Answer:
option A
Explanation:
The correct answer is option A
given,
mass of ball and bean bag is same
golf ball rebound with certain velocity where as the bean bag stops.
Impulse = Change in momentum
I = m v_f - m v_i
When the golf ball rebound there will be negative velocity on the ball.
Which will add up and increase the impulse of the golf ball.
But in the case of beam bag velocity after the collision is zero the impulse will be less.
The golf ball experiences the greater impulse from the ground.
What is impulse?Impulse refers to the force acting over time to change the momentum of an object. It is represented by J and usually expressed in Newton-seconds or kg m/s so we can conclude that the golf ball experiences the greater impulse from the ground because of the structure and elasticity.
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Objects with masses of 130 kg and a 430 kg are separated by 0.300 m.(a) Find the net gravitational force exerted by these objects on a 35.0 kg object placed midway between them.(b) At what position (other than infinitely remote ones) can the 35.0 kg object be placed so as to experience a net force of zero?
Final answer:
The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, and for a 35.0 kg object to experience a net force of zero, it must be at a point where gravitational pulls from surrounding masses are balanced.
Explanation:
To solve the student’s question regarding the gravitational forces, we will use Newton's Law of Universal Gravitation which states the gravitational force (F) between two masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers. It is mathematically expressed as F = G * (m1 * m2) / r², where G is the gravitational constant (6.674 × 10-11 N·m²/kg²).
(a) To find the net gravitational force exerted on the 35.0 kg object placed midway between the two objects with masses of 130 kg and 430 kg and separated by 0.300 m, we calculate the gravitational force from each mass separately and then find the vector sum which in this case will just be the difference, as the object is placed in the middle.
(b) For the 35.0 kg object to experience a net force of zero, it has to be placed at a point where the gravitational force due to both masses equals each other. This can be found by setting the gravitational forces from each mass to the 35.0 kg object equal and solving for the distance from one of the masses. There is only one point between the two masses where this can occur as per the symmetry of the system.
You push very hard on a heavy desk, trying to move it. You do work on the desk:
a. only if it doesn't move.
b. whether or not it moves, as long as you are exerting a force.
c. only if it starts moving.
d. never—it does work on you.
Answer:
(C) Only if it starts moving
Explanation:
We know that work done is given by
[tex]W=F.d=Fdcos\Theta[/tex]
So there are two case in which work done is zero
First case is that when force and displacement are perpendicular to each other
And other case is that when there is no displacement
So for work to be done there must have displacement, if there is no displacement then there is no work done
So option (c) will be the correct option
Work in a physical sense is done on a desk only if the desk moves when you push it. Simply exerting force is not considered work without displacement in the direction of the force. So the correct option is C.
Explanation:When considering whether you do work on a heavy desk by pushing it, it's important to understand the scientific definition of work. In physics, work is defined as the transfer of energy that occurs when a force is applied over a distance. Therefore, you do work on an object if, and only if, the object moves in the direction of the force. So the correct answer to the question is:
c. only if it starts moving.
Simply exerting a force on an object does not constitute work unless the object is displaced. For example, if you continue to push against a wall and it does not move, despite your effort and energy consumption, physically no work is done because there is no displacement.
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When the speed of your car is doubled, by what factor does its kinetic energy increase?
a. 4
b. √2
c. 8
d. 2
Answer:
a. 4
Explanation:
Hi there!
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where:
m = mass of the car.
v = speed of the car.
Let´s see how would be the equation if the velocity is doubled (2 · v)
KE2 = 1/2 · m · (2 · v)²
Distributing the exponent:
KE2 = 1/2 · m · 2² · v²
KE2 = 1/2 · m · 4 · v²
KE2 = 4 (1/2 · m · v²)
KE2 = 4KE
Doubling the velocity increased the kinetic energy by 4.
Final answer:
The kinetic energy of a car increases by a factor of four when its speed is doubled due to the relationship KE = 1/2 m v^2, where kinetic energy is proportional to the square of the velocity.
Explanation:
When the speed of your car is doubled, the kinetic energy increases by a factor of four. This is because kinetic energy is proportional to the square of the velocity. The equation for kinetic energy (KE) is KE = 1/2 m v^2 where m is the mass and v is the velocity. If you double the velocity (v), the new kinetic energy will be 1/2 m (2v)^2 = 1/2 m (4v^2) = 4 times the original kinetic energy. Therefore, the correct answer to the question is a. 4.
The fastest time for the Sears Tower (now Willis Tower) stair climb (103 flights, or 2232 steps) is about 20 min.
a) Estimate the mechanical power in watts for a top climber. Assume the mass of climber is 80 kg and the heigth of the step is 0.18 m.Express your answer to two significant figures and include the appropriate units.
b) If the body is 20% efficient at converting chemical energy into mechanical energy, approximately how many joules and kilocalories of chemical energy does the body expend during the stair climb? Note: 1 food calorie = 1 kilocalorie = 4186 J.
Final answer:
To estimate the mechanical power in watts for a top climber, you can use the formula for power. The work done in climbing the stairs can be calculated as the product of the force applied and the distance covered. The joules of chemical energy expended during the stair climb can be calculated using the formula and converted to kilocalories.
Explanation:
To estimate the mechanical power in watts for a top climber, we can use the formula for power: P = W/t, where P is power, W is work, and t is time. The work done in climbing the stairs can be calculated as the product of the force applied and the distance covered. The force applied can be calculated as the product of the mass of the climber (80 kg) and the acceleration due to gravity (9.8 m/s²). The distance covered can be calculated as the product of the number of steps (2232) and the height of each step (0.18 m). Using these values, we can calculate the work done and then divide it by the time taken (20 min converted to seconds) to get the power in watts.
To calculate the joules of chemical energy expended during the stair climb, we can use the formula: E = P × t, where E is energy, P is power, and t is time. Using the power value calculated in part (a), and the time taken for the stair climb, we can calculate the energy expended in joules. To convert joules to kilocalories, we can divide the energy value in joules by the conversion factor of 4186 joules per kilocalorie.
For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?
To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity
At constant temperature and pressure, the change in Gibbs free energy is defined as
[tex]\Delta G = \Delta H - T\Delta S[/tex]
Where,
H = Entalpy
T = Temperature
S = Entropy
When the temperature is less than that number it is negative meaning it is a spontaneous reaction. [tex]\Delta G[/tex] is also always 0 when using single element reactions. In numerical that implies [tex]\Delta G = 0[/tex]
At the equation then,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
[tex]0 = \Delta H - T\Delta S[/tex]
[tex]\Delta H = T\Delta S[/tex]
[tex]T = \frac{\Delta H}{\Delta S}[/tex]
[tex]T = \frac{-93.8kJ}{-156.1J/K}[/tex]
[tex]T = \frac{-93.8*10^3J}{-156.1J/K}[/tex]
[tex]T = 600.89K[/tex]}
Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
What is the relation between the standard free Gibbs energy and the spontaneity of a reaction?When ΔG° < 0, the reaction is spontaneous.When ΔG° = 0, the reaction is not spontaneous nor nonspontaneous.When ΔG° > 0, the reaction is nonspontaneous.We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.
Given the standard enthalpy and entropy of the reaction, we can calculate that temperature using the following expression.
ΔG° = ΔH° - T . ΔS°
0 = -93.8 kJ - T . (-156.1 J/K)
T = 601 K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
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Grandfather clocks are designed so they can be adjusted by moving the weight at the bottom of the pendulum up or down. Suppose you have a grandfather clock at home that runs fast. Which of the following adjustments of the weight would make it more accurate?
A. Raise the weight.
B. Decrease the amplitude of swing by a small amount.
C. Add more mass to the weight.
D. Lower the weight.
E. Remove some mass from the weight.
To fix a grandfather clock that runs fast, you should lower the weight on the pendulum (option D), as this will increase the pendulum's period and slow down the clock's ticking rate.
Explanation:If you have a grandfather clock at home that runs fast, you should make the pendulum swing slower to correct the time. To do this, you would lower the weight on the pendulum. The period of a pendulum, which determines the tick-tock rate of the clock, is directly affected by the length of the pendulum, not the weight, with a longer pendulum resulting in a slower tick-tock. Therefore, the correct adjustment would be option D: Lower the weight.
Adding or removing mass does not affect the period, so options C and E would not be effective. While decreasing the amplitude of swing might have a very small effect to speed up the pendulum (option B), this is not the standard method for adjusting the timekeeping of a clock and would not provide a consistent solution. Thus, the most accurate way to adjust the clock is by changing the length of the pendulum directly by lowering the weight.
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 52 m/s (terminal speed), that his mass (including gear) was 86 kg, and that the force on him from the snow was at the survivable limit of 1.2 ✕ 105 N.What is the minimum depth of snow that would have stopped him safely?
Answer:
s = 0.9689 m
Explanation:
given,
Height of fall of paratroopers = 362 m
speed of impact = 52 m/s
mass of paratrooper = 86 Kg
From from snow on him = 1.2 ✕ 10⁵ N
now using formula
F = m a
a = F/m
[tex]a = \dfrac{1.2 \times 10^5}{86}[/tex]
[tex]a =1395.35\ m/s^2[/tex]
Using equation of motion
v² = u² + 2 a s
[tex]s =\dfrac{v^2}{2a}[/tex]
[tex]s =\dfrac{52^2}{2\times 1395.35}[/tex]
s = 0.9689 m
The minimum depth of snow that would have stooped him is s = 0.9689 m
An individual experiences a deep cut that severs the radial artery near the elbow. Ignoring air resistance, approximately how high will the blood spurt? (Hints: the specific gravity of blood is 1.050 g cm-3 and the specific gravity of mercury is 13.6 g cm-3 )
The blood will spurt to a height of 0.077 meters or 7.7 centimeters.
Height of blood spurtThe height to which blood spurts from a severed artery can be calculated using Torricelli's law, which is based on the principles of fluid dynamics.
The formula for the height h that blood spouts from the artery can be derived from Torricelli's law:
h = (2 * ΔP) / (ρ * g),
where:
ΔP is the change in pressure,ρ is the density of the fluid (blood),g is the acceleration due to gravity.To find ΔP, we can calculate the pressure difference between the cut artery and the atmosphere using the following formula:
ΔP = ρ * g * h.
Given that the specific gravity of blood is 1.050 g cm⁻³, we can find its density:
ρ_blood = [tex](1.050 g cm^{-3}) * (1 g / cm^3) = 1.050 g / cm^3.[/tex]
The specific gravity of mercury is 13.6 g[tex]cm^{-3[/tex], which corresponds to its density:
ρ_mercury = 13.6 g / [tex]cm^3[/tex].
Now, we can find the pressure difference:
ΔP = [tex](1.050 g / cm^3) * (9.81 m/s^2) * h[/tex],
where g is the acceleration due to gravity (approximately 9.81 m/s²).
We'll assume that the height (h) is in meters. We want to find h, so we'll solve for it:
h = ΔP / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, plug in the values:
h = ΔP / ((1.050 g / cm³) * (9.81 m/s²)).
h = (ρ_mercury * g * h) / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, solve for h:
h = (13.6 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex]) * h / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
h = (13.6 / 1.050) * h.
h = 12.952 * h.
h = h / 12.952.
1 = 1 / 12.952.
h = 1 / 12.952.
h ≈ 0.077 m.
So, the blood will spurt to a height of approximately 0.077 meters or 7.7 centimeters.
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The blood spurt from a deep cut is determined by the blood pressure, which needs to exceed the vein pressure for fluid entry, and by the pressure gradient from the heart to smaller vessels. These factors, combined with the density disparity between blood and mercury, influence the height of the blood spurt.
Explanation:To determine how high the blood will spurt when an individual experiences a deep cut that severs the radial artery near the elbow, we first need to understand the pressure at which the fluid, in this case blood, enters the vascular system. It must exceed the blood pressure in the vein, which is approximately 18 mm Hg above atmospheric pressure.
The blood pressure of a young adult is represented by 120 mm Hg at systolic (maximum output of the heart) and 80 mm Hg at diastolic (due to the elasticity of arteries maintaining pressure between heartbeats). Thus, the density of mercury which raises the mercury fluid in the manometer is 13.6 times greater than water, meaning the height of the fluid will be 1/13.6 of that in a water manometer.
The blood leaves the heart with a pressure of about 120 mm Hg but its pressure continues to decrease as it goes from the aorta to smaller arteries to small veins. The pressure differences in the circulation system are due to the blood flow as well as the position of the person.
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Two castings are proposed for a study. One casting is a cube 10.16 cm (4 in.) on a side; the other is a rectangle 10.64 cm (16 in.) tall × 10.16 cm (4 in.) wide × 2.45 cm (1 in.) thick. Both are cast from the same metal at the same temperature, and the mold material is the same in both cases. If the cube-shaped casting solidifies in 14.5 min, how rapidly do we expect the rectangular casting to solidify?
To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
Where,
[tex]V_c[/tex] = Volume cube
[tex]SA_c[/tex] = Superficial Area from Cube
[tex]V_r[/tex] = Volume Rectangle
[tex]SA_r[/tex]= Superficial Area from Rectangle
Our values are given as (I will try to develop the problem in English units for ease of calculations),
[tex]V_c = 4^3 = 64in^3[/tex]
[tex]SA_c = 6*4^2=96in^2[/tex]
[tex]SA_r = 2*(1*16+4*1+16*4)=168in^2[/tex]
[tex]V_r = 4*1*16 = 64in^3[/tex]
Applying the Chvorinov equation we have to,
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = 0.3265[/tex]
The stipulated time for the cube is 14.5 then,
[tex]T_r = 0.3265*14.5[/tex]
[tex]T_r = 4.735min[/tex]
Chapter 38, Problem 001
Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.
(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?
Answer:
a) [tex]\lambda=2.95x10^{-6}m[/tex]
b) infrared region
Explanation:
Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".
Part a
Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:
[tex]E=hf[/tex]
Where h is the Planck's Constant. By the other hand the know that [tex]c=f\lambda[/tex] and if we solve for f we have:
[tex]f=\frac{c}{\lambda}[/tex]
If we replace the last equation into the E formula we got:
[tex]E=h\frac{c}{\lambda}[/tex]
And if we solve for [tex]\lambda[/tex] we got:
[tex]\lambda =\frac{hc}{E}[/tex]
Using the value of the constant [tex]h=4.136x10^{-15} eVs[/tex] we have this:
[tex]\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m[/tex]
[tex]\lambda=2.95x10^{-6}m[/tex]
Part b
If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of [tex]10^{-6}m[/tex]
A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The car then
a. Need more information about the rate of the leak.
b. maintains a constant speed.
c. speeds up.
d. slows down.
The answer is option d. slows down.
Explanation:The answer to the question is option d. slows down.
When the milk begins to leak out of the bottom of the railroad tank car, the car will experience a decrease in its mass as the milk is being lost. According to the Law of Conservation of Momentum, if the mass of an object decreases without any external forces acting on it, the object will slow down.
Therefore, as the milk leaks out of the tank car, its speed will slow down over time.
A 70-kg astronaut is space walking outside the space capsuleand is stationary when the tether line breaks. As a means of returning to the capsule he throws his 2.0 kg space hammer at a speed of 14 m/s away from the capsule. At what speed does the astronaut move toward the capsule?
Answer:
0.4 m/s
Explanation:
Law of conservation of momentum tell us that the change in momentum of the hammer will be equal to the change in momentum of the astronaut
change in momentum of hammer = change in momentum of astronaut
2 kg (14 m/s - 0 m/s) = 70 kg * (v-0)
v = 0.4 m/s
Answer:
The speed of the astronaut toward the capsule is [tex]v_{a}=0.4\frac{m}{s}[/tex]
Explanation:
We have a system of two "particles" which are the astronaut and the hammer.
Initially, they are together and their relative velocities are zero, therefore the initial linear momentum is zero.
As there are no external forces to this system, the momentum is constant. This means that the initial momentum is equal to the final momentum:
[tex]0=p_{i}=p_{f}=m_{h}v_{h}-m_{a}v_{a}[/tex]
where the mass and velocity with h subscript corresponds to the hammer, and the ones with a subscript corresponds to the astronaut.
Then, we clear the velocity of the astronaut, and calculate
[tex]m_{h}v_{h}-m_{a}v_{a}\Leftrightarrow v_{a}=\frac{m_{h}}{m_{a}}v_{h}\Leftrightarrow v_{a}=\frac{2kg}{70kg}*14\frac{m}{s}=0.4\frac{m}{s}[/tex]
which is the speed of the astronaut toward the capsule.
(I) A 52-kg person riding a bike puts all her weight on each pedal when climbing a hill. The pedals rotate in a circle of radius 17 cm. (a) What is the maximum torque she exerts? (b) How could she exert more torque?
Answer:
a. 86.7 Nm
b. b. By making sure her foot is on the outermost edge of the pedal aat all times
Explanation:
a. Torque = force * perpendicular distance
= (52 kg * 9.81 m/s²) * 0.17 m
= 86.7204 Nm
b. Torque can be increased by increasing the force exerted peperndicular to the distance from the fulcrum or by increasing the distance itself. Since the person cannot increase their weight, they have to maximize the distance from the fulcrum
The maximum torque is 86.7204 Nm and the rider can increase this by putting int leg on the edge of the paddle.
What is Torque?It is defined as the force that causes the rotation of an object on an axis. It is measured in Nm.
[tex]\tau = rF\sin\theta[/tex]
[tex]\tau[/tex] = torque
[tex]r\\[/tex] = radius
[tex]F[/tex] = force = mg = (52 kg x 9.81 m/s²) = 510.12 N
= angle between F and the lever arm
Torque = force x perpendicular distance
Put the values,
Torque= 510.12 N x 0.17 m
Torque = 86.7204 Nm
B) Since the torque is directly proportional to the force and perpendicular distance. Hence, the rider must increase the perpendicular distance in order to increase the torque.
Therefore, the maximum torque is 86.7204 Nm and the rider can increase this by putting int leg on the edge of the paddle.
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Two people carry a heavy electric motor by placing it on a light board 2.45 m long. One person lifts at one end with a force of 410.0 N , and the other lifts at the opposite end with a force of 645.0 N . You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Playing on a seesaw. Part A What is the weight of the motor? Express your answer in newtons.
Answer:
W=1055N
Explanation:
In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)
In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:
[tex]\sum F=0[/tex]
so we can do the sum now:
[tex]F_{1}+F_{2}-W=0[/tex]
when solving for the Weight W, we get:
[tex]W=F_{1}+F_{2}[/tex]
and now we can substitute the given data, so we get:
W=410N+645N
W=1055N
In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energy decreases by 4.07 x 104 J.
Determine each of the following quantities (including the algebraic sign):
(a)W,
(b)U
(c)Q.
Answer:
Explanation:
(a) Work done, W = 1.82 x 10^4 J
(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)
(c) According to the first law of thermodynamics
Q = W + U
Q = 1.82 x 10^4 - 4.07 x 10^4
Q = - 2.25 x 10^4 J
The work done (W), the change in internal energy (ΔU), and heat transferred (Q) are 1.82 x 10^4 J, -4.07 x 10^4 J, and -2.25 x 10^4 J respectively as per the First Law of Thermodynamics.
Explanation:The values listed in the question are for the work done (W), and the change in internal energy (U). In physics, these concepts are investigated under a principle called the First Law of Thermodynamics, which is basically a version of the law of conservation of energy as applied for thermal processes. As per this law, the change in internal energy (U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W).
Hence, the formula can be given as: ΔU = Q - W.
If we substitute the given values:
-W = Work done = 1.82 x 10^4 J and ΔU = internal energy decreases = -4.07 x 10^4 J
Now, the formula can be rewritten as:
:-Q = ΔU + W = -4.07 x 10^4 J + 1.82 x 10^4 J = -2.25 x 10^4 J
Values:
(a) W = Work done = 1.82 x 10^4 J (Work done by the system is positive.)
(b) ΔU = Change in internal energy = -4.07 x 10^4 J (Decrease in internal energy is negative.)
(c) Q = Heat transferred = -2.25 x 10^4 J (As per the convention, heat added to the system is positive. Here, it's negative, which means heat is lost from the system.)
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Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a θ = 58.4° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J
The upward lift on an aircraft wing is directly proportional to the atmospheric density. All other factors aside, will the upward lift on an aircraft wing increase or decrease as it gains altitude? Explain, using concepts discussed in chapter 1 regarding the vertical structure of our atmosphere.
Answer:
upward lift on an aircraft wing decreases as it gains altitude.
Explanation:
The lift on an airplane wing is generated due to the the difference in the pressure on the top of the wing and the bottom of the wing in accordance with the Bernoulli's Principle.The pressure on the lower part of the wing is higher due to the low velocity stream of air than on the upper part of the wing.The governing equation of the Bernoulli's Principle is:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} +z=constant[/tex]
where:
P = pressure of the fluid
g = acceleration due to gravity
[tex]\rho=[/tex] density of fluid
v = velocity of the fluid
z = height of fluid from the datum
But the lift force on the wings depends upon several aerodynamic factors given mathematically as:
[tex]L=cl. \rho.A.\frac{v^2}{2}[/tex]
where:
cl = experimental constant
[tex]\rho=[/tex] density of air
A = area of wing
v = velocity of the air
As we move up in the atmosphere the density of air reduces and thus the force of lift will eventually decrease, that is the reason why airplanes have a flight ceiling, an altitude above which it cannot fly.
In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the spring of 9.50 cm is recorded. With 1.00 kg hanging on the spring, a second elongation of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)
To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.
By definition we know that Newton's second law is defined as
[tex]F = ma[/tex]
m = mass
a = Acceleration
By Hooke's law force is described as
[tex]F = k\Delta x[/tex]
Here,
k = Gravitational constant
x = Displacement
To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.
The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so
[tex]k\Delta x = mg[/tex]
So for state 1 we have that with 0.2kg there is an elongation of 9.5cm
[tex]k (9.5-l)=0.2*g[/tex]
[tex]k (9.5-l)=0.2*9.8[/tex]
For state 2 we have that with 1Kg there is an elongation of 12cm
[tex]k (12-l)= 1*g[/tex]
[tex]k (12-l)= 1*9.8[/tex]
We have two equations with two unknowns therefore solving for both,
[tex]k = 3.136N/cm[/tex]
[tex]l = 8.877cm[/tex]
In this way converting the units,
[tex]k = 3.136N/cm(\frac{100cm}{1m})[/tex]
[tex]k = 313.6N/m[/tex]
Therefore the spring constant is 313.6N/m
What is the pressure (in N/m2) inside an alveolus having a radius of 2.22 ✕ 10−4 m if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as created by a spherical bubble.
Answer:
The pressure inside the bubble is 666.67 [tex]N/m^{2}[/tex]
Solution:
As per the question:
Radius, R = [tex]2.22\times 10^{- 4}\ m[/tex]
Now,
Given that the surface tension of the wall is the same as that of soapy water.
The air trapped inside the bubble exerts pressure on the soap bubble which is given by:
Gauge Pressure, P = [tex]\frac{4T}{r}[/tex]
Also, the surface tension of the soapy water, [tex]T_{s} = 0.0370\ N/m^{2}[/tex]
To calculate the pressure inside the alveolus:
[tex]P_{i} = \frac{4T}{R} = \frac{4T_{s}}{R}[/tex]
[tex]P_{i} = \frac{4\times 0.0370}{2.22\times 10^{- 4}} = 666.67\ N/m^{2}[/tex]
A machine carries a 33.6 kg package from an initial position of r0 = (0.502 + 0.751 + 0.207) m at t0 = 0 s to a final position of r1 = (7.82 + 2.17 + 7.44) m at t1 = 11.9 s. The constant force applied by the machine on the package is F = (21.5 + 42.5 + 63.5) N.
Answer:
The work done on the package by the machine's force is 676.94 J.
Explanation:
Given that,
Mass of package = 33.6 kg
Initial position [tex]r_{0}=(0.502i+0.751j+0.207k)\ m[/tex]
Final position [tex]r_{1}=(7.82i+2.17j+7.44k)\ m[/tex]
Final time = 11.9 s
Force [tex]F=(21.5i+42.5j+63.5k)[/tex]
Suppose we need to find the work done on the package by the machine's force
We need to calculate the displacement
Using formula of displacement
[tex]d=r_{1}-r_{0}[/tex]
Put the value into the formula
[tex]d=(7.82i+2.17j+7.44k)-(0.502i+0.751j+0.207k)[/tex]
[tex]d=7.318i+1.419j+7.233k[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=F\dotc d[/tex]
[tex]W=(21.5i+42.5j+63.5k)\dotc(7.318i+1.419j+7.233k)[/tex]
[tex]W=676.94\ J[/tex]
Hence, The work done on the package by the machine's force is 676.94 J.
Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?
The measurement will be significantly affected.
Recall that the relationship between linear velocity and angular velocity is subject to the formula
[tex]v = \omega r[/tex],
Where r indicates the radius and [tex]\omega[/tex] the angular velocity.
As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.
Mounting larger-diameter tires on a truck leads to a systematic error in the speedometer reading, whereby the speedometer will underreport the truck's true linear speed due to an increase in linear distance covered per tire rotation at a constant angular velocity.
If larger-diameter tires are mounted on a truck, the speedometer reading will not be correct. This is due to the relationship between the angular velocity of the tires and the linear speed of the truck. Since the speedometer measures the angular speed of the tires to determine the truck's linear speed, changing to larger tires will cause a systematic error in the reading. For a fixed angular velocity, larger tires will cover a greater linear distance because of their larger circumference. Thus, the speedometer will display a speed lower than the truck's true linear speed. This affects the accuracy of the speedometer.
To further illustrate, consider a truck tire rotating with an angular velocity α. The linear (tangential) velocity v at the surface of the tire is given by v = rα, where r is the tire radius. With larger tires, for the same angular velocity α, r is bigger, hence v is larger. Therefore, when the tires are larger, the speedometer, which is calibrated for the standard tire size, underestimates the truck's speed because it is based on an incorrect assumption about the tire circumference.
Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?
Answer:
253.65259 seconds
Explanation:
f' = Approaching frequency = 460 Hz
f = Receding frequency = 410 Hz
v = Speed of sound in air = 343 m/s
v' = Speed of truck
Doppler effect
[tex]\frac{f'}{f}=\frac{v+v'}{v-v'}\\\Rightarrow 460\left(343-v'\right)=410\left(343+v'\right)\\\Rightarrow 157780-460v'-157780=140630+410v'-157780\\\Rightarrow v'=19.712\ m/s[/tex]
The distance is 5 km
Time = Distance / Speed
[tex]Time=\frac{5000}{19.712}=253.65259\ s[/tex]
Time will it take to hear the jet from your position is 253.65259 seconds
A coil of 1000 turns encloses an area of 25cm2. It is rotated in 0.010 s from a position where its plane is perpendicular to Earth’s magnetic field to one where its plane is parallel to the field. If the strength of the field is 6.0×10−5T, what is the average emf induced in the coil?
The average induced emf in the coil, when it is rotated in Earth's Magnetic field, is 15 Volts.
Explanation:The question is about the change in magnetic flux, which induces an electromotive force (emf) in a loop according to Faraday's law. The formula to calculate the change in magnetic flux is ΔФ = B × A × N, where B is the magnetic field, A is the area, and N is the number of coil turns.
Given B = 6.0×10⁻⁵T, A = 25cm² = 2.5 × 10⁻³ m² (since 1 m² = 10,000 cm²), and N=1000 turns, the change in magnetic flux equals to 0.15 Wb. According to Faraday's law, which is |emf| = |dФ/dt|, and dФ/dt is the rate of change of magnetic flux, thus |emf| = |0.15 / 0.010|= 15 V. Therefore, the average emf induced in the coil is 15 Volts.
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A perceived increase in the volume of sound is best explained by ________.
A. an increase in the height of the sound wave
B. an increase in the width of the sound wave
C. a combination of multiple sound waves
D. a decrease in the width of the sound wave
Answer:
A. an increase in the height of the sound wave
Explanation:
Volume of the sound is measured by its intensity
So here we know that
intensity is directly depends on the square of the amplitude of the sound wave
so here as intensity depends on the loudness of the sound given as
[tex]L = 10 Log(\frac{I}{I_o})[/tex]
so here as the loudness of sound will increase then the intensity will increase and hence the amplitude of the sound will also increase
So correct answer will be
A. an increase in the height of the sound wave
A curve of radius 80 m is banked at 45 deg.
Suppose that an ice storm hits, and the curve is effectively frictionless.
What is the safe speed with which to take the curve without either sliding up or down?
Answer:
[tex]v=28m/s[/tex]
Explanation:
The vertical component of the normal force must cancel out with the weight of the car taking the curve:
[tex]N_y=W[/tex]
[tex]Ncos\theta=mg[/tex]
(Notice it has to be cos and not sin, because the angle [tex]\theta[/tex] is the slope, for null slope [tex]N_y=Ncos(0)=N[/tex], as it should be).
The horizontal component of the normal force must be the centripetal force, that is:
[tex]N_x=F_{cp}[/tex]
[tex]Nsen\theta=ma_{cp}[/tex]
[tex](\frac{mg}{cos\theta})sin\theta=m\frac{v^2}{r}[/tex]
[tex]gtan\theta=\frac{v^2}{r}[/tex]
[tex]v=\sqrt{grtan\theta}=\sqrt{(9.8m/s^2)(80m)tan(45^{\circ})}=28m/s[/tex]
To find the safe speed on a banked curve without sliding up or down, calculate the square root of a specific formula involving the radius, acceleration due to gravity, and the tangent of the angle.
Explanation:To find the safe speed with which to take the curve without sliding up or down, we need to consider the force components acting on the vehicle. The gravitational force can be split into two components: the vertical component and the horizontal component. The vertical component helps keep the vehicle on the road, while the horizontal component provides the centripetal force needed to keep the vehicle moving in a curved path. By setting up an equation involving the gravitational force, the normal force, and the friction force, we can calculate the safe speed as the square root of the product of the radius of the curve, the acceleration due to gravity, and the tangent of the banking angle.
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A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly according to the relation PV1.5 = constant until it reaches a final temperature of 360 K. Calculate the work input during the process, in kJ.
Answer:
427.392 kJ
Explanation:
m = Mass of gas = 4.5 kg
Initial temperature = 200 K
Final temperature = 360 K
R = Mass specific gas constant = 296.8 J/kgK
[tex]\gamma[/tex] = Specific heat ratio = 1.5
Work done for a polytropic process is given by
[tex]W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ[/tex]
The work input during the process is -427.392 kJ
Two particles of equal masses (m = 5.5x10-15 kg) are released from rest with a distance between them is equal to 1 m. If particle A has a charge of 12 μC and particle B has a charge of 60 μC, what is the speed of particle B at the instant when the particles are 3m apart?
To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.
By Coulomb's law, electrical energy is defined as
[tex]EE = \frac{kq_1q_2}{d}[/tex]
Where,
EE = Electrostatic potential energy
q= charge
d = distance between the charged particles
k = Coulomb's law constant
While kinetic energy is defined as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m= mass
v = velocity
There by conservation of energy we have that
EE= KE
There is not Initial kinetic energy, then
[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2[/tex]
[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2[/tex]
[tex]v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}[/tex]
[tex]v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}[/tex]
Replacing with our values we have,
[tex]v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}[/tex]
[tex]v_f = 2.802*10^7m/s[/tex]
Therefore the speed of particle B at the instat when the particles are 3m apart is [tex]2.802*10^7m/s[/tex]
A student stands on a horizontal platform that is free to rotate around a vertical axis. He holds two weights in his outstretched hands. Another student gives him a push and starts the platform rotating at 0.500 rev/s. The student then pulls the weights in close to his chest. The moment of inertia with the weights extended is 2.40 kgrn2; the moment of inertia with the weights close to the axis is 0.904 kgm2. Ignore any frictional effects. What is his new rate of rotation?
Answer:
1327.43362 rev/s
Explanation:
[tex]I_i[/tex] = Initial moment of inertia = 2.4 kgm² (arms outstretched)
[tex]I_f[/tex] = Final moment of inertia = 0.904 kgm² (arms close)
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity = 0.5 rev/s
Here the angular momentum is conserved
[tex]I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\frac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\frac{2.4\times 500}{0.904}\\\Rightarrow \omega_f=1327.43362\ rev/s[/tex]
The new rate of rotation is 1327.43362 rev/s
What is mass wasting? Select one: a. Chemical and physical weathering of dirt and rock b. Movement of regolith down a slope under the force of gravity c. Weathering of regolith by rainfall d. Oversaturation of loose rock and soil by water e. Flash flooding by a stream
Answer:
B. Movement of regolith down a slope under the force of gravity
It is also known as slope movement or mass movement.
A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effective value of 110 V is applied to the primary coil. At the secondary coil, an alternating voltage with an effective value of 550 V is obtained. A current of 29 A is supplied to the primary coil of the transformer. Calculate the maximum effective current in the secondary coil. The maximum effective current in the secondary coil is A.
Answer:
[tex]I_s=5.8A[/tex]
Explanation:
Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:
[tex]P_p=P_s[/tex]
So:
[tex]V_p*I_p=V_s*I_s[/tex]
Where:
[tex]V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil[/tex]
Solving for [tex]I_s[/tex]
[tex]I_s=\frac{V_p*I_p}{V_s}[/tex]
Replacing the data provided:
[tex]I_s=\frac{110*29}{550} =5.8A[/tex]