After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

Answers

Answer 1

Answer:

It will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

We can assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. We also assume that air resistance can be neglected.

Assuming that the package was dropped from rest at a height of h, the time it takes for the package to reach sea level can be calculated using the equation:

h = (1/2) * g * t²

where g is the acceleration due to gravity (9.8 m/s²) and t is the time it takes for the package to reach sea level.

Solving for t, we get:

t = sqrt(2h/g)

To convert the initial velocity of the package from km/hour to m/s, we can use the conversion factor:

1 km/hour = 0.2778 m/s

Therefore, the initial velocity of the package is:

v0 = 342 km/hour * 0.2778 m/s/km/hour = 95.0 m/s

if the package was dropped from a height of 5000 meters, the time it takes for the package to reach sea level is:

h = 5000 m

t = sqrt(2h/g) = sqrt(2*5000/9.8) = 32.0 seconds

Therefore, it will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

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Answer 2

Answer:

Final answer:

The time a dropped package takes to reach sea level from a plane is determined by its vertical motion. If the package retains only horizontal velocity when released, the time taken would be calculated using the height from which the object is dropped. However, to give a numerical value of time, we need to know the exact height.

Explanation:

The time it takes for the package dropped from the plane to reach sea level is determined exclusively by the package's vertical motion, assuming the package does not face air resistance. Specifically, the time of flight for a projectile launched and landing at the same elevation is governed by the equation: t = 2*v/g, where v represents the initial vertical velocity and g is the acceleration due to gravity. From the scenario, it seems the package retains only horizontal velocity when released since it's dropped down directly rather than being thrown downward, hence rendering initial vertical velocity as zero. Simply put, the package only begins to accelerate in the vertical direction once it's dropped, meaning the time taken would be calculated using the equation: t = √(2h/g), h being the height from which the object is dropped.

In the provided context, unfortunately, we need the height from which the package is dropped to give a specific numerical value of the time in seconds. If we knew the height of the plane at the time the package was dropped, we'd recalculate the time in seconds more precisely.

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

Answers

Answer:

789.8 W

Explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=(2c)/(3b).$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left ((dx)/(dt)\right )_(t=t_o)=0.$
$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Applying both these conditions,

$\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).$

For $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = (2c)/(3b)$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.$

Here,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Thus, the particle reach its maximum x value at time $\rm t_o = (2c)/(3b).$

A typical automobile under hard braking loses speed at a rate of about 7.2 m/s2; the typical reaction time to engage the brakes is 0.55 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.6 m. What maximum speed does this imply for a car in the school zone?

Answers

Answer:

4.3 m/s

Explanation:

a = rate at which the automobile loses speed = - 7.2 m/s²

v₀ = initial maximum speed of automobile

t' = reaction time for applying the brakes = 0.55 s

d = distance available for stopping the vehicle = 3.6 m

d' = distance traveled while applying the brakes = v₀ t' = (0.55) v₀

v = final speed after the vehicle comes to stop = 0 m/s

Using the equation

v² = v₀² + 2 a (d - d' )

0² = v₀² + 2 (- 7.2) (3.6 - (0.55) v₀)

v₀ = 4.3 m/s

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?

Answers

Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

$module = √(horizontal^2 + vertical^2)$

$module = √(460.0639 + 1.1929)$

$module = 21.4769$

$angle = arc\ tangent(vertical/horizontal)$

$angle = arc\ tangent(1.0922/21.4491)$

$angle = 2.915\°$

So the resultant direction is 21.48 km 2.92° north of east.

Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through experimental data, the students determine that the mass of the object is distributed radially. They determine that the radius of the object as a function of its mass is given by the equation r = km², where k = 3. Which of the following is a correct expression for the rotational inertia of the object?

(A) m3
(B) 1.8 m3
(C) 3.6 m3
(D) 6 m3
(E) 9 m3

Answers

Answer:

Explanation:

$r=km^2\n$ = $3m^2$

Since the object is a solid sphere, the equation for rotational inertia is:

$I = (2)/(5)mr^2$

$I=(2)/(5)m(3m^2)^2=(2)/(5)*9m^5=3.6m^5$

Final answer:

The provided question seems to have a discrepancy as the calculated value of rotational inertia for a spherical object with a given mass-radius relationship is 4.5M³, which does not match any of the supplied answer choices.

Explanation:

The question is asking for the correct expression for the rotational inertia of a spherically shaped object with mass distribution given by the radius as a function of mass (r = km² where k = 3). The rotational inertia, or moment of inertia, for a solid sphere is given by the formula ⅒MR², where M is the mass of the sphere, and R is its radius. Considering that R is defined by r = km², we substitute R with km² in the formula:

I = ⅒M(km²)² = ⅒Mk²m⁴ = ⅒Mk²M²

Since k = 3, we further simplify the expression:

I = ⅒M(3M)² = ⅒(3²)M³ = ⅒ × 9M³ = 4.5M³

However, none of the options (A) to (E) match the value 4.5M³, which indicates there may be an error in the supplied options or an error within the initial assumptions or question parameters. It's important to recheck the given data and the calculation steps to ensure accuracy. If the question and the parameters are indeed accurate as stated, additional information or clarification would be necessary.

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