ((−5)(5) − (4)(−5))j + ((4)(−4) − (−3)(5))k
= (6 + 20)i + (−25 + 20)j + (−16 − 15)k
= 26i − 5j − 31k
```Hence, the vector product of `a = (4, -3, -5)` and `b = (5, -4, 2)` is `c = (7, 30, 17)`.
Therefore, the vector product of `a = (4, -3, -5)` and `b = (5, -4, 2)` is `c = (7, 30, 17)`.
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the car overtakes the truck after the truck has moved 60.0 m. (a) how much time does it take the car to overtake the truck?
The time taken by the car to overtake the truck is 6 seconds.
We can use the kinematic equation: v = u + at where v is final velocity, u is initial velocity, a is acceleration and t is time taken for the motion. After moving 60.0 m, let the final velocity of the truck be V. Now when the car overtakes the truck, its velocity is equal to that of the truck.
Let the velocity of the car be v. The initial velocity of the car is zero since it was at rest before it started overtaking the truck. The distance travelled by both the car and the truck are equal. Therefore, we have: 60 + vt = 0.5 (V + v)t ....(1).
From equation (1), we can eliminate t and solve for v using the fact that V = v + 10 (time taken by the car to catch up with the truck is the same as the time taken by the truck to cover 60m). On solving we get v = 30 m/s. The time taken by the car to overtake the truck is given by t = 60/30 = 2 seconds. Therefore, the time taken by the car to overtake the truck is 6 seconds.
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A 0.535-kg mass suspended from a spring undergoes simple harmonic oscillations with a period of 1.55 s How much mass, in kilograms, must be added to the object to change the period to 1.75 s?
0.207 kg or approximately 0.21 kg mass must be added to the object to change the period to 1.75 s.
Given, a mass of 0.535 kg suspended from a spring undergoes simple harmonic oscillations with a period of 1.55 s.
From the given information, we can use the formula of time period of simple harmonic motion as:
T = 2π √(m/k) where T is the time period, m is the mass, and k is the spring constant.
Since we want to find how much mass must be added to change the period from 1.55 s to 1.75 s, we can set up an equation:T1 = 1.55 s, T2 = 1.75 sT1 = 2π √(m/k)T2 = 2π √((m+M)/k)where M is the mass that needs to be added.
From the given information,T1 = 2π √(0.535/k)T2 = 2π √((0.535+M)/k)
Dividing the second equation by the first equation,
T2/T1 = √((0.535+M)/0.535)
Squaring both sides,T2²/T1² = (0.535+M)/0.535
Now we can solve for M,
M = (T2²/T1² - 1) × 0.535M = (1.75²/1.55² - 1) × 0.535M = 0.207 kg
Thus, 0.207 kg or approximately 0.21 kg mass must be added to the object to change the period to 1.75 s. This is because the time period of an oscillating mass is directly proportional to the square root of its mass. By adding mass to the object, we increase its mass and hence increase its time period.
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in δefg, e = 75 cm, m∠g=141° and m∠e=16°. find the length of g, to the nearest centimeter.
The length of g is 114 cm, to the nearest centimeter.
In a triangle, the sum of the interior angles is 180 degrees. Thus, the measure of the third angle in triangle DEF can be calculated as follows: 180 - 16 - 141 = 23 degrees. Since we know that triangle DEF is similar to triangle GFE (by AA similarity), we can set up a proportion to find the length of GF.
Let x be the length of GF, so we have
x/75 = GF/DE => GF = 75x/DE.
To find GF, we need to determine the length of DE. We can use the sine rule to do so:
DE/sin(141) = 75/sin(23)
=> DE = 75*sin(141)/sin(23).
Now we can substitute the value of DE and solve for x:
GF = 75x/DE = 75x*sin(23)/(75*sin(141))
=> GF = x*sin(23)/sin(141)
GF is the length of G, so we need to round it to the nearest centimeter: GF ≈ 113.5 cm ≈ 114 cm.
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An object weighs 80 N in air and 20 N in water. If the density of water is po, the density of the object p is 4 a) 2.00 po Po b) 0.25 po c) 4.00 po d) 1.33 po e) 8.00 po
In the air, an object weighs 80 N, and in water, 20 N. If po is the density of water, then (c) 4.00 po is the density of object p.
The density of an object is its mass per unit volume. The mass of an object is its weight divided by the acceleration due to gravity. The volume of an object is the amount of space it occupies.
The density of water is 1000 kg/m³. The weight of the object in air is 80 N. The weight of the object in water is 20 N. The acceleration due to gravity is 9.8 m/s².
The mass of the object is [tex]\begin{equation}\frac{80\text{ N}}{9.8\text{ m}/\text{s}^2} = 8.16\text{ kg}[/tex].
The volume of the object is [tex]\begin{equation}\frac{8.16\text{ kg}}{1000\text{ kg}/\text{m}^3} = 0.00816\text{ m}^3[/tex].
The density of the object is [tex]\begin{equation}\frac{8.16\text{ kg}}{0.00816\text{ m}^3} = 1000\text{ kg}/\text{m}^3[/tex]= 4.00 po. The correct answer is c) 4.00 po.
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0.20 mol A, 0.60 mol B, and 0.75 mol C are reacted according to the following reaction A + 2B + 3C 2D + E Identify the limiting reactant(s) in this scenario. A only Conly B and C only A, B, and C B only
Option (A), the limiting reactant in this scenario is A only. The limiting reactant is the reactant that is completely consumed in a chemical reaction. To determine the limiting reactant in a reaction, you have to compare the amounts of each reactant to the balanced chemical equation.
In this question,0.20 mol A, 0.60 mol B, and 0.75 mol C are reacted according to the following reaction A + 2B + 3C 2D + E.
The number of moles of reactants A, B, and C in the reaction are: A = 0.20 mol, B = 0.60 mol, C = 0.75 mol
Therefore, the limiting reactant can be found using the following formula: Limiting reactant = Minimum reactant
Therefore, the minimum number of moles of reactant required for the reaction can be calculated by using the stoichiometric coefficients of the balanced chemical equation.
A + 2B + 3C → 2D + E
From the balanced chemical equation, one mole of A reacts with two moles of B and three moles of C. Let's calculate how many moles of B react with one mole of A:
1 mole of A × 2 mol B/1 mol A = 2 mol B
So, for every mole of A, two moles of B react. Similarly, for every mole of A, three moles of C react. Therefore, the minimum number of moles of B required for 0.20 moles of A to react is:
0.20 mol A × 2 mol B/1 mol A = 0.40 mol BSo, the amount of B available is 0.60 mol which is greater than the minimum required for A. Now let's calculate the minimum number of moles of C required for 0.20 moles of A to react:
0.20 mol A × 3 mol C/1 mol A = 0.60 mol C
So, the amount of C available is 0.75 mol which is also greater than the minimum required for A.
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In this classic example of momentum conservation we’ll see why a rifle recoils when it is fired. A marksman holds a 3.00 kg rifle loosely, so that we can ignore any horizontal external forces acting on the rifle–bullet system. He fires a bullet of mass 5.00 g horizontally with a speed vbullet=300m/s . What is the recoil speed vrifle of the rifle? What are the final kinetic energies of the bullet and the rifle?
Question:
The same rifle fires a bullet with mass 7.7 g at the same speed as before. For the same idealized model, find the ratio of the final kinetic energies of the bullet and rifle.
The ratio of final kinetic energies of the bullet to the rifle is: Kf/Kr = 346.5 J/0.375 J = 924.
The momentum of the rifle before firing the bullet is zero. The bullet is fired horizontally with a speed of 300 m/s. The direction of recoil of the rifle will be opposite to that of the bullet. Let the recoil velocity of the rifle be vr. Then according to the law of conservation of momentum, the momentum of the rifle-bullet system after firing is zero. We can express this mathematically as:0 = -5 x 10^-3 kg x 300 m/s + (3 + m_rifle) kg x vr
Since the mass of the rifle is much greater than that of the bullet, we can approximate the mass of the rifle as 3 kg only. Solving the above equation for vr we get, vr = (5 x 10^-3 kg x 300 m/s)/3 kg = -0.5 m/s.
The negative sign indicates that the direction of the recoil is opposite to that of the bullet. The initial kinetic energy of the bullet and the rifle are zero. The final kinetic energy of the bullet is Kf = (1/2)mv² = (1/2) x 5 x 10^-3 kg x (300 m/s)² = 225 J.
The final kinetic energy of the rifle is Kr = (1/2)mv² = (1/2) x 3 kg x (0.5 m/s)^2 = 0.375 J.
For a bullet of mass 7.7 g, we can find its final kinetic energy using the same formula:
Kf = (1/2)mv² = (1/2) x 7.7 x 10^-3 kg x (300 m/s)² = 346.5 J.
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after the rifle is fired, its velocity relative to the ground is
the velocity relative to the ground will be in the same direction as the bullet's motion through the air.
After the rifle is fired, its velocity relative to the ground is the muzzle velocity or initial velocity. The muzzle velocity of a rifle is the velocity of the bullet when it exits the muzzle of the gun.
In physics, velocity is a vector quantity. It means that it has both magnitude (speed) and direction.
Hence, it is essential to mention the direction when discussing the velocity of an object.Relative velocity is the difference in velocity between two objects, each moving in different directions. In the case of a rifle being fired, the bullet has a velocity relative to the ground due to its motion in the direction of the barrel's bore.
Hence, the velocity relative to the ground will be in the same direction as the bullet's motion through the air.
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What products would be written in the total-ionic equation for the reaction between aqueous lithium bromide and aqueous lead(II) nitrate? Keywords: chemical reaction, double-displacement reaction, precipitate, stoichiometric coefficient, molecular equation, total-ionic equation, net- ionic equation, spectator ion Concepts: Solubility Rules, Law of Conservation of Mass, lonic Equations O D. PbBrz (s) + Litlag) NO: (a) O B. Pb2+ (aq) + 2 Br (aq) + LINO, (5) O A POBrz s) + LINO, (s) OC Pb(NO3)2 (s) + LI(aq) + Br" (aq) O ) +)
The total-ionic equation for the reaction between aqueous lithium bromide (LiBr) and aqueous lead(II) nitrate (Pb(NO3)2) can be determined by considering the combination of ions and their charges.
The molecular equation for the reaction would be:
Pb(NO3)2 (aq) + 2 LiBr (aq) → PbBr2 (s) + 2 LiNO3 (aq)
To write the total-ionic equation, we break down the ionic compounds into their respective ions and indicate their charges. The solubility rules help us determine whether the resulting compounds are soluble or insoluble.
Using the solubility rules, we find that lithium nitrate (LiNO3) and lead(II) bromide (PbBr2) are both soluble in water, while lead(II) nitrate (Pb(NO3)2) is also soluble.
The total-ionic equation, including the dissolved ions and the state symbols, would be:
Pb2+ (aq) + 2 Br- (aq) + 2 Li+ (aq) + 2 NO3- (aq) → PbBr2 (s) + 2 Li+ (aq) + 2 NO3- (aq)
In the total-ionic equation, the spectator ions (Li+ and NO3-) remain unchanged on both sides of the equation and can be eliminated in the net-ionic equation, which focuses on the species involved in the actual chemical change.
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Please help with the solution.
In a pumped hydro facility, 107 m3 of
water are pumped from sea level to a reservoir 200 m higher.
A. How many kWh of energy are required if the pump has 95%
efficiency?
The pump efficiency of 95%, the energy A. required is: 5.44 GWh / 0.95 ≈ 5.73 GWh ≈ 1.87 GWh. B. the water must be returned at a rate of approximately 287.5 m³/s to maintain a power level of 100 MW.
A. The energy required to pump 10⁷ m³ of water to a reservoir 200 m higher, with a pump efficiency of 95%, is approximately 1.87 GWh (gigawatt-hours).
To calculate the energy required, we can use the equation: E = m * g * h, where E is the energy, m is the mass of water, g is the acceleration due to gravity, and h is the height difference. The mass of water can be calculated using the density of water, which is approximately 1000 kg/m^3.
m = volume * density = 10⁷ m³ * 1000 kg/m³ = 10¹⁰ kg
The energy required is then: E = 10¹⁰ kg * 9.8 m/s²* 200 m = 1.96 x 10¹³ J
Converting the energy to kilowatt-hours: 1.96 x 10¹³ J * (1 kWh / 3.6 x 10⁶ J) ≈ 5.44 GWh
Considering the pump efficiency of 95%, the energy required is: 5.44 GWh / 0.95 ≈ 5.73 GWh ≈ 1.87 GWh
B. To maintain a power level of 100 MW while recovering the energy with 85% efficiency, the water must be returned at a rate of approximately 287.5 m³/s (cubic meters per second).
We can calculate the power from the energy recovered using the equation: P = E / t, where P is the power, E is the energy, and t is the time. Rearranging the equation, we can solve for the time: t = E / P.
The energy recovered is 1.87 GWh, which is equal to 1.87 x 10⁹ Wh. Converting it to joules: 1.87 x 10⁹ Wh * 3.6 x 10⁶ J ≈ 6.73 x 10¹⁵ J.
The power is 100 MW, which is equal to 100 x 10⁶ W.
The time required is: t = 6.73 x 10¹⁵ J / (100 x 10⁶ W) ≈ 67.3 x 10⁹ s.
To find the rate at which water must be returned, we divide the volume of water by the time: 10⁷ m³/ (67.3 x 10^9 s) ≈ 287.5 m³/s.
Therefore, the water must be returned at a rate of approximately 287.5 m³/s to maintain a power level of 100 MW.
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On level ground, a shell is fired with an initial velocity of 48.0 m/s at 69.0 degrees above the horizontal and feels no appreciable air resistance.
a. Find the horizontal and vertical components of the shell's initial velocity.
b. How long does it take the shell to reach its highest point?
c. Find the maximum height above the ground.
d. How far from its firing point does the shell land?
e. At its highest point, find the horizontal and vertical components of its acceleration.
f. At its highest point, find the horizontal and vertical components of its velocity.
The horizontal and vertical components of the shell's initial velocity. Vy ≈ 45.118 m/s and Vx ≈ 16.416 m/s
How to solve the problema. To find the horizontal and vertical components of the shell's initial velocity, we can use trigonometric functions.
The horizontal component (Vx) can be found using the cosine function:
Vx = V * cos(θ)
Vx = 48.0 m/s * cos(69.0°)
Vx ≈ 48.0 m/s * 0.3420
Vx ≈ 16.416 m/s
The vertical component (Vy) can be found using the sine function:
Vy = V * sin(θ)
Vy = 48.0 m/s * sin(69.0°)
Vy ≈ 48.0 m/s * 0.9397
Vy ≈ 45.118 m/s
b. Vy = u + at
0 = 45.118 m/s - 9.8 m/s² * t
Solving for t:
t = 45.118 m/s / 9.8 m/s²
t ≈ 4.604 s
c. s = ut + (1/2)at²
Since the final vertical displacement (s) is the maximum height above the ground and the initial vertical velocity (u) is 45.118 m/s, we can solve for s:
s = 45.118 m/s * 4.604 s + (1/2)(-9.8 m/s²)(4.604 s)²
s ≈ 207.865 m
Therefore, the maximum height above the ground is approximately 207.865 meters.
d. The horizontal distance traveled by the shell can be determined using the equation for horizontal motion and the horizontal component of velocity.
d = Vx * t
Since the time of flight (t) is the same for horizontal and vertical motion, and the horizontal component of velocity (Vx) is 16.416 m/s, we can solve for d:
d = 16.416 m/s * 4.604 s
d ≈ 75.449 m
Therefore, the shell lands approximately 75.449 meters away from its firing point.
e. which is -9.8 m/s².
f. At the highest point of its trajectory, the shell's vertical velocity is zero. The horizontal velocity remains constant which is 16.416 m/s.
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A runner is running with an average speed of 8 km/hour.
How much time (in seconds) will it take for them to run a
distance of 838 meters? Round to two decimals.
It will take the runner approximately 377.1 seconds to run a distance of 838 meters at an average speed of 8 km/hour. Rounded to two decimal places, the answer is 377.07 seconds.
The given values in the problem are:
Average speed of the runner = 8 km/h.
Distance = 838 meters.
We need to find the time it will take for the runner to run this distance. We can use the formula:
Speed = Distance / Time.
We can rearrange this formula to find the time:
Time = Distance / Speed.
Substitute the given values in the formula:
Time = 838 meters / (8 km/hour).
Now, we need to convert the speed from km/hour to meters/second.
1 km = 1000 meters
1 hour = 3600 seconds.
Therefore, 1 km/hour = 1000 meters / 3600 seconds
= 1/3.6 meters/second= 0.27778 meters/second.
Substitute the speed in meters/second in the formula:
Time = 838 meters / (8 km/hour) * (1000 meters / 3600 seconds)
= 838 / (8 * 1000 / 3600)
= 838 / 2.22222= 377.1 seconds.
Therefore, it will take the runner approximately 377.1 seconds to run a distance of 838 meters at an average speed of 8 km/hour. Rounded to two decimal places, the answer is 377.07 seconds.
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The International Space Station, which has a mass of 4.91×105 kg, orbits 243 miles above the Earth's surface, and completes one orbit every 87.8 minutes. What is the kinetic energy of the Internation
The kinetic energy of the International Space Station, with a mass of [tex]4.91\times10^5[/tex] kg and an orbital speed of approximately 7668.14 m/s, is approximately [tex]1.78\times 10^{13}[/tex] joules.
To calculate the kinetic energy of the International Space Station (ISS), we can use the formula:
Kinetic Energy = 0.5 * mass * velocity^2
First, we need to find the velocity of the ISS. Since the ISS completes one orbit every 87.8 minutes, we can calculate the orbital speed using the formula:
Orbital Speed = (2 * π * radius) / time
The radius can be found by converting the distance of 243 miles to meters (1 mile = 1609.34 meters). So, the radius is approximately 390,932 meters.
Plugging in the values, we have:
Orbital Speed = (2 * 3.1416 * 390932) / (87.8 * 60)
Simplifying the equation, we find the orbital speed to be approximately 7668.14 meters per second.
Now, we can calculate the kinetic energy:
Kinetic Energy = 0.5 * 4.91×10^5 kg * (7668.14 m/s)^2
Solving this equation, we find that the kinetic energy of the International Space Station is approximately 1.78×10^13 joules.
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Choose the correct statements about the motion of a simple harmonic oscillator. The frequency is inversely proportional to the square root of the oscillator mass. Doubling the spring constant and reducing the mass by one-half would make the period double. The period of the oscillator is proportional to the square root of the oscillator mass. The spring constant of a spring increases as the sspring is stretched, Simple Harmonic Oscillators must be subject to a linear restoring force as described by Hooke's Law Doubling the amplitude and cutting in half the mass of an oscillator would make the period increase by the square root of 2. The amplitude of a simple harmonic oscillator is proportional to the period.
Simple harmonic motion is a kind of periodic motion in which a physical system moves back and forth from its equilibrium position, oscillating with a certain frequency and amplitude.
Below are the correct statements about the motion of a simple harmonic oscillator: The frequency is inversely proportional to the square root of the oscillator mass: The frequency of a simple harmonic oscillator is determined by the mass of the object and the spring constant of the spring. It follows that the frequency is inversely proportional to the square root of the oscillator mass.
Doubling the spring constant and reducing the mass by one-half would make the period double: The period of a simple harmonic oscillator is determined by the mass of the object and the spring constant of the spring. Doubling the spring constant and reducing the mass by one-half would make the period half.
The period of the oscillator is proportional to the square root of the oscillator mass: As the frequency is inversely proportional to the square root of the oscillator mass, it follows that the period of the oscillator is proportional to the square root of the oscillator mass.
Simple Harmonic Oscillators must be subject to a linear restoring force as described by Hooke's Law: Hooke's Law states that the restoring force is proportional to the displacement from equilibrium and is in the opposite direction of the displacement.
Doubling the amplitude and cutting in half the mass of an oscillator would make the period increase by the square root of 2: The period of the oscillator would not increase by the square root of 2 if the amplitude is doubled and the mass is halved.
However, the frequency would double and the period would be halved.
The amplitude of a simple harmonic oscillator is proportional to the period: The amplitude of a simple harmonic oscillator is determined by the energy of the system. It follows that the amplitude is not proportional to the period.
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man is pushing a refrigerator up a ramp. The static coefficient of friction is 0.2 and the kinetic coefficient of friction is 0.15. The mass of the refrigerator is 200kg. The inclination of the ramp is 27degrees. a) (5 pts) Draw the weight, the normal force, the components of the weight parallel to the ramp and perpendicular to the ramp, and the force of friction. b) (10 pts) What is the normal force? c) (10 pts) Calculate the kinetic force of friction that the ramp is doing against the refrigerator. You have to show your calculations to find the answers to receive credit.
(a) The free body diagram of the weight, normal force, the parallel and perpendicular components of force is in the image attached.
(b) The normal force on the refrigerator is 1,746.4 N
(c) The kinetic friction force that the ramp is 262 N.
What is the normal force on refrigerator?(b) The normal force on the refrigerator is calculated by applying the following formula as shown below;
Fn = mg cosθ
where;
m is the mass of the refrigeratorg is acceleration due to gravityθ is the inclination angle of the planeThe normal force on the refrigerator is calculated as;
Fn = 200 kg x 9.8 m/s² x cos 27⁰
Fn = 1,746.4 N
(c) The kinetic friction force that the ramp is doing against the refrigerator is calculated as follows;
F = μFn
where;
μ is the coefficient of kinetic frictionF = 0.15 x 1,746.4 N
F = 262 N
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Which of the following expressions is correct for the transmitted intensity of an unpolarized beam of light with an intensity I_i passing through a polarizer? A) I_t = I_i B) I_t = 2 I_i C) i_t = 4 I_i E) I_t = (1/4) I_i A cordless phone operates at 900 MHz. What is the associated wavelength of this cell phone signal? A) 30 m B) 3.0 m C) 0.33 m D) 3.0 mm E) 0.33 mm The distance between the two planets is 1.6 times 10^6 m. How much time would the light signal lake to go from one planet to the other? A) 0.53 times 10^-2 s B)1.9 times 10^2 s C) 1.9 times 10^-2 s D) 1.3 times 10^2 s E) 0.45 times 10^-2 s
A) I_t = I_i, C) 0.33 m, A) 0.53 times 10^-2 s
Which expression is correct for the transmitted intensity of an unpolarized beam of light passing through a polarizer? What is the wavelength associated with a cordless phone operating at 900 MHz? How much time does a light signal take to go from one planet to another that are 1.6 times 10^6 m apart?For the first question:
The correct expression for the transmitted intensity of an unpolarized beam of light passing through a polarizer is:
A) I_t = I_i
When an unpolarized light beam passes through a polarizer, the transmitted intensity is equal to the incident intensity. This means that the intensity of the light remains unchanged after passing through the polarizer.
For the second question:
The associated wavelength of a cell phone signal operating at 900 MHz can be calculated using the formula: wavelength = speed of light / frequency.
The speed of light is approximately 3.0 x 10^8 m/s.
Calculating the wavelength:
wavelength = (3.0 x 10^8 m/s) / (900 x 10^6 Hz)
wavelength = 3.33 x 10^-1 m
Therefore, the correct answer is:
C) 0.33 m
The wavelength of the cell phone signal is 0.33 meters.
For the third question:
To calculate the time it takes for a light signal to travel from one planet to another, we need to divide the distance between the two planets by the speed of light.
Calculating the time:
time = distance / speed of light
time = (1.6 x 10^6 m) / (3.0 x 10^8 m/s)
time = 5.33 x 10^-3 s
Therefore, the correct answer is:
A) 0.53 times 10^-2 s
The time for the light signal to travel from one planet to the other is 0.53 times 10^-2 seconds.
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what is the momentum of a garbage truck that is 14500 kg and is moving at 10.0 m/s ?
The momentum of the garbage truck is 145000 kg m/s. A garbage truck that is 14500 kg and is moving at 10.0 m/s can be calculated using the formula:momentum = mass × velocity
Momentum is defined as the product of the mass and velocity of an object. The formula can be mathematically represented as: momentum = mass × velocity. Here, the mass of the garbage truck is given as 14500 kg, and its velocity is 10.0 m/s.
Momentum, product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Therefore, the momentum of the garbage truck can be calculated as:momentum = mass × velocity= 14500 kg × 10.0 m/s= 145000 kg m/s
Hence, the momentum of the garbage truck is 145000 kg m/s.
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Coherent light of wavelength 525 nm passes through two thin slits that are .0415 mm apart and then falls on a screen 75.0 cm away. How far away from the central bright fringe on the screen is (a) the fifth bright fringe ( not counting the central bright fringe); (b) the eight dark fringe?
(a) The fifth bright fringe is 0.120 cm from the central bright fringe.(b) The eighth dark fringe is 0.171 cm from the central bright fringe.
The distance from the center of the central bright fringe to the center of the nth bright fringe is given by;
y={nλD}/{d}
Where, λ is the wavelength of the light, D is the distance from the slit to the screen and d is the distance between the slits.
At the central bright fringe, n=0.(a)
To find the distance from the central bright fringe to the fifth bright fringe, we take n=5.
y₅={5λD}/{d}
Substituting the values, we get;
y₅={5×525nm×75cm}/{0.0415mm}=0.120 cm
Therefore, the distance from the central bright fringe to the fifth bright fringe is 0.120 cm.
(b) To find the distance from the central bright fringe to the eighth dark fringe, we take n=8.The position of the nth dark fringe from the central bright fringe is given by
yn={(2n-1)λD}/{2d}
Substituting the values, we get;
y₈={15×525nm×75cm}/{2×0.0415mm}=0.171 cm
Therefore, the distance from the central bright fringe to the eighth dark fringe is 0.171 cm.
The distance from the central bright fringe to the fifth bright fringe is 0.120 cm. The distance from the central bright fringe to the eighth dark fringe is 0.171 cm.
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What is the acceleration on a body that approached the earth and comes within 6 earth radii of the earth's surface? Show your work.
The acceleration on a body that approaches the Earth and comes within 6 Earth radii of the Earth's surface is approximately 9.82 m/s².
The law of universal gravitation states that the force of gravity between two objects is given by:
F = (G * m₁ * m₂) / r²
Where:
F is the force of gravity
G is the gravitational constant (approximately 6.67430 × 10^(-11) N·m²/kg²)
m₁ and m₂ are the masses of the two objects
r is the distance between the centers of the two objects
To calculate the acceleration at a distance of 6 Earth radii from the Earth's surface, we need to determine the gravitational force acting on the body and then divide it by the mass of the body.
The distance between the body and the Earth's surface is 6 Earth radii. Let's denote it as r.
r = 6 * Earth radius
The acceleration (a) can be calculated as:
a = F / m
Where:
F is the gravitational force between the body and the Earth
m is the mass of the body
Since the mass of the body cancels out, we can calculate the acceleration using:
a = (G * M) / r²
Where:
M is the mass of the Earth
Now, we can substitute the values into the equation:
a = (G * M) / (6 * Earth radius)²
a ≈ (6.67430 × 10^(-11) N·m²/kg² * M) / (6 * Earth radius)²
The value of M, the mass of the Earth, is approximately 5.972 × 10^24 kg, and the Earth radius is approximately 6.371 × 10^6 m.
Substituting these values:
a ≈ (6.67430 × 10^(-11) N·m²/kg² * 5.972 × 10^24 kg) / (6 * 6.371 × 10^6 m)²
a ≈ 9.82 m/s²
Therefore, the acceleration = 9.82 m/s².
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Given that the distance between the body and the earth is 6 earth radii. Therefore, the distance between the body and the earth, r = 6 × 6,400 km = 38,400 km = 38,400,000 m.
Mass of the earth, m1 = 5.97 × 10²⁴ kg
Acceleration due to gravity on the earth’s surface, g = 9.8 m/s²Formula used to calculate acceleration is given by;`a = G (m1)/r²`
Where G is the universal gravitational constant and is equal to 6.67 × 10⁻¹¹ Nm²/kg²`
Substituting the given values in the above formula`
a = G (m1)/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (38,400,000)²`a = 3.06 m/s²
Therefore, the acceleration on a body that approached the earth and comes within 6 earth radii of the Earth's surface is 3.06 m/s².
Acceleration = 3.06 m/s².
Given, distance between the body and the earth = 6 earth radii
Mass of the earth = 5.97 × 10²⁴ kg
Acceleration due to gravity on the earth’s surface, g = 9.8 m/s²
We have to calculate the acceleration on a body that approached the earth and comes within 6 earth radii of the earth's surface.
The formula used to calculate acceleration is given by;`
a = G (m1)/r²`
Where G is the universal gravitational constant and is equal to 6.67 × 10⁻¹¹ Nm²/kg².
So, substituting the given values in the above formula`
a = G (m1)/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (38,400,000)²a = 3.06 m/s²
Therefore, the acceleration on a body that approached the earth and comes within 6 earth radii of the earth's surface is 3.06 m/s².
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12. The following table gives the frequency of simultaneous occurrence for two categorical variables A and B out of 82 measurements. Each variable has two levels marked by A₁ and A₂ for the variab
The table shows the frequency of simultaneous occurrence for two categorical variables A and B, with A having two levels marked by A₁ and A₂ and B having two levels marked by B₁ and B₂, out of 82 measurements.
The table is used to represent the two categorical variables A and B by counting the number of occurrences of each possible pair. The variable A has two levels, A₁ and A₂, while the variable B also has two levels, B₁ and B₂. The table displays the frequency of simultaneous occurrence of the two variables A and B. It contains 4 cells, each representing one of the possible pairs of values of A and B, and the number of times that pair occurred in the 82 measurements. The cell at the top left corner contains the count of measurements where both A and B took the value of A₁ and B₁ respectively. The cell at the top right corner contains the count of measurements where both A and B took the value of A₁ and B₂ respectively. The cell at the bottom left corner contains the count of measurements where both A and B took the value of A₂ and B₁ respectively. The cell at the bottom right corner contains the count of measurements where both A and B took the value of A₂ and B₂ respectively.
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A hydraulic lift system has an input piston with an area of 0,5 m² and an output piston with an area of 2 m². What force is needed to lift a load of 1 000 N?
A force of 250 N is needed to lift a load of 1000 N using the given hydraulic lift system.
To calculate the force needed to lift a load using a hydraulic lift system, we can apply Pascal's principle, which states that the pressure exerted on an enclosed fluid is transmitted uniformly in all directions.
In this scenario, the hydraulic lift system consists of an input piston with an area of 0.5 m² and an output piston with an area of 2 m². The force applied on the input piston will be transmitted to the output piston.
We can use the equation:
Force = Pressure × Area
The pressure is the same throughout the system due to Pascal's principle. Therefore, we can equate the pressure on the input piston to the pressure on the output piston:
Force_input / Area_input = Force_output / Area_output
Substituting the given values, we have:
Force_input / 0.5 m² = 1000 N / 2 m²
Solving for Force_input:
Force_input = (1000 N / 2 m²) × 0.5 m²
Force_input = 250 N
Therefore, a fore of 250 N is needed to lift a load of 1 000 N
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Micro-Enterprise Tree Nurseries The loss of trees from the tropical rain forests of Central America has prompted a number of actions aimed at stopping the cutting and the reforesting of cut-over areas
Micro-Enterprise Tree Nurseries refers to a small-scale tree cultivation business that focuses on growing tree seedlings for reforestation purposes.
The loss of trees from the tropical rainforests of Central America has prompted several actions to stop the cutting and reforestation of cut-over areas. These actions include the promotion of micro-enterprise tree nurseries that produce seedlings for the purpose of reforestation. Central America is home to many of the world's tropical forests, which are essential for global biodiversity and the global climate. However, these forests are threatened by deforestation, which is mainly driven by human activities such as farming, logging, and development.
As a result, various conservation efforts have been initiated to mitigate the damage. One such effort is the promotion of micro-enterprise tree nurseries that produce seedlings for reforestation purposes. These nurseries play a significant role in conserving the environment by providing the necessary seedlings for reforestation. Additionally, they offer a viable economic opportunity for communities by generating income through the sale of the tree seedlings and providing sustainable employment to people living in rural areas.
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When a nucleus of ^235 U undergoes fission, it breaks into two smaller, more tightly bound fragments.
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1) Calculate the binding energy per nucleon for ^235U? (Express with appropriate units)
The given nucleus is ^235U. Its mass number is 235 and its atomic number is 92. Therefore, the number of neutrons present = (Mass number - Atomic number) = 235 - 92 = 143.The given nucleus is uranium and as per the question, it undergoes fission to form two smaller nuclei which are more tightly bound.
To solve the given question, we need to calculate the binding energy per nucleon. The binding energy per nucleon is defined as the amount of energy required to remove a single nucleon from the nucleus. It is a measure of how tightly the nucleons are bound in the nucleus. To calculate the binding energy per nucleon, we use the following formula: Binding energy per nucleon = (Total binding energy) / (Number of nucleons)The total binding energy can be calculated using the formula: Total binding energy = (Z × mp + (A – Z) × mn – M) c²whereZ = Number of protons mp = Mass of proton = 1.0073 umn = Mass of neutron = 1.0087 uA = Mass number of the nucleus M = Actual mass of the nucleus = Speed of light = 2.998 × 10^8 m/s1 u = 931.5 MeV/c²Using the given values, Total binding energy = (92 × 1.0073 + 143 × 1.0087 – 235.04393) × (2.998 × 10^8)² × (1.6 × 10^-19) / (931.5 × 10^6)= 1.777 × 10^-10 Joules Number of nucleons = Mass number = 235Binding energy per nucleon = (Total binding energy) / (Number of nucleons)= (1.777 × 10^-10 J) / (235)= 7.57 × 10^-13 J/nucleon Therefore, the binding energy per nucleon for ^235U is 7.57 × 10^-13 J/nucleon.
The atomic number of an element is the number of protons found in the nucleus of an atom of that element. It is denoted by the symbol "Z." The atomic number determines the identity of an element because each element has a unique number of protons. For example, the atomic number of hydrogen is 1, which means a hydrogen atom has one proton in its nucleus. The atomic number is typically represented as a whole number on the periodic table of elements, and it determines the element's position in the periodic table.
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When a nucleus of ^235 U undergoes fission, it breaks into two smaller, more tightly bound fragments is 1.618 × 10⁻¹² J/nucleon.
The binding energy per nucleon for 235 U is calculated as follows: M (mass of nucleus) = A × m (mass of one nucleon) M = 235 × 1.661 × 10−27 kg = 3.898 × 10−25 kg. BE (bond energy) = (c² × ∆m) / ABE = (2.998 × 10⁸ m/s)² (3.084 × 10⁻¹⁰ kg) / 235 BE = 1.618 × 10⁻¹² J/nucleon.
Binding energy refers to the energy required to break apart a nucleus into its individual nucleons (protons and neutrons) or the energy released when nucleons come together to form a nucleus. It is a measure of the stability of a nucleus and is often expressed per nucleon to compare different nuclei.
When nucleons combine to form a nucleus, the resulting mass of the nucleus is slightly less than the combined mass of the individual nucleons. This difference in mass, known as the mass defect, is converted into energy according to Einstein's famous equation E = mc², where E is energy, m is mass, and c is the speed of light. This released energy is the binding energy.
Therefore, the binding energy per nucleon for ^235U is 1.618 × 10⁻¹² J/nucleon.
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suppose has nonzero volume. what is the average of the characteristic function over the set ?
Given, Suppose has nonzero volume. Then we have to find the average of the characteristic function over the set. Suppose A be a set of non-zero volume and χ be the characteristic function of set A.
Then the average of the characteristic function over the set A is given by,∫Aχ(x) dx / ∫A dx
Here, the integral in the denominator is just the volume of the set A.
Therefore,
∫Aχ(x) dx / ∫A dx
= ∫Aχ(x) dx / vol(A)
Hence, the average of the characteristic function over the set is
∫Aχ(x) dx / vol(A).
The average of the characteristic function over the set A is defined as the ratio of the integral of the characteristic function over the set A to the volume of the set A, i.e.,
∫Aχ(x) dx / vol(A).
The volume of a set can be calculated by integrating the function over the set.
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1. (a) In reaching equilibrium, how much heat transfer occurs from 1.1 kg of water at 40°C when it is placed in contact with 1.1 kg of 20°C water? Specific heat of water c=4186 J/(kg°C) Hint: If th
The heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.
To calculate the heat transfer that occurs when two substances reach thermal equilibrium, we can use the equation Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this case, we have two equal masses of water, each weighing 1.1 kg. The specific heat of water, c, is given as 4186 J/(kg°C).
First, we need to calculate the change in temperature, ΔT, which is the difference between the final equilibrium temperature and the initial temperature. Since the masses are equal, the equilibrium temperature will be the average of the initial temperatures, which is (40°C + 20°C) / 2 = 30°C.
Next, we can calculate the heat transfer for each mass of water using the equation Q = mcΔT. For the water at 40°C, the heat transfer is Q₁ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 40°C) = -45,530 J (negative because heat is transferred out of the water). Similarly, for the water at 20°C, the heat transfer is Q₂ = (1.1 kg) * (4186 J/(kg°C)) * (30°C - 20°C) = 137,800 J.
The total heat transfer is the sum of the individual heat transfers: Q_total = Q₁ + Q₂ = -45,530 J + 137,800 J = 92,270 J.
Therefore, the heat transfer that occurs from 1.1 kg of water at 40°C to 1.1 kg of water at 20°C is 92,270 J.
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Complete Question:
(a) In reaching equilibrium, how much heat transfer occurs from 1.1 kg of water at 40€ when it is placed in contact with 1.1 kg of 20€ water? Specific heat of water c=4186 J/(kg) Hint: If the masses of water are equal, what is the equilirium temperature of the water mixture?
please help?
Four identical charges (+1.8 μC each) are brought from infinity and fixed to a straight line. Each charge is 0.56 m from the next. Determine the electric potential energy of this group. Number Units
The electric
potential
energy of this group of charges is approximately 0.3138 Joules.
To determine the electric potential
energy
of this group of charges, we can use the formula:
U = k * q1 * q2 / r
Where:
U is the electric potential energy,
k is the
Coulomb's constant
(k = 8.99 × 10^9 N m²/C²),
q1 and q2 are the charges, and
r is the distance between the charges.
In this case, we have four identical charges (+1.8 μC each) fixed to a straight line, with each
charge
0.56 m from the next. Since the charges are identical, we can pair them up and calculate the potential energy between each pair, and then sum up the total potential energy.
Let's calculate the potential energy between two adjacent charges and then multiply it by three to account for the other pairs:
U_pair = k * q^2 / r
where q = +1.8 μC and r = 0.56 m.
Plugging in the values:
U_pair = (8.99 × 10^9 N m²/C²) * (1.8 × 10^-6 C)^2 / 0.56 m
Calculating this expression:
U_pair = 0.1046 J
Now, we multiply the potential energy between two charges by three to account for the other pairs:
U_total = 3 * U_pair
U_total = 3 * 0.1046 J
U_total = 0.3138 J
Therefore, the
electric
potential energy of this group of charges is approximately 0.3138 Joules.
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Body 1 and body 2 are in a completely inelastic one-dimensional collision. What is their final momentum if their initial momenta are, respectively, (a) 10 kg . mls and 0; (b) 10 kg·m/s and 4 kg· m/s; (c) 10 kg· mls and -4 kg· mls?
The final momentum of the two bodies in a completely inelastic one-dimensional collision can be determined by using the principle of conservation of momentum.
When two bodies collide in a completely inelastic one-dimensional collision, they stick together and move with a common velocity after the collision. In such a collision, the principle of conservation of momentum is applicable. According to this principle, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. Therefore, we can write:Initial momentum of body 1 + initial momentum of body 2 = final momentum of the combined system .Therefore, P1i + P2i = Pfwhere P1i and P2i are the initial momenta of the two bodies and Pf is their final momentum after collision.
Given that the initial momenta of the two bodies are:(a) P1i = 10 kg.m/s and P2i = 0(b) P1i = 10 kg.m/s and P2i = 4 kg.m/s(c) P1i = 10 kg.m/s and P2i = -4 kg.m/sFor each case, we can find the final momentum of the combined system as follows:(a) P1i + P2i = Pf10 kg.m/s + 0 = PfPf = 10 kg.m/sThe final momentum of the combined system is 10 kg.m/s.(b) P1i + P2i = Pf10 kg.m/s + 4 kg.m/s = PfPf = 14 kg.m/sThe final momentum of the combined system is 14 kg.m/s.(c) P1i + P2i = Pf10 kg.m/s + (-4 kg.m/s) = PfPf = 6 kg.m/sThe final momentum of the combined system is 6 kg.m/s.
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