Answer:
242 lts de fuel
Explanation:
We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.
If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds or 110 * 2.2 = 242 Kgs
Now by rule of three
If 1 lt weight 1 kg.
x 242 kgs
x = 242 lts de fuel
To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).
To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.
In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.
So, the calculation is as follows:
Excess weight: 110 pounds.
Weight of fuel per gallon: 6 pounds/gallon.
Number of gallons to be drained = Excess weight / Weight of fuel per gallon.
Number of gallons to be drained = 110 pounds / 6 pounds/gallon.
Number of gallons to be drained = 18.33 gallons (approximately).
Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.
Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.
Answer:
n=5, l=1, m(l) = -1, m(s)= + 1/2
Explanation:
Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.
(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5
(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).
Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)
(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.
Here, our value is -1 that is m(l)= -1
(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.
A set of quantum numbers for an electron in a 5p orbital could be: principal quantum number (n) = 5, angular momentum quantum number (l) = 1, magnetic quantum number (m) = 0, spin quantum number (m_s) = -1/2.
Explanation:
An electron in a 5p orbital can be described using four quantum numbers. Knowing that p orbitals correspond to the angular momentum quantum number l=1, and the principal quantum number n for the fifth energy level is 5, we can deduce the following. The magnetic quantum number, m, ranges from -l to +l, and thus it can have the values -1, 0, or +1 representing different p orbitals in the same shell. The spin quantum number, m_s, can either be -1/2 or +1/2, indicating the orientation of the electron's spin. Thus, an example of a set of four quantum numbers that could be assigned to an electron occupying a 5p orbital could be: (5, 1, 0, -1/2).
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A planar electromagnetic wave of frequency 8.5×1014Hz is propagating in a vacuum in the +y direction. At a certain point P and at a given instant, the electric field of the wave is given by E⃗ =(0.041V/m)z ^. What is the magnetic vector of the wave at the point P at that instant?
Answer:
[tex]B = 1.37 \times 10^{-10} T[/tex]
Explanation:
As we know that in electromagnetic waves the two fields will induce each other
So here we have
[tex]B = \frac{E}{c}[/tex]
we have
[tex]E = 0.041 V/m[/tex]
[tex]c = 3\times 10^8 m/s[/tex]
so we have
[tex]B = \frac{0.041}{3 \times 10^8}[/tex]
[tex]B = 1.37 \times 10^{-10} T[/tex]
In which one of the following situations is zero net work done?
a) a bunch of bananas are placed on a spring scale in the supermarket.
b) a horse pulls a wagon at a constant velocity
c) a sky diver falls from an airplane before opening her parachute
d) a snowball rolls down a hill e a skateboarder steps on a skateboard and begins to roll
Answer:
b) a horse pulls a wagon at a constant velocity
Explanation:
As we know that ,work done is the dot product of force vector and displacement vector.
W= F.d
W=work
F=Force
d=Displacement
We also know that
F = m a
m= mass ,a = acceleration
When velocity is constant then rate of change in the velocity will be zero,then we can say that acceleration will be zero.
When a= 0 Then F= 0
W= F.d ( F=0)
W = 0
Therefore option b i correct because horse is going with constant velocity.
Zero net work is done when a bunch of bananas is placed on a spring scale in the supermarket, as there is no displacement of the bananas despite the force exerted by the scale.
Explanation:The situation in which zero net work is done is when a bunch of bananas is placed on a spring scale in the supermarket. In this scenario, the force exerted by the scale is upward and counteracts the weight of the bananas, but since there is no displacement or movement of the bananas, the work done is zero. The concept of work in physics is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Since the bananas don't move, the displacement is zero, resulting in zero work.
In the other scenarios provided:
A horse pulls a wagon at a constant velocity has a force in the direction of motion, so work is done.A sky diver falls from an airplane before opening her parachute is experiencing a gravitational force and moving in the direction of that force, hence work is done.A snowball rolls down a hill is also subject to a force (gravity) and motion (down the hill), which means work is done.A skateboarder stepping on a skateboard and beginning to roll is exerting a force that causes the skateboard to move, doing work in the process.Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature is −9.0∘C. (a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 33.0∘C? (b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 33.0∘C?
Answer:
a) Space = [tex] 6.05 x 10^{-3} m = 0.605 cm [/tex]
b) Stress= [tex] -100.8 x 10^{6} Pa [/tex]
Explanation:
1) Data Given
[tex] L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree [/tex]
2) Calculate the space using Linear thermal expansion formula
We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel. The formula is given by:
[tex] \Delta L = L_i \alpha_{steel} \Delta T [/tex]
We have everything except the [tex] \alpha_{steel} [/tex] , so we look for this on a book and we find that [tex] \alpha_{steel} = 1.2 x 10^{-5} C^{-1} [/tex], so we can replace.
[tex] \Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm [/tex]
3) Calculate the stress of the steel
The Stress is the ratio of applied force F to a cross section area - defined as
[tex] \sigma = \frac{F_n}{A} [/tex]
Since we don't have the force and the Area, we need to look for another way to find the stress.
For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:
[tex] Y =\frac{F L}{A \Delta L} [/tex] (1)
Solving [tex] \frac{F}{A} [/tex] from the previous formula we have this:
[tex] \frac{F}{A} [/tex] = (Y Δ L)/L (2)
From the Linear thermal expansion formula we can solve like this
[tex] \frac{\Delta L}{L} [/tex] = α ΔT (3)
And replacing equation (3) into equation (2) we have:
[tex] \frac{F}{A} [/tex] = Y α ΔT (4)
We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)
[tex] \frac{F}{A} [/tex] = [tex] 20x10^{10} [/tex] Pa (1.2x10^-5 C^-1) (42C) = [tex] 100.8 *10^{6} [/tex] Pa
That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.
The rail segment's length change due to thermal expansion can be calculated using the formula ∆L = αL0∆T, which gives the space to be left between rails. When the rails are constrained, thermal stress can be determined using the formula σ= Eα∆T, demonstrating the stress in the steel rails during the summer.
Explanation:Calculating the expansion of steel train tracksTo solve this physics problem, one must first calculate the change in length of the steel rails due to thermal expansion, which introduces the concept of thermal stress—a stress created through thermal expansion or contraction of materials. The formula used is ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion for steel (around 12 × 10⁻⁶ °C⁻¹), L0 is the initial length of the steel rails (12 meters), and ∆T is the change in temperature (33 - (-9) = 42 degrees Celsius). The calculation will yield the amount of space to be left between the rails, providing the answer to the first part of the question.
Next, to determine the stress in the rails when no gap is left for expansion (assuming they are constrained), the formula is σ= Eα∆T, where E represents Young's modulus (200 × 10⁹ N/m² for steel), and the other variables remain as earlier defined. After calculation, the result will exhibit the thermal stress exerted on the steel rails during the summer. Here, Young's modulus reflects the relationship between stress and strain in the material, indicating how much deformation will happen within the steel rails due to thermal stress.
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Which of the following is an example of potential rather than kinetic energy?
a. a crawling beetle foraging for food
b. light flashes emitted by firefly water
c. rushing over Niagara Falls
d. a molecule of glucose
A molecule of glucose is an example of potential energy rather than kinetic energy.
What is potential energy and kinetic energy?Potential energy is a form of stored energy that is dependent on the relationship between different system components. Example are when a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy.
The energy an object has as a result of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.
From the given option, crawling beetle foraging for food, light flashes emitted by firefly water, rushing over Niagara Falls and a molecule of glucose. Molecule of glucose have potential rather than kinetic energy
because there is no motion in the molecule of glucose.
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A molecule of glucose is an example of potential energy because it has energy stored within its chemical bonds waiting to be used, whereas the other examples are instances of kinetic energy in action.
Potential vs. Kinetic Energy
When discussing energy, it is vital to understand the difference between potential energy and kinetic energy. Potential energy is stored energy, waiting to be released, whereas kinetic energy is the energy of motion. An example of potential energy is a molecule of glucose, which contains energy in its chemical bonds that is not currently in motion but has the possibility to be used in the future. This contrasts with kinetic energy, which would include objects or substances that are actively moving or releasing energy, such as a crawling beetle, light flashes from a firefly, or water rushing over Niagara Falls.
To answer the student's question, the example representing potential energy rather than kinetic energy is:
A molecule of glucose because it has chemical energy stored within its bonds.The examples of a crawling beetle, light from a firefly, and rushing water over Niagara Falls all represent kinetic energy as they are all in motion and actively using or releasing energy.
A bobsled, moving at 32 m/s, decelerates to 22 m/s at a rate of 4.8 m/s2. Determine the distance traveled by the bobsled during this time.
Answer:
56.25 m
Explanation:
Cinematics describes the variables involved in movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time and displacement (or its scalar equivalent, distance).
We know the bobsled starts at 32m/s and ends at 22m/s with acceleration [tex]-4.8m/sec^2[/tex]. The acceleration is negative because the bobsled is breaking of losing speed
The formula relating these three variables is
[tex]v_f^2=v_o^2+2ax[/tex]
Solving for x
[tex]x=\frac{v_f^2-v_o^2}{2a}[/tex]
[tex]x=-\frac{22^2-32^2}{2(-4.8)}[/tex]
[tex]x=\frac{540}{9.6}[/tex]
[tex]x=56.25\ m[/tex]
Meaning of geosynchronous satellite
Meaning of geosynchronous satellite:
A satellite above Earth placed in a geosynchronous circular orbit, inclined at an angle with the equatorial plane of Earth and orbiting at an orbital period equal to that of earth’s rotational period is termed as a geosynchronous satellite. Geosynchronous orbit is a High Earth orbit at a distance of 42,164 kilometres from the centre of earth such that the gravitational pull from earth is fair enough for the object in orbital motion to match with the speed of earth’s rotational period (24 hrs)
As an observer from ground, a geosynchronous satellite appears to be at a stationary point, at any part of the day; since it's velocity synchronizes with the angular velocity that of earth. A geosynchronous satellite orbiting in an equatorial plane without any inclinations is termed as a geostationary satellite.
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of gasoline occupies 0.003 80 m3. How many extra kilograms of gasoline would you receive if you bought 8.50 gal of gasoline at 0°C rather than at 21.7°C from a pump that is not temperature compensated?
Answer: 0.4911 kg
Explanation:
We have the following data:
[tex]\rho_{0\°C}= 730 kg/m^{3}[/tex] is the density of gasoline at [tex]0\°C[/tex]
[tex]\beta=9.60(10)^{-4} \°C^{-1}[/tex] is the average coefficient of volume expansion
We need to find the extra kilograms of gasoline.
So, firstly we need to transform the volume of gasoline from gallons to [tex]m^{3}[/tex]:
[tex]V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}[/tex] (1)
Knowing density is given by: [tex]\rho=\frac{m}{V}[/tex], we can find the mass [tex]m_{1}[/tex] of 8.50 gallons:
[tex]m_{1}=\rho_{0\°C}V[/tex]
[tex]m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg[/tex] (2)
Now, we have to calculate the factor [tex]f[/tex] by which the volume of gasoline is increased with the temperature, which is given by:
[tex]f=(1+\beta(T_{f}-T_{o}))[/tex] (3)
Where [tex]T_{o}=0\°C[/tex] is the initial temperature and [tex]T_{f}=21.7\°C[/tex] is the final temperature.
[tex]f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))[/tex] (4)
[tex]f=1.020832[/tex] (5)
With this, we can calculate the density of gasoline at [tex]21.7\°C[/tex]:
[tex]\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)[/tex]
[tex]\rho_{21.7\°C}=745.207 kg/m^{3}[/tex] (6)
Now we can calculate the mass of gasoline at this temperature:
[tex]m_{2}=\rho_{21.7\°C}V[/tex] (7)
[tex]m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})[/tex] (8)
[tex]m_{2}=24.070 kg[/tex] (9)
And finally calculate the mass difference [tex]\Delta m[/tex]:
[tex]\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg[/tex] (10)
[tex]\Delta m=0.4911 kg[/tex] (11) This is the extra mass of gasoline
Final answer:
To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we can use the formula for volume expansion and the density of gasoline.
Explanation:
To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we need to calculate the difference in volume and then convert it to kilograms using the density of gasoline. The average coefficient of volume expansion is given as 9.60 x 10^-4 (°C)^-1. We can calculate the change in volume using the formula ΔV = V₀ * β * ΔT, where ΔV is the change in volume, V₀ is the initial volume, β is the coefficient of volume expansion, and ΔT is the temperature difference in Celsius. In this case, the initial volume is 0.00380 m³ and the temperature difference is 21.7 - 0 = 21.7 °C. Substituting the values, we get ΔV = 0.00380 * 9.60 x 10^-4 * 21.7. Now, to convert the change in volume to kilograms, we multiply it by the density of gasoline, which is 730 kg/m³. So, the extra kilograms of gasoline received is ΔV * 730.
A roller-coaster car rolls down a frictionless track, reaching speed v0 at the bottom. If you want the car to go twice as fast at the bottom, by what factor must you increase the height of the track
Answer:
h should become four times.
Explanation:
The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two
mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]
You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher
When a man stands on a bathroom scale here on Earth, it reads 640 N . Assume each planet to be a perfect sphere with the following parameters.
Planet Mass, kg Radius, m
Mars 6.419×1023 3.396×106
Venus 4.869×1024 6.052×106
Saturn 5.685×1026 6.027×107
Part A: What would his mass be on Mars, Venus, and Saturn?
Part B: What is the magnitude of the gravitational force Mars would exert on the man if he stood on its surface?
Part C: What is the magnitude of the gravitational force Venus would exert on the man if he stood on its surface?
Part D: What is the magnitude of the gravitational force Saturn would exert on the man if he stood on its surface?
Part A: Mass on Mars, Venus, and Saturn: ≈65.21kg
Part B: Gravitational Force on Mars: ≈246.5N
Part C: Gravitational Force on Venus: ≈610.9N
Part D: Gravitational Force on Saturn: ≈572.5N.
Part A: Mass on Different Planets
The weight of an object (the reading on a scale) is given by the formula:
Weight = Mass × Acceleration due to gravity
The acceleration due to gravity on a planet can be calculated using the formula:
Acceleration due to gravity = G × (Planet Mass) / (Planet Radius)²
Where G is the universal gravitational constant.
Given that the weight on Earth is 640 N, we can rearrange the weight formula to solve for mass:
Mass = Weight / Acceleration due to gravity on Earth
The acceleration due to gravity on Earth is approximately 9.81 m/s², and the universal gravitational constant is approximately
6.674 × [tex]10^{-11[/tex] N(m/kg)².
Now we can calculate the mass on different planets:
Mars:
Acceleration due to gravity on Mars = (6.674 × [tex]10^{-11[/tex]) × (6.419 × [tex]10^{23[/tex]) / (3.396 × [tex]10^6[/tex])²
Mass on Mars = 640 N / Acceleration due to gravity on Mars
Venus:
Acceleration due to gravity on Venus = (6.674 × [tex]10^{-11[/tex]) × (4.869 × [tex]10^{24[/tex]) / (6.052 × [tex]10^6[/tex])²
Mass on Venus = 640 N / Acceleration due to gravity on Venus
Saturn:
Acceleration due to gravity on Saturn = (6.674 × [tex]10^{-11[/tex]) × (5.685 × [tex]10^{26[/tex]) / (6.027 × [tex]10^7[/tex])²
Mass on Saturn = 640 N / Acceleration due to gravity on Saturn
So, we get, Mass on Mars, Venus, and Saturn: ≈65.21kg.
Part B: Gravitational Force on Mars
The magnitude of the gravitational force between two objects can be calculated using the formula:
Gravitational Force = (G × Mass of the man × Mass of Mars) / (Radius of Mars)²
Gravitational Force on Mars: ≈246.5N.
Part C: Gravitational Force on Venus
Similar to Part B, use the formula:
Gravitational Force = (G × Mass of the man × Mass of Venus) / (Radius of Venus)²
Gravitational Force on Venus: ≈610.9N.
Part D: Gravitational Force on Saturn
Similar to Parts B and C, use the formula:
Gravitational Force = (G × Mass of the man × Mass of Saturn) / (Radius of Saturn)²
Gravitational Force on Saturn: ≈572.5N.
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The mass of the man on Mars is 171.77 kg, on Venus is72.19 kg, and on Saturn is 61.30 kg. The magnitude of the gravitational force on Mars is 639.78 N, on Venus is 640.52 N, and on Saturn is 640.77 N.
Part A: The mass of the man on Mars can be calculated using the formula:
Weight on Mars = mass on Mars × acceleration due to gravity on Mars
Mass on Mars = Weight on Mars ÷ acceleration due to gravity on Mars
Using the given parameters, we can calculate the mass of the man on Mars, Venus, and Saturn.
Mass on Mars = 640 N ÷ 3.726 m/s^2 = 171.77 kg
Mass on Venus = 640 N ÷ 8.87 m/s^2 = 72.19 kg
Mass on Saturn = 640 N ÷ 10.44 m/s^2 = 61.30 kg
Part B: The magnitude of the gravitational force on Mars can be calculated using the formula:
Gravitational force = Mass of the man × acceleration due to gravity on Mars
Using the calculated mass on Mars from Part A and the gravitational constant of Mars, we can calculate the magnitude of the gravitational force on Mars.
Gravitational force on Mars = 171.77 kg × 3.726 m/s^2 = 639.78 N
Part C: The magnitude of the gravitational force on Venus can be calculated using the same formula as in Part B:
Gravitational force on Venus = 72.19 kg × 8.87 m/s^2 = 640.52 N
Part D: The magnitude of the gravitational force on Saturn can also be calculated using the same formula:
Gravitational force on Saturn = 61.30 kg × 10.44 m/s^2 = 640.77 N
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Taipei 101 (a 101-story building in Taiwan) is sited in an area that is prone to earthquakes and typhoons, both of which can lead to dangerous oscillations of the building. To reduce the maximum amplitude, the building has a tuned mass damper, a 660,000 kg mass suspended from 42 m long cables that oscillate at the same natural frequency as the building. When the building sways, the pendulum swings, reaching an amplitude of 75 cm in strong winds or tremors. Damping the motion of the mass reduces the maximum amplitude of the oscillation of the building.a. What is the period of oscillation of the building?b. During strong winds, how fast is the pendulum moving when it passes through the equilibrium position?
Answers:
a) 13 s
b) 0.362 m/s
Explanation:
We have the following data:
[tex]m=660000 kg[/tex] is the mass of the mass damper
[tex]L=42 m[/tex] is the length of the pendulum
[tex]A=75 cm \frac{1m}{100 cm}=0.75 m[/tex] is the amplitude
a) Period of oscillation:This can be solved by the following equation:
[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex] (1)
Where:
[tex]T[/tex] is the period
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]T=2 \pi \sqrt{\frac{42 m}{9.8 m/s^{2}}}[/tex] (2)
[tex]T=13 s[/tex] (3)
b) Maximum Velocity:The velocity in a pendulum is maximum [tex]V_{max}[/tex] when the pendulum is in its mean position and the amplitude is maximum. So, the equation in this case is:
[tex]V_{max}=A \frac{2 \pi}{T}[/tex] (4)
[tex]V_{max}=0.75 m \frac{2 \pi}{13 s}[/tex] (5)
[tex]V_{max}=0.362 m/s[/tex]
Object A has mass mA = 8 kg and initial momentum pA,i = < 15, -8, 0 > kg · m/s, just before it strikes object B, which has mass mB = 11 kg. Just before the collision object B has initial momentum pB,i = < 2, 7, 0 > kg · m/s.
A) Consider a system consisting of both objects A and B. What is the total initial momentum of this system, just before the collision?
B) The forces that A and B exert on each other are very large but last for a very short time. If we choose a time interval from just before to just after the collision, what is the approximate value of the impulse applied to the two-object system due to forces exerted on the system by objects outside the system?
C) Therefore, what does the Momentum Principle predict that the total final momentum of the system will be, just after the collision?
D) Just after the collision, object A is observed to have momentum pA,f = < 13, 4, 0 > kg · m/s. What is the momentum of object B just after the collision?
Answer:
A)
<17, - 1, 0>
B)
0 Ns
C)
<17, - 1, 0>
D)
<4, - 5, 0>
Explanation:
A)
[tex]p_{A,i}[/tex] = initial momentum of object A = 15 i - 8 j + 0 k
[tex]p_{B,i}[/tex] = initial momentum of object B = 2 i + 7 j + 0 k
Total initial momentum of the system is given as the sum of initial momenta of A and B , hence
[tex]p_{i}[/tex] = [tex]p_{A,i}[/tex] + [tex]p_{B,i}[/tex]
[tex]p_{i}[/tex] = (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)
[tex]p_{i}[/tex] = 17 i - j + 0 k
B)
[tex]F_{ext}[/tex] = Net external force on the two objects = 0 N
[tex]t[/tex] = duration of the collision
[tex]I[/tex] = Impulse
Impulse is given as
[tex]I = F_{ext} t \\I = (0) t \\I = 0 Ns[/tex]
C)
[tex]p_{f}[/tex] = final momentum of the system
we know that the Impulse is nothing but change in momentum of the system of objects, hence
[tex]I = p_{f} - p_{i}\\0 = p_{f} - p_{i}\\p_{f} = p_{i}[/tex]
[tex]p_{f}[/tex] = 17 i - j + 0 k
D)
[tex]p_{A,f}[/tex] = final momentum of object A = 13 i + 4 j + 0 k
[tex]p_{B,f}[/tex] = final momentum of object B = ?
Total final momentum of the system is given as
[tex]p_{f} = p_{A,f} + p_{B,f}[/tex]
17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + [tex]p_{B,f}[/tex]
[tex]p_{B,f}[/tex] = 4 i - 5 j + 0 k
So final momentum of the object is <4, - 5, 0>
The total initial momentum of this system just before the collision is 17 i - j + 0 k.
What is Impulse?
This is defined as a term which quantifies the overall effect of a force acting over time.
Total initial momentum of this system just before the collision:Initial momentum of A = 15 i - 8 j + 0 k
initial momentum of B = 2 i + 7 j + 0 k
Total = sum of initial momentum of A and B
= (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)
= 17 i - j + 0 k
The approximate value of the impulseImpulse = Ft where F is force and t is time
Net external force on the two objects = 0 N
Impulse = (0)t
= 0Ns
The final momentum of the systemImpulse = change in momentum
= 17 i - j + 0 k
The momentum of object B just after the collisionFinal momentum of object A = 13 i + 4 j + 0 k
Final momentum of object B = ?
Total final momentum = Final momentum of object A + Final momentum of object B
17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + Final momentum of object B
= 4 i - 5 j + 0 k
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In a PA projection of the chest being used for cardiac evaluation, the heart measures 15.2 cm between its widest points. If the magnification factor is known to be 1.3, what is the actual diameter of the heart?
A. 9.7 cm
B. 11.7 cm
C. 19.7 cm
D. 20.3 cm
The size of the real heart is B) 11.7 cm
Explanation:
In lenses, mirror and other optical systems, the magnification factor is given by:
[tex]M=\frac{y'}{y}[/tex]
where
y' is the size of the image
y is the size of the real object
In this problem, we have:
M = 1.3 is the magnification factor
y' = 15.2 cm is the size of the image (the image of the heart)
Solving for y, we find the actual size of the heart:
[tex]y=\frac{y'}{M}=\frac{15.2 cm}{1.3}=11.7 cm[/tex]
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A scientist is examining an unknown solid. Which procedure would most likely help determine a chemical property of the substance?
a. heating it to find the temperature at which it melts
b. exposing it to a flame to see if it catches on fire
c. hitting it with a hammer to determine if it breaks or bends
d. placing it in water to find out whether or not it dissolves
Answer:
b. exposing it to a flame to see if it catches on fire
Explanation:
The Procedure will most likely help to determine a chemical property of substance is : exposing material to a flame to see if it catches on fire Chemical property is the characteristic that a substance has that differentiate it from another substance. The most common charatcteristics that most scientists wanted to know are : - It's flammability - It's radioactivity - Its toxicity By throwing the object into fire, we will easily find out these 3 characteristics
Hence the correct answer is b. exposing it to a flame to see if it catches on fire.
Answer:
Exposing it to a flame to see if it catches on fire
Explanation:
This is known as the flame test; The flame test is used to visually determine the identity of an unknown metal ions in a compound.
With the help of the flame test, the scientist can make use it as a qualitative test that guides him in making decision when trying to pinpoint the identity of the solid.
When the solid is exposed to flame, there may be a characteristic color given off that is visible to the naked eye (Note: Not all solids give flame colors).
The colors observed during the flame test result from the excitement of the electrons caused by the increased temperature. When the atoms of the electron are excited, for instance by, their electrons are able to move from their ground state to higher energy levels.
With this, the scientist can draw a valid conclusion.
A 22 kg solid door is 220 cm tall, 91 cm wide. a) What is the door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge? b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge?
Answer
given,
mass of the solid door = 22 Kg
dimension of door = 220 cm x 91 cm
moment of inertia about the hinge
[tex]I = \dfrac{1}{3}Mr^2[/tex]
r is the distance from the one edge which is equal to 91 cm or 0.91 m
[tex]I = \dfrac{1}{3}\times 22 \times 0.91^2[/tex]
[tex]I = 6.073\ kg m^2[/tex]
Moment of inertia about center for rectangular gate is equal to
[tex]I_{CM} = \dfrac{1}{12}Mr^2[/tex]
moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge
[tex]I = I_{CM} + MR^2[/tex]
[tex]I = I_{CM} + M(\dfrac{91}{2}- 15)^2[/tex]
[tex]I = \dfrac{1}{12}Mr^2+ M(0.0930)[/tex]
[tex]I = \dfrac{1}{12}\times 22 \times 0.91^2+ 22 \times (0.093)[/tex]
[tex]I = 3.56\ Kg m^2[/tex]
The question asks about the moment of inertia of a door with given dimensions, rotating about an axis 15 cm from one edge. The moment of inertia is calculated using the formula for a rectangular slab and then using the Parallel Axis Theorem, allowing for the axis to be moved from the center of mass to 15 cm from the edge.
Explanation:The question pertains to the computation of the moment of inertia of the door about a vertical axis inside the door, 15 cm from one edge twice. The moment of inertia regarding an axis can be deduced from the moment of inertia regarding a parallel axis that passes through the center of mass. The formula for the moment of inertia of a rectangular slab about an axis through its center and perpendicular to the slab is Icm = M(H² + W²)/12, where M is the door’s mass, H is the height, and W is the width.
For the calculation specifically 15 cm from the one edge, the Parallel Axis Theorem will need to be applied which declares I = Icm + MD², where D is the distance from the center of mass to the new axis, which considering this scenario is W/2 + 15 cm. After the computation of Icm and replacing the necessary values in, the moment of inertia for the both asked instances are procured.
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What is the potential difference between the terminals of the battery?
Answer:
Explained
Explanation:
Potential difference in a battery across its terminals is the force that pushes the charge to move around the circuit. This means that work id neede to be done to move the charge. And potential difference is the work done per unit charge. The unit of potential difference is V and defined as one joule per coulomb.
V= W/q J/C
A power station with an efficiency of 0.4 generates 108 W of electric power and dissipates 1.3×108 J of heat energy each second to the cooling water that flows through it, which increases its temperature by 3.0 Celsius degrees.
find what mass of warmed water flows through the plant each
Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]\frac{dm}{dt} = 10352 kg/s[/tex]
Explanation:
As we know that thermal energy given to the water to raise its temperature is given as
[tex]Q = ms\Delta T[/tex]
now rate of energy given to the system is
[tex]\frac{dQ}{dt} = \frac{dm}{dt} s \Delta T[/tex]
so we have
[tex]\frac{dQ}{dt} = 1.3 \times 10^8[/tex]
s = 4186 J/kg C
now we have
[tex]1.3 \times 10^8 = \frac{dm}{dt} (4186) 3[/tex]
[tex]\frac{dm}{dt} = 10352 kg/s[/tex]
With the given data, the mass of the warmed water is approximately 1.0×10⁴ kg each second.
To determine the mass of water that is warmed as it flows through the power plant, we can use the following steps:
Calculate the amount of heat transferred to the water:
The power station dissipates 1.3×10⁸ J of heat each second.
Use the specific heat capacity formula:
The formula to calculate heat transfer is
Q = m * c * ΔT,where:
Q = heat energy (1.3×10⁸ J)m = mass of water (kg)c = specific heat capacity of water (4,186 J/kg°C)ΔT = change in temperature (3.0°C)Rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)Plugging in the values, we get:
m = (1.3×10⁸ J) / (4,186 J/kg°C * 3.0°C)m ≈ 1.04×10⁴ kgTherefore, the mass of warmed water flowing through the plant each second is approximately 1.0×104 kg.
A steel sphere with radius 1.0010 cm at 41.0°C must slip through a brass ring that has an internal radius of 1.0000 cm at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at 41.0°C, can just slip through? Coefficient of linear expansion α for brass is 19.0 × 10−6 K−1.
Answer:
[tex]\Delta T = 52.6 ^o C[/tex]
Explanation:
As we know that radius of the brass ring is given as
[tex]R_{brass} = 1.0000 cm[/tex]
radius of the sphere is given as
[tex]R_{sphere} = 1.0010 cm[/tex]
now by thermal expansion formula we know that
[tex]L = L_o(1 + \alpha \Delta T)[/tex]
so we will have
[tex]1.0010 = 1.0000(1 + (19\times 10^{-6})\Delta T)[/tex]
so we have
[tex]\Delta T = 52.6 ^o C[/tex]
The weights of bags of baby carrots are normally distributed, with a mean of 32 ounces and a standard deviation of 0.32 ounce. Bags in the upper 4.5% are too heavy and must be repackaged. What is the most a bag of baby carrots can weigh and not need to be repackaged?
Answer:
[tex]x= 32.544[/tex]
Explanation:
given,
mean weight of bag (μ) = 32
standard deviation (σ) = 0.32
percentage of bag heavier = 4.5%
weight of the bag less than 4.5 % = 100 - 4.5
= 95.5%
we have to determine the z- value according to 95.5% or 0.955
using z-table
z-value = 1.70
now, using formula
[tex]Z = \dfrac{x-\mu}{\sigma}[/tex]
[tex]1.70 = \dfrac{x-32}{0.32}[/tex]
[tex]x-32 = 1.70\times {0.32}[/tex]
[tex]x= 32.544[/tex]
The most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.
To determine the most a bag of baby carrots can weigh and not need to be repackaged, we need to find the weight corresponding to the upper 4.5% of the normal distribution.
The weights are normally distributed with a mean of 32 ounces and a standard deviation (σ) of 0.32 ounces.
We first find the z-score that corresponds to the top 4.5% of the distribution.
The cumulative probability up to this z-score is 1 - 0.045 = 0.955.Using a z-table or calculator,
the z-score for 0.955 is approximately 1.645.Next, we use the z-score formula:
Z = (X - μ) / σSolving for X (the weight we need):
1.645 = (X - 32) / 0.32Multiplying both sides by 0.32:
1.645 * 0.32 = X - 320.5264 = X - 32Adding 32 to both sides:
X = 32 + 0.5264 X = 32.5264Thus, the most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.
A gymnast of mass 51.00 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.100 m above the trampoline and, when she lands, her feet stretch the trampoline down 61.00 cm. How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]
Answer:
[tex]x = 6.7 cm[/tex]
Explanation:
Here we can use energy conservation for two positions of the gymnast
When she is at the lowest position of the trampoline then its potential energy will convert into gravitational potential energy at the top
So we will have
[tex]\frac{1}{2}kx^2 = mg(H + x)[/tex]
so we have
[tex]\frac{1}{2}k(0.61^2) = 51 \times 9.81(2.1 + 0.61)[/tex]
[tex]k = 7287.5 [/tex]
now when she stands on it
then by force balance we will have
[tex]mg = kx[/tex]
[tex]51 \times 9.81 = 7287.5 \times x[/tex]
[tex]x = 6.7 cm[/tex]
In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 9.2 rev/s in 7.3 s. You lift the lid of the washer and notice that the tub decelerates and comes to a stop in 12.0 s. Assuming that the tub rotates with constant angular acceleration while it is starting and stopping, determine the total number of revolutions undergone by the tub during this entire time interval.
Answer:
90 revolutions
Explanation:
t = Time taken
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Number of rotation
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{9.2-0}{7.3}\\\Rightarrow a=1.26027\ rev/s^2[/tex]
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{9.2^2-0^2}{2\times 1.26027}\\\Rightarrow \theta=33.5801\ rev[/tex]
Number of revolutions in the 7.3 seconds is 33.5801
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-9.2}{12}\\\Rightarrow a=-0.76\ rev/s^2[/tex]
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-9.2^2}{2\times -0.76}\\\Rightarrow \theta=55.68421\ rev[/tex]
Number of revolutions in the 12 seconds is 55.68421
Total total number of revolutions is 33.5801+55.68421 = 89.26431 = 90 revolutions
The washer undergoes an equivalent of approximately 75.36 revolutions during its start-stop cycle, assuming constant angular acceleration.
Explanation:The problem involves the concepts of angular speed and acceleration in physics. Firstly, convert the angular speed from rev/s to rad/s. 1 rev = 2π rad, so 9.2 rev/s = 57.96 rad/s. The angular acceleration for the washer starting is the change in angular speed over time, so (57.96 rad/s - 0 rad/s) / 7.3 s = 7.94 rad/s².
For the washer stopping, the acceleration is (0 rad/s - 57.96 rad/s) / 12.0 s = -4.83 rad/s².
To find total revolutions we use the equation θ = θ0 + ω0t + 0.5αt². θ0 and ω0 are zero since washer starts at rest. The total revolutions for the starting process: 0.5 * 7.94 rad/s² * (7.3 s)² = 212.45 rad = 33.79 rev. For the stopping process with similar computations get 41.57 rev. Hence, total revolutions = 33.79 rev + 41.57 rev = 75.36 rev.
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A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitude of the motion for which the block does not separate from the plate? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
maximum amplitude = 0.13 m
Explanation:
Given that
Time period T= 0.74 s
acceleration of gravity g= 10 m/s²
We know that time period of simple harmonic motion given as
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]0.74=\dfrac{2\pi}{\omega}[/tex]
ω = 8.48 rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
10= 8.48² A
A=0.13 m
maximum amplitude = 0.13 m
The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m
Given data :
period ( T ) = 0.74 secs
acceleration due to gravity ( g ) = 9.8 m/s62
Determine the maximum amplitude of the motionTime period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]
First step : solve for w
w = 2π / T
= 2π / 0.74
= 8.49 rad/sec
Next step : determine the maximum amplitude ( A )
a = w² A
where ; a = maximum acceleration = 9.8 , w = 8.49
therefore
A = a / w²
= 9.8 / ( 8.49 )²
= 0.136 ≈ 0.14 m
Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.
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Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surface of the disk and do not slip. Ladybug 1 is halfway between ladybug 2 and the axis of rotation. How does the rotational speed of ladybug 1 compare to ladybug 2?
The two ladybugs have same rotational (angular) speed
Explanation:
The rotational (angular) speed of an object in circular motion is defined as:
[tex]\omega=\frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement
t is the time interval considered
Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle [tex]\theta[/tex] in the same time [tex]t[/tex]: this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.
On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:
[tex]v=\omega r[/tex]
where r is the distance from the axis: and since the two ladybugs are located at different [tex]r[/tex], they have different linear speed.
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The rotational speed of ladybug 1 is slower than that of ladybug 2.
When a rotating disk slows down at a constant rate, the angular velocity decreases over time. In this scenario, the ladybugs are at rest with respect to the surface of the disk, which means they are not experiencing any relative motion with respect to the disk.
Since ladybug 1 is halfway between ladybug 2 and the axis of rotation, it is closer to the center of the disk than ladybug 2. As the disk slows down, the distance from the axis of rotation to ladybug 1 decreases at a faster rate compared to the distance from the axis of rotation to ladybug 2.
As the distance from the axis of rotation decreases for ladybug 1 at a faster rate than for ladybug 2, ladybug 1 needs to decrease its rotational speed more to maintain the constant product of angular velocity and moment of inertia.
Thus, ladybug 1 has a slower rotational speed compared to ladybug 2.
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One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is shot west at 0.80 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90∘ and sending it north at 1.37 m/s .
Answer:
Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.
Explanation:
By conservation of the linear momentum:
[tex]m*V_{1o}+m*V_{2o}=m*V_{1f}+m*V_{2f}[/tex] Since both masses are the same, and expressing the equation for each axis x,y:
[tex]V_{1ox}+V_{2ox}=V_{1fx}[/tex] eqX
[tex]0=V_{1fx}+V_{2fx}[/tex] eqY
From eqX: [tex]V_{1fx}=1m/s[/tex]
From eqY: [tex]V_{1fy}=-1.37m/s[/tex]
The module is:
[tex]V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s[/tex]
The angle is:
[tex]\theta=atan(-1.37/1)=-53.87\°[/tex] This is 53.87° due south of east
The zone of earthquakes and volcanoes surrounding the pacific ocean is called
Answer:
The Ring of Fire
Explanation:
The ring of fire is also called the Circum-Pacific Belt, it is a path along the pacific ocean consisting of active volcanoes and frequent earthquakes.
It has a length of approximately 40,000 kilometers. It lies on the edge of tectonic plates where the in-earth vibrations and geothermal energies are prone to erupt out.
Ring of fire inhibits about 75% o the earth's volcanoes and 95% of earthquakes occur in this region.
Technician A says that the cylinders should be bored and/or honed to match the size of the pistons to be used. Technician B says that bolt holes should be chamfered and cleaned with a thread chaser before assembly. Who is right?
A) Technician A only
B) Technician B only
C) Both technicians A and B
D) Neither technician A nor B
Final answer:
Both Technician A and Technician B are correct.
Explanation:
Technician A is correct. Boring and honing the cylinders is done to achieve the proper size, shape, and finish required for the installation of the pistons. This ensures a precise fit between the cylinders and pistons, allowing for optimal performance and reduced wear.
Technician B is also correct. Chamfering the bolt holes helps to ensure smooth insertion of the bolts and prevents damage to the threads. Cleaning with a thread chaser removes any debris or contaminants that could compromise the integrity of the assembly.
Therefore, the correct answer is C) Both technicians A and B.
A spring with a force constant of 5.5 N/m hasa relaxed length of 2.54 m. When a mass isattached to the end of the spring and allowedto come to rest, the vertical length of thespring is 3.52 m.Calculate the elastic potential energystored in the spring.Answer in units of J.
Answer:
E = 2.6411 J
Explanation:
we know that elastic potential energy could be calculated by:
[tex]E = \frac{1}{2}kx^2[/tex]
where k is the constant of the spring and x is the deform of the spring.
Now x can be calculated as:
x = 3.52m-2.54m
x = 0.98 m
Finally, replacing the data, we get:
[tex]E = \frac{1}{2}(5.5 N/m)(0.98m)^2[/tex]
E = 2.6411 J
Suppose we wanted to represent all of these objects on the 1-to-10-billion scale from Chapter 1, on which the Sun is about the size of a grapefruit. Approximately how large in diameter would the star Aldebaran be on this scale?
Answer:
6.62 m
About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran
Explanation:
First we know,
Diameter of the Sun - 1.39 million km (Ds)
Diameter of Aldbaran - 61.40 million km (Da)
Thus,
[tex]\frac{Da}{Ds}[/tex] = [tex]\frac{61.40}{1.39}[/tex] ≈ 44.14
We know that in this scale the size of the sun is that of a grapefruit.
The average diameter of a grapefruit is about 15cm
Thus Ds in this scale is 15cm
Thus Da in this scale,
Da = Ds *44.14
Da = 0.15 * 44.14 ≈ 6.62 m
Thus the diameter of Aldebaran in this scale is 6.62 m
A lady perspires. How much liquid water (as a percentage of the mass of the lady) must evaporate to lower the temperature of said lady by 5°C? [Assume that the specific heat of the human body is approximately that of water, 4200 J K-1kg-1.]
Answer:
0.929%
Explanation:
[tex]m[/tex] = mass of water evaporated
[tex]M[/tex] = mass of the lady
[tex]c[/tex] = specific heat of human body = 4200 JK⁻¹kg⁻¹
[tex]L[/tex] = Latent heat of evaporation of water = 2260000 Jkg⁻¹
[tex]\Delta T[/tex] = Drop in temperature = 5 C
Using conservation of heat
[tex]m L = Mc \Delta T \\m (2260000) = M (4200) (5)\\m (2260000) = M (21000)\\m = 0.00929 M \\\frac{m}{M} = 0.00929\\\frac{m(100)}{M} = (0.00929) (100)\\\\\frac{m(100)}{M} = 0.929[/tex]
The 1.30-kg head of a hammer has a speed of 7.3 m/s just before it strikes a nail and is brought to rest Estimate the temperature rise of a 14-g iron nail generated by 8.0 such hammer blows done in quick succession. Assume the nail absorbs all the energy. The specific heat of iron is 450 J/kg⋅C∘.
Answer:
The rise in temperature is [tex]43.98^{\circ}C[/tex]
Solution:
As per the question:
Mass of hammer, M = 1.30 kg
Speed of hammer, v = 7.3 m/s
Mass of iron, [tex]m_{i} = 14\ g[/tex]
No. of blows, n = 8
Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]
Now,
To calculate the temperature rise:
Transfer of energy in a blow = Change in the Kinetic energy
[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]
For 8 such blows:
[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]
Now, we know that:
[tex]Q = m_{i}C_{i}\Delta T[/tex]
[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]
[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]