Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g
Answers
Answer:
a = 5.53 g , a = -15g
Explanation:
This is an exercise in kinematics.
a) Let's look for the acceleration
as part of rest v₀ = 0
v = v₀ + a t
a = v / t
a = 282 / 5.2
a = 54.23 m / s²
in relation to the acceleration of gravity
a / g = 54.23 / 9.8
a = 5.53 g
b) let's look at the acceleration to stop
va = 0
0 = v₀ -2 a y
a = vi / y
a = 282/2 1
a = 141 m /s²
a / G = 141 / 9.8
a = -15g
Related Questions
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?
A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?
A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C
Answers
Answer:
a)
492 kJ
b)
Consistent
Explanation:
Q = Heat stored by woman from food = 600 k J
η = Efficiency of woman = 18% = 0.18
Q' = heat transferred to the environment
heat transferred to the environment is given as
Q' = (1 - η) Q
Inserting the values
Q' = (1 - 0.18) (600)
Q' = 492 kJ
b)
Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm
Final answer:
A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.
Explanation:
The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.
To calculate the energy lost as heat:
- Determine the total energy metabolized, which is 6.00 × 10² kJ.
- Multiply this total energy by the percentage of energy lost as heat (100% - efficiency), which gives: (6.00 × 10² kJ) * (100% - 18%) = 492 kJ.
The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.
Learn more about Energy Efficiency and Heat Transfer here:
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What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answers
Answer:
a) F = -1.82 10⁻¹⁵ N, b) K = 9.1 10⁻¹⁶ J
Explanation:
a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation
F = k q Q / r²
as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual
let's calculate
F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²
F = -1.82 10⁻¹⁵ N
b) they ask us for kinetic energy
let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / r
we substitute
F = m v² / r
v = √ (F r / m)
v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)
v = √ (0.2 10⁻¹⁶)
v = 0.447 10⁸ m / s
kinetic energy is
K = ½ m v²
K = ½ 9.1 10⁻³¹ (0.447 10⁸)²
K = 0.91 10⁻¹⁵ J
K = 9.1 10⁻¹⁶ J
B. Oceanic crust is made out of much less dense material than continental crust
C. Continental crust is continually renewed through convection in the earth's mantle
D. Continental crust eats oceanic crust for breakfast
Answers
Answer:
A. Oceanic crust is continually recycled through convection in the earth's mantle
Explanation:
The oceanic plate is constantly being recycled through the forces of convection within the earth's mantle.
New oceanic plate are formed mid-oceanic ridge for example. As the magma cools and solidifies, they are moved away continually.
This is not the case for the continental curst.
Answers
Answer:
s = 6.25 10⁻²² m
Explanation:
Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by
p = q s
s = p / q
let's calculate
s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹
s = 0.625 10⁻²¹ m
s = 6.25 10⁻²² m
We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.
B) d/√2
C) d/4
D) 2d
E) d/2
Answers
Answer:b
Explanation:
Given
Force of attraction is F when charges are d distance apart.
Electrostatic force is given by
where k=constant
and
are charges
d=distance between them
In order to double the force i.e. 2F
divide 1 and 2 we get
Answers
Answer:
10ms
Explanation:
The bird must travel the length of the train (150m), with a combined speed of 15m/s this means it will take 10s to cross an accumulated 150ms.
Answer:
The time taken by the parrot to cross the train is = 10 m/s
Explanation:
given:
A train of 150 m length is going toward north direction at a speed of 10 m/s
and a parrot is flying towards south direction parallel to the railway track with a speed of 5 m/s
find;
The time taken by the parrot to cross the train
using the distance over speed relation formula t = d / v
where:
v = velocity
d = distance
t = time
v = 10 m/s + 5 m/s = 15 m/s (combine velocity)
d = 150 m
t = ?
plugin values into the formula
t = 150
15
t = 10 m/s
therefore,
The time taken by the parrot to cross the train is = 10 m/s