A system where friction is considered and calculated

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

Maximum height reached by the helicopter:

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

height fallen freely by Austin:

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

Velocity just before opening the parachute:

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

Time taken by the helicopter to fall:

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

Now the height fallen in the remaining time using parachute:

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

Now the height of Austin above the ground when the helicopter crashed on the ground:

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

2 years ago

Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

barxatty [35]

For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles

5 0

2 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

Charge of an Electron

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

2 years ago

Describe a life cycle of a star

vagabundo [1.1K]

Planetary Nebula are the outer layers of a star that are lost when the star changes from a red giant to a white dwarf. A star is a luminous globe of gas producing its own heat and light by nuclear reactions (nuclear fusion). They are born from nebulae and consist mostly of hydrogen and helium gas. Is this what you needed?

8 0

3 years ago

Read 2 more answers

Whats is the resistance of a bulb with a current of 60a in a circuit with a 12v battery

lara [203]

Answer:

60a in a circuit with a 12v battery

Explanation:

60a in a circuit with a 12v battery

6 0

2 years ago