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Hitman42 [59]
3 years ago
8

A 50 kg parachutist jumps out of an airplane at a height of 1000m. The parachute opens, and the jumper lands on the ground with

a speed of 5 m/s. What is the average force of air resistance during the descent?

Physics
1 answer:
valina [46]3 years ago
7 0

Answer:

The force is equal to  489.87 [N]

Explanation:

We need to make a free body diagram whe we can see the forces acting over the parachutist. In the attached image we can see the sketch of the free body diagram.

First we use the analysis of forces equal to mass by acceleration to be able to determine the equation with the term of the resistance force.

Then we have to find the acceleration to be able to apply it in the equation of the resistance force of the parachute. The acceleration can be found using the following kinematics equation

\\v_{f}^{2}=v_{0}^{2}+2*a*y\\ where\\v_{f} = final velocity = 5 [m/s]\\v_{o}= initial velocity = 0 [m/s]\\y=elevation =1000[m]\\

Note: here the acceleration is positive because the direction of the acceleration is towards the movement

Therefore the resistance force is 489.87 [N] or 49.93[kg-f]

Note: We can see that the resistance force of the parachute is almost equal to the equivalent of the weight of the parachutist, although its value is lower in this way it is explained by the fact that the system is not in balance and the person descends.

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Answer:

C. 3.375 m/s^2

Explanation:

The acceleration of an object can be found using the equation:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

a=\frac{84-30}{16}=3.375 m/s^2

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3 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r
Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
which when integrated from 0 to r is 
total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
and when r = R our total charge is 
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
and after substituting ρ0 = 3Q / πR³ we have 
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B) E = kQ/d² 
since the distribution is symmetric spherically 

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
so 
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is 
and after substituting for ρ0 is 
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4)) 
which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

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3 years ago
Question 18 (2 points)
fomenos

Answer:

A.Moving electric charges (electrons in a circuit) creates a magnetic field and

a magnetic field can cause an electric charge to move (electricity).

Explanation:

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3 years ago
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ale4655 [162]

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A piece of iron of mass 200g and tempreture 300°C is dropped into 1.00 kg of water of tempreture 20°C. Predict the final equilib
Alekssandra [29.7K]

Answer:

The final equilibrium T_{f} = 25.7[°C]

Explanation:

In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.

That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.

The temperature of thermal equilibrium will be T_f.

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Q_{iron} = Q_{water}

Heat transfer can be found by means of the following equation.

Q_{iron}=m*C_{piron}*(T_{i}-T_{f})

where:

Qiron = Iron heat transfer [kJ]

m = iron mass = 200 [g] = 0.2 [kg]

T_i = Initial temperature of the iron = 300 [°C]

T_f = final temperature [°C]

Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})

Cp_iron = 437 [J/kg*°C]

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3 years ago
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