The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver was achieved in a jump from a height of 39,000 m. At this elevation, the density of the air is only 4.3% of the surface density. Estimate the terminal speed of a skydiver at this elevation. Suppose that the density of air at sea level is 1.2 kg/m3, the mass of the skydiver is 75 kg , and the cross section area of the skydiver is 0.72 m2.
Answer: The terminal velocity of the skydiver is 281.30 m/s.
Explanation:
Mass of the sky diver = 75 kg
Density of the air at height of 39,000 m =4.3% of the surface density of the air
= [tex]1.2 kg/m^3\times \frac{4.3}{100}=0.0516 kg/m^3[/tex]
Cross sectional area of the sky diver , A = [tex]0.72 m^2[/tex]
Coefficient of drag ,C = 0.5
The terminal speed is given by the formula:
[tex]v_T=\sqrt{\frac{2mg}{c\rho A}}[/tex]
[tex]v_T=\sqrt{\frac{2\times 75 kg\times 9.8 m/s^2}{0.5\times 0.0516 km/m^3\times 0.72 m^2}}=281.30 m/s[/tex]
The terminal velocity of the skydiver is 281.30 m/s.
The terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
Further Explanation:
When a body falls from the sky it will accelerate continuously in the downward direction by the force of gravity results in increment in the acceleration. Another force is acting on the body is friction force due to air or the particles present in the atmosphere in the in the upward direction or opposite to the direction of motion also called drag force. The drag force is directly proportional to the square of the velocity of falling body. Therefore, with increase in the velocity the drag force also increasing which is proportional to equal to the square of the velocity of falling body. And time will come where the drag force is equal to the weight or the force of gravity results in zero resultant force means zero acceleration. At this point the speed become constant and reached its maximum velocity. This speed is called the terminal velocity.
Given:
The height of the jump is [tex]39000\text{ m}[/tex].
The density air is [tex]4.3\%[/tex] of the surface density.
The density of air at sea level is [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex].
The mass of the skydiver is [tex]75\text{ kg}[/tex].
The cross sectional area of the skydiver is [tex]0.72{\text{ }}{{\text{m}}^2}[/tex].
Concept:
The expression for the terminal speed is:
[tex]\fbox{\begin\\{v_t} =\sqrt {\dfrac{{2mg}}{{\rho A{C_d}}}}\end{minispace}}[/tex] …… (1)
Here, [tex]{v_t}[/tex] is the terminal speed, [tex]m[/tex] is the mass of the skydiver, [tex]g[/tex] is the acceleration due to gravity, [tex]\rho[/tex] is the density, [tex]A[/tex] is the cross sectional area and [tex]{C_d}[/tex] is the drag coefficient.
The term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] and the drag coefficient of a human in free fall is [tex]0.5[/tex].
The density of the air at the altitude of [tex]39000\text{ m}[/tex] is:
[tex]\rho=4.3\%\times{\rho _{air}}[/tex]
Here,[tex]\rho[/tex] is the density of the air at the altitude of [tex]39000\text{ m}[/tex] and [tex]{\rho _{air}}[/tex] is the surface density of the air.
Substitute [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]{\rho _{air}}[/tex] in the above equation.
[tex]\begin{aligned}\rho&=4.3\%\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=\frac{{4.3}}{{100}}\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=0.0516{\text{ kg/}}{{\text{m}}^3}\\\end{aligned}[/tex]
Substitute [tex]0.0516{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]\rho[/tex], [tex]75\text{ kg}[/tex] for [tex]m[/tex], [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex], [tex]0.72{\text{ }}{{\text{m}}^2}[/tex] for [tex]A[/tex] and [tex]0.5[/tex] for [tex]{C_d}[/tex] in equation (1).
[tex]\begin{aligned}{v_t}&=\sqrt{\frac{{2\left({75{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right)}}{{\left({0.0516{\text{ kg/}}{{\text{m}}^3}}\right)\left({0.72{\text{}}{{\text{m}}^2}} \right)\left( {0.5} \right)}}}\\ &=\sqrt{\frac{{1471.5}}{{0.018576}}}{\text{ m/s}}\\&=\sqrt{79215.116}{\text{ m/s}}\\&=281.452{\text{ m/s}}\\ \end{aligned}[/tex]
Therefore, the terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
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Answer Details:
Grade: High school
Subject: Physics
Chapter: Kinematics
Keywords:
Air, less, dense, higher, elevation, skydivers, reach high, terminal speed, recorded, achieved, jump, height, 39,000 m, density, air, 4.3%, surface density, estimate, sea, level, , mass, 75 kg, cross section, area, , 281.452 m/s.
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Answer:d
Explanation:
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You've always wondered about the acceleration of the elevators in the 101 story-tall empire state building. one day, while visiting new york, you take your bathroom scale into the elevator and stand on them. the scales read 130 lb as the door closes. the reading varies between 110 lb and 160 lb as the elevator travels 101 floors. part a what is the magnitude of the acceleration as the elevator starts upward?
The acceleration of the elevator can be calculated using Newton's second law of motion and the recorded changes in weight on the scale while standing in the elevator. When the elevator begins its ascent, the apparent weight increases due to the increased force needed to move upward, which can be converted to calculate the acceleration.
Explanation:The subject of your question involves understanding the physics concepts of acceleration and force. When standing in an elevator that is accelerating upwards, the scale reads a higher weight because the scale must exert more force to move you upwards, resulting in a feeling of being heavier.
To determine the acceleration of the elevator, we first need to comprehend the concept of net force, which in this scenario, is the difference between your apparent weight (the reading on the scale) and your actual weight. From Newton's second law, we know that net force is equal to mass times acceleration (F=ma).
So, if your weight on the scale varies between 110 lb and 160 lb, then the difference is 50lb. Converting this weight difference to force (gravity being ~9.8 m/s^2), we get approximately 222.41 Newtons.
Given that your regular weight is 130lb, converting this to mass gives us approximately 59 kg (1lb ≈ 0.453592 kg). So, we can now find the acceleration (a) using the formula a = F/m. Substituting the values, we get the acceleration as approximately 3.76 m/s² as the elevator starts upward. Note that this is an ideal value, actual acceleration would be less due to factors like air resistance, friction, etc.
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The law of conservation of mass states that _______________________________. matter cannot be created or destroyed atoms conserve mass when they fuse to form molecules mass can be created if a chemical reaction occurs
The law of conservation of mass, established by French chemist Antoine Lavoisier, states that matter is neither created nor destroyed in a chemical reaction, ensuring that the mass remains unchanged in a closed system.
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What kind of mouse is a battery-powered mouse that uses radio waves or bluetooth technology to communicate with a device?
A wireless mouse is a battery-powered mouse that uses radio waves or bluetooth technology to communicate with a device.
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You push an object 0.032 m by exerting 0.010 N in force. The work is the force times the distance. How much work have you done expressed in ergs?
A. 3200 ergs
B. 0.00032 ergs
C. 0.32 ergs
D. 32 ergs
Answer:
Work done, W = 3200 ergs
Explanation:
It is given that
Force acting on the object, F = 0.01 N
Distance covered by the object, d = 0.032 m
Let W is the work done by the object. The product of force and the distance is called the work done by the object. Mathematically, it is given by :
[tex]W=F\times d[/tex]
[tex]W=0.01\ N\times 0.032\ m[/tex]
W = 0.00032 Joules
Since, [tex]1\ joule=10^7\ ergs[/tex]
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A stone is thrown upward from ground level. the initial speed is 352 feet per second. in how many seconds will the stone hit the ground?
Final answer:
Use the kinematic equation for projectile motion, S = ut + 0.5at², to find the time it takes for a stone thrown upward at 352 feet per second to hit the ground considering the acceleration due to gravity as 32.2 feet per second squared. Calculate the time to the peak, and then double it to find the total time.
Explanation:
The student has asked about the time it will take for a stone thrown upward at an initial speed of 352 feet per second to hit the ground. To find the time it takes for the stone to hit the ground, we can use the kinematic equations of motion that describe the motion of projectiles under the influence of gravity. Since the acceleration due to gravity is 32.2 feet per second squared downward, we can use the equation S = ut + 0.5at², where S is the displacement (which would be zero, as the stone returns to the ground level), u is the initial velocity, a is the acceleration due to gravity, and t is the time.
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Answer:
The correct answer is B) Fewer peaches will be sold.
Explanation:
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A water rocket can reach a speed of 75 m/s in 0.050 seconds from launch. What is its average acceleration?
Answer:
[tex]a = 1500 m/s^2[/tex]
Explanation:
Initial speed of the rocket is ZERO as it starts from rest
[tex]v_i = 0[/tex]
final speed of the rocket is 75 m/s
[tex]v_f = 75 m/s[/tex]
so we will say that average acceleration is defined as rate of change in velocity
so it is given as
[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]
[tex]a = \frac{75 - 0}{0.05}[/tex]
[tex]a = 1500 m/s^2[/tex]
The average acceleration of the water rocket is 1500 m/s².
Explanation:To find the average acceleration of the water rocket, we can use the formula:
average acceleration = change in velocity / change in time
Given that the rocket reaches a speed of 75 m/s in 0.050 seconds, the change in velocity is 75 m/s and the change in time is 0.050 seconds. Plugging these values into the formula, we get:
average acceleration = 75 m/s / 0.050 seconds = 1500 m/s²
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Explanation: I have had the same question on my High School Physics test, and got the answer correct.